HC Verma Solutions for Class 11 Physics Chapter 10 – Rotational Mechanics

Page No 192:

Question 1:

Can an object be in pure translation as well as in pure rotation?

Answer:

Yes, such motion is possible if the translation takes place along the axis of rotation. This type of motion is called screw motion.

Question 2:

A simple pendulum is a point mass suspended by a light thread from a fixed point. The particle is displaced towards one side and then released. It makes small oscillations. Is the motion of such a simple pendulum a pure rotation? If yes, where is the axis of rotation?

Answer:

Yes, it is an example of pure rotation.
The axis of rotation passes through the pivot of the pendulum, perpendicular to the plane containing the pendulum.

Question 3:

In a rotating body,

α=αr and ν=ω r. Thus

αα=νω. Can you use the theorems of ration and proportion studied in algebra so as to write

α+αα-α=ν+ων-ω

Answer:

No, we cannot use componendo-dividendo theorem of proportion here. This is because

α and a, and v and

ω are dimensionally different. Therefore, ​v +

ωand/or  â€‹

α + a are not possible.

Question 4:

A ball is whirled in a circle by attaching it to a fixed point with a string. Is there an angular rotation of the ball about its centre? If yes, is this angular velocity equal to the angular velocity of the ball about the fixed point?

Answer:

Yes, there is an angular rotation of the ball about its centre.
Yes, angular velocity of the ball about its centre is same as the angular velocity of the ball about the fixed point.

Explanation:
Let the time period of angular rotation of the ball be T.
Therefore, we get:
Angular velocity of the ball about the fixed point =

2πTAfter one revolution about the fixed centre is completed, the ball has come back to its original position. In this case, the  point at which the ball meets with the string is again visible after one revolution. This means that it has undertaken one complete rotation about its centre.
The ball has taken one complete rotation about its centre. Therefore, we have:

Angular displacement of the ball = â€‹

2πTime period = T
So, angular velocity is again â€‹

2πT. Thus, in both the cases, angular velocities are the same.

Question 5:

The moon rotates about the earth in such a way that only one hemisphere of the moon faces the earth (figure 10-Q1). Can we ever see the “other face” of the moon from the earth? Can a person on the moon ever see all the faces of the earth?
Figure

Answer:

No, we cannot see the other face of the Moon from the Earth.

Yes, a person on the Moon can see all the faces of the Earth.

Explanation: Angular velocity of the Moon about its own axis of rotation is same as its angular velocity of revolution about the Earth. This means that its rotation time period equals its revolution time period. So, we can see only one face of the Moon from the Earth.

However, angular velocity of the Earth about its axis is not same as the angular velocity of Moon about the Earth. So, all the faces of the Earth is visible from the Moon.

Question 6:

The torque of the weight of any body about any vertical axis is zero. If it always correct?

Answer:

No, its not always correct.

Figure 1
Explanation: If the centre of mass of the body is not on the same vertical line as the normal reaction R of the body, a net torque acts on the body about its vertical axis. In fig. 1,  R and CM lies in the same vertical line. Thus, there is no torque about any vertical axis

Figure 2

But in fig. 2, as R and CM do not lie along the same vertical line, there exists a torque about the vertical axis.

Question 7:

The torque of a force

→Fabout a point is defined as

→Γ=→r×→F. Suppose

→r, →Fand

→Γare all nonzero. Is

r×→Γ||→Falways true? Is it ever true?

Answer:

No,→ r×→τ|| →Γ is not true.In fact, it is never true. This is because:→ r×→τ=→ r×→ r×→FApplying vector triple product, we get:→ r×→ r×→F=→ r.→F→ r-→ r.→ r→F∵→ r.→ r=r2=→ r.→F→ r-r2→FIf → r.→F=0; that is,  →r⊥→F, then:→ r×→Γ=-r2→FWe know that r2 is never negative and → r×→Γ=-r2→.FThis implies that both vectors may be antiparallel to each other but not parallel.

Question 8:

A heavy particle of mass m falls freely near the earth’s surface. What is the torque acting on this particle about a point 50 cm east to the line of motion? Does this torque produce any angular acceleration in the particle?

Answer:

We know that:τ→=r→×F→Given:r→=-0.5i^ mF→=-mgj^The torque becomes: τ→=0.5-i^×mg-j^ τ→=0.5 mg k^  ∵ i^×j^= k^ No, there will be no angular acceleration on the particle due to the torque.Angular acceleration is given by α=τI. As the particle here moves in a straight line, the centre of rotation lies at a distance infinity (r=∞); so, moment of inertia (I = mr2) of the particle is infinity. ∴  α = 0

Question 9:

If several forces act on a particle, the total torque on the particle may be obtained by first finding the resultant force and then taking torque of this resultant. Prove this. Is this result valid for the forces acting on different particles of a body in such a way that their lines of action intersect at a common point?

Answer:

Let  f1→ , f2 →, f3→ ,… fn→  be the forces acting on a point P.Let O be the point along which torques moments will be taken.Let:  f1→ + f2 →+ f3→ +…+ fn→=R→        …1Moments of force torque fi→ about O will be:τ1→=OP →×f1→The sum of the torques about O will be:M→= OP →×f1→+OP →×f2→+…+OP →×fn→⇒ M→= OP →×f1→+f2→+f3→ +…+fn→⇒ M→=OP →×R→    From 1Thus, we see that the torque of the resultant force R→ of the forces  f1→ , f2 →, f3→, …, fn→  gives the sum of the moments of the torques.

Question 10:

If the sum of all the forces acting on a body is zero, is it necessarily in equilibrium? If the sum of all the forces on a particle is zero, is it necessarily in equilibrium?

Answer:

No, if the sum of all the forces acting on a body is zero, the body is not necessarily in equilibrium. To be in equilibrium, the sum of torque acting on the body must be zero too (see the fig.). In the above case, although the sum of the forces acting on the body is zero (F1→+-F1→=0). Still, the body will rotate along OP→. So, it won’t remain in equilibrium.

Question 11:

If the angular momentum of a body is found to be zero about a point, is it necessary that it will also be zero about a different point?

Answer:

No, angular momentum is dependent on the position vector of the particle, angle between the radius vector and the linear velocity of the particle. So, there may be finite angular momentum along any different point even if it is zero at a particular point.If angular momentum is zero along O’ but finite along O.

Question 12:

If the resultant torque of all the forces acting on a body is zero about a point, is it necessary that it will be zero about any other point?

Answer:

No, it is not necessary that the torque about any other point be zero if it is zero about one point.

Let F  →be the resultant force due to all the forces acting on the plane of the body. Therefore, torque due to force F→ at any pointwill be the resultant torque. Now, we see that the torque due to F →at point Q will be zero because Q lies on the line of support of the force F but the torque due to force F→ will not be zero along the point P.

Question 13:

A body is in translational equilibrium under the action of coplanar forces. If the torque of these forces is zero about a point, is it necessary that it will also be zero about any other point?

Answer:

Yes , if the torque due to forces in translation equillibriumis zero about a point, it will be zero about other point in the plane.

Let us consider a planner lamina of some mass, acted upon byforces F1→, F2→, … Fi→, etc.Let a force Fi→ act on a ith particle and torque due to Fi→ be zeroat a point Q.Since the body is in translation equillibrium, we have:ΣFi→=0    …1Again, torque about P is zero. Therefore, we have:Σrpi→×Fi→=0   …2Now, torque about point Q will be: Σr→Qi×Fi→=Σr→QP+r→Pi×Fi→   From fig. 13=Σr→QP×Fi→+rPi→×Fi→=ΣrQP→×Fi→+ΣrPi→×Fi→=rQP→×ΣFi→+0   From 2=rQP→×0    From 1=0Thus, F→ is zero about any other point Q.

Question 14:

A rectangular brick is kept on a table with a part of its length projecting out. It remains at rest if the length projected is slightly less than half the total length but it falls down if the length projected is slightly more than half the total length. Give reason.

Answer:

The centre of mass CM of a rectangular block lies in the middle of the block. When the block is projected less than half of its length (CM being over the table), no net force acts on it. Thus, no net torque acts upon the body. But if the block is  projected more than half of its length outside the table (CM being outside the table), gravitational force acts along the CM of the block. This force produces a moment along the edge of the table. This rotates the block, and as a result, it falls down.

Question 15:

When a fat person tries to touch his toes, keeping the legs straight, he generally falls. Explain with reference to figure (10-Q2).
Figure

Answer:

When the man tries to touch his toe, he exerts force along the hand downwards. This force produces amoment along the Centre of Mass CM of the man as shown in the figure. This moment makes him rotateand, thus, he falls down after losing the balance.

Question 16:

A ladder is resting with one end on a vertical wall and the other end on a horizontal floor. If it more likely to slip when a man stands near the bottom or near the top?

Answer:

The ladder is more likely to slide when the man stands near the top. This is because when the man stands near the top, it creates more torque compared to the torque caused by the weight of man near the bottom. When the man stands near the bottom, the Centre of Gravity of the ladder is shifted to C’ from C. Now, the couple due to forces m+Mg and N makes the ladder fall. We see that due to its shift from C to C’, the moment arm of the couple decreases from r to r’; hence, the couple decreases.

When the man stands near the top of the ladder, the Centre of Mass shifts from C to C’. This increases the moment arm of the couple and from r to r’.Increase in moment arm increases the couple and, thus, the ladder easily falls.

Question 17:

When a body is weighed on an ordinary balance we demand that the arum should be horizontal if the weights on the two pans are equal. Suppose equal weights are put on the two pans, the arm is kept at an angle with the horizontal and released. Is the torque of the two weights about the middle point (point of support) zero? Is the total torque zero? If so, why does the arm rotate and finally become horizontal?

Answer:

When the balance is kept at an angle, there is a net extra torque given to one of its arm. When the extra torque is removed, the balance becomes torque free and sum of all the torque acting on it is zero.

But balance kept at an angle has got a greater potential energy compared to the balance kept horizontal. The potential energy acquired is due to the initial torque applied on it. This displaces the balance by an angle. As soon as the body is set free to rotate, the body tends to have the lowest potential energy. Thus, potential energy starts converting in to kinetic energy, but on the other side, kinetic energy converts into potential energy when the other arm of the balance is raised. This energy transformation oscillates the balance. But in this process, friction with the air and fulcrum dissipates energy converting into heat. Finally, the balance loses the energy and becomes horizontal, or attains equilibrium.

Question 18:

The density of a rod AB continuously increases from A to B. Is it easier to set it in rotation by clamping it at A and applying a perpendicular force at B or by clamping it at B and applying the force at A?

Answer:

It will require more force to set the bar into rotation by clamping at A and then clamping at B.

Explanation: Since the rod has mass density increasing towards B, the Center of Mass (CM) of the rod is near B.  If the rod is clamped along A, the distance of CM of the rod from the pivot will be greater when the rod is clamped along B. Greater distance of CM from the Center of rotation increases the moment of inertia of the rod and hence more torque will be necessary to rotate the bar about A. Greater torque implies greater force will be necessary to rotate it.
FA = Force required to rotated the rod clamped at A
RA= Distance of CM from pivot A
M = Mass of the rod
FB = Force required to rotate the rod clamped at B
RB= Distance of CM from pivot B
We have RA>RB.
We have to find the torque required to rotate rod clamped at A to produce angular acceleration a.
TA = MRA2a = RAFA
=> FA = MRAa
We have to find torque required to rotate rod clamped at B to produce angular acceleration a.
TB = MRB2a = RBFB
=> FB = MRBa
On comparing, since RA>RB, we get:
FA>FB

Question 19:

When tall building are constructed on earth, the duration of day-night slightly increases. Is it true?

Answer:

Yes, because tall buildings have their CM much above the ground. It increases moment of inertia of the Earth. As the Earth’s rotation does not involve torque, its angular momentum is constant. Thus, an increase in MI leads to lower angular velocity of the Earth about its axis of rotation. This means length of night and day will increase. However, the increase is very small. 

Question 20:

If the ice at the poles melts and flows towards the equator, how will it affect the duration of day-night?

Answer:

Ice caps near the poles concentrate the mass of water at the poles through which axis of rotation of the Earth passes. If the ice melts, water will spread across the globe due to hydrostatic equilibrium and tend to move to the equatorial areas of the Earth due to centrifugal force of rotation. Mass, now being distributed more along the equator, will increase MI of the Earth and this in turn will decrease the angular velocity of the Earth. Decrease in angular velocity will increase the duration of day-night.

Question 21:

A hollow sphere, a solid sphere, a disc and a ring all having same mass and radius are rolled down on an inclined plane. If no slipping takes place, which one will take the smallest time to cover a given length?

Answer:

The body with the smallest moment of inertia will roll down taking the smallest time. Here, the solid sphere has the lowest moment of inertia among all the other bodies. So, it will roll down taking the least time.

Question 22:

A sphere rolls on a horizontal surface. If there any point of the sphere which has a vertical velocity?

Answer:

Some points on the equator of the sphere has got vertical velocity with respect to the direction of motion of the sphere.

Page No 193:

Question 1:

Let

→Abe a unit vector along the axis of rotation of a purely rotating body and

→Bbe a unit vector along the velocity of a particle P of the body away from the axis. The value of

→A. →Bis
(a) 1
(b) −1
(c) 0
(d) None of these.

Answer:

(c) 0

For a purely rotating body, the axis of rotation is always perpendicular to the velocity of the particle.
Therefore, we have:

A→.B→ = 0

Question 2:

A body is uniformly rotating about an axis fixed in an inertial frame of reference. Let

→Abe a unit vector along the axis of rotation and

→Bbe the unit vector along the resultant force on a particle P of the body away from the axis. The value of

→A . →Bis
(a) 1
(b) −1
(c) 0
(d) none of these.

Answer:

(c) 0

The unit vector along the axis of rotation and the unit vector along the resultant force on the particle are perpendicular to each other in a uniform rotation.
Therefore, we have:

A→.B→=0

Question 3:

A particle moves with a constant velocity parallel to the X-axis. Its angular momentum with respect to the origin
(a) is zero
(b) remains constant
(c) goes on increasing
(d) goes on decreasing.

Answer:

(b) remains constant

For angular momentum, we have:

L→=mr→×v→v→=vi^ and r→=xi^+yj^So, L→=-mvyk^m, v and y are constant; therefore, angular momentum remains constant.

Question 4:

A body is in pure rotation. The linear speed

νof a particle, the distance r of the particle from the axis and the angular velocity

ωof the body are related as

ω=νr. Thus
(a)

ω=νr(b)

ω∝r(c)

ω=0(d)

ωis independent of r.

Answer:

(d)

ωis independent of r

In a pure rotation, angular velocity of all the particles remains same and does not depend on the position of the particle from the axis of rotation.

Question 5:

Figure (10-Q3) shows a small wheel fixed coaxially on a bigger one of double the radius. The system rotates about the common axis. The strings supporting A and B do not slip on the wheels. If x and y be the distance travelled by A and B in the same time interval, then
(a) x = 2 y
(b) x = y
(c) y = 2 x
(d) none of these.
Figure

Answer:

(c) y = 2 x

It is given that angular velocity is same for both the wheels.
Therefore, we have:
vA =

ωR
vB =

ω2R
x = vAt =

ωRt    … (i)
y = vBt =

ω(2R)t    … (ii)
From equations (i) and (ii), we get:
y = 2 x

Question 6:

A body is rotating uniformly about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is
(a) vertical
(b) horizontal and skew with the axis
(c) horizontal and intersecting the axis
(d) none of these.

Answer:

(c) horizontal and intersecting the axis

Because resultant force on a particle of the body rotating uniformly is always perpendicular to the rotation axis and pass through it.

Question 7:

A body is rotating nonuniformity about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is
(a) vertical
(b) horizontal and skew with the axis
(c) horizontal and intersecting the axis
(d) none of these.

Answer:

(b) horizontal and skew with the axis

The resultant force on a particle of the body rotating non-uniformly is always horizontal and skew with the rotation axis because net torque on the body is non-zero.

Question 8:

Let

→Fbe a force acting on a particle having position vector

→r. Let

→Γbe the torque of this force about the origin, then
(a)

r→.Γ→=0 and F→.Γ→=0(b)

r→.Γ→=0 but F→.Γ→≠0(c)

r→.Γ→≠0 but F→.Γ→=0(d)

r→.Γ→≠0 and F→.Γ→≠0

Answer:

(a)

r→.Γ→=0 and F→.Γ→=0We have:

τ→=r→×F→Thus,

τ→is perpendicular to

r→and

F→.
Therefore, we have:

r→.τ→=0 and F→.τ→=0

Question 9:

One end of a uniform rod of mass m and length l is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity

ω. The force exerted by the clamp on the rod has a horizontal component
(a)

mω2l(b) zero
(c) mg
(d)

12mω2t.

Answer:

(d)

12mω2l.

Consider a small portion of rod at a distance x from the clamped end (as shown in fig.) with width dx and mass dm.

Centripetal force on this portion =

ω2xdm

dm=mlldxForce on the whole rod = F =

∫0lω2xmldx ∴ F =

12mω2l

Question 10:

A uniform rod is kept vertically on a horizontal smooth surface at a point O. If it is rotated slightly and released, it falls down on the horizontal surface. The lower end will remain
(a) at O
(b) at a distance less than l/2 from O
(c) at a distance l/2 from O
(d) at a distance larger than l/2 from O.

Answer:

(c) at a distance l/2 from O.

It is given that there is no force along x-axis.
COM of rod will remain and will not shift along x-axis (horizontal direction).
Force gravity is acting along y-axis (vertical direction). So, COM will shift along the y-axis by l/2 distance and COM of horizontal rod is at a distance l/2 from one end. Therefore, lower end of the rod will remain at a distance l/2 from O.

Question 11:

A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t/4. The relation between the moments of inertia IA and IB is
(a) IA > IB
(b) IA = IB
(c) IA < IB
(d) depends on the actual values of t and r.

Answer:

(c) IA < IB

Moment of inertia of circular disc of radius r:
I =

12mr2Mass = Volume × Density
Volume of disc =

πr2tHere, t is the thickness of the disc.
As density is same for both the rods, we have:
Moment of inertia,

I∝thickness×radius4 IAIB=t.r4t44r4<1⇒IAIB<1

⇒IA<IB

Question 12:

Equal torques act on the disc A and B of the previous problem, initially both being at rest. At a later instant, the linear speeds of a point on the rim of A and another point on the rim of B are

νAand

νBrespectively. We have
(a)

νA > νB(b)

νA = νB(c)

νA < νB(d) the relation depends on the actual magnitude of the torques.

Answer:

(a)

νA > νB τ=Iα magnitudeFor equal torque, we have:

IAαA=IBαBIA < IB

αA>αB    …(i)

Now,

ω=αtOr,

vr=αt

νA > νB   (using (i))

Question 13:

A closed cylindrical tube containing some water (not filling the entire tube) lies in a horizontal plane. If the tube is rotated about a perpendicular bisector, the moment of inertia of water about the axis
(a) increases
(b) decreases
(c) remains constant
(d) increases if the rotation is clockwise and decreases if it is anticlockwise.

Answer:

(a) increases

Moment of inertia of a mass is directly proportional to the square of the distance of mass from the axis of rotation.
Therefore, we have:

I∝r2

As the tube is rotated, water is collected at the end of tube because of centrifugal force and distance from the rotation axis increases. Hence, moment of inertia increases.

Question 14:

The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire through the centre is
(a)

Mr2(b)

12Mr2(c)

14Mr2(d)

25Mr2.

Answer:

(a)

Mr2Consider an element of length, dl = rdθ.

dm=Mπrdl=Mπrrdθ
MOI of semicircular wire =

∫0πr2dm

I=∫0πr2mπrrdθ⇒I=mr2

Question 15:

Let I1 an I2 be the moments of inertia of two bodies of identical geometrical shape, the first made of aluminium and the second of iron.
(a) I1 < I2
(b) I1 = I2
(c) I1 > I2
(d) relation between I1 and I2 depends on the actual shapes of the bodies.

Answer:

(a) I1 < I2

In the given case, we have:
MOI

∝Density
The density of iron is more; therefore, I2 will be greater.

Question 16:

A body having its centre of mass at the origin has three of its particles at (a,0,0), (0,a,0), (0,0,a). The moments of inertia of the body about the X and Y axes are 0⋅20 kg-m2 each. The moment of inertia about the Z-axis
(a) is 0⋅20 kg-m2
(b) is 0⋅40 kg-m2
(c) is

0·202kg-m2
(d) cannot be deduced with this information.

Answer:

(d) cannot be deduced with this information.

Ix = m2a2 + m3a2 = 0.20    …(i)
Iy = m1a2 + m3a2 = 0.20    …(ii)
Iz = m1a2 + m2a2                …(iii)
We have three equations and four variables. So, Iz cannot be deduced with the given information.

Question 17:

A cubical block of mass M and edge a slides down a rough inclined plane of inclination θ with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude
(a) zero

(b) Mga

(c) Mga sinθ

(d)

12 Mga sinθ.

Answer:

(d)

12 Mga sinθLet N be the normal reaction on the block.

From the free body diagram of the block, it is clear that the forces N and mgcosθ pass through the same line. Therefore, there will be no torque due to N and mgcosθ. The only torque will be produced by mgsinθ.

∴ τ→=F→×r→                        Since a is the edge of the cube, r=a2.Thus, we have:τ=mgsinθ×a2    =12mgasinθ

Page No 194:

Question 18:

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become
(a)

ωMM+m(b)

ωMM+2 m(c)

ωM-2 mM+2 m(d)

ωM+2 mM.

Answer:

(b)

ωMM+2 mNo external torque is applied on the ring; therefore, the angular momentum will be conserved.

Iω=I’ω’⇒ω’=IωI’    …i

I=Mr2I’=Mr2+2mr2On putting these values in equation (i), we get:

ω’=ωMM+2 m

Question 19:

A person sitting firmly over a rotating stool has his arms stretched. If he folds his arms, his angular momentum about the axis of rotation
(a) increases
(b) decreases
(c) remains unchanged
(d) doubles.

Answer:

(c) remains unchanged

Rate of change of angular momentum of the body is directly proportional to the net external torque acting on the body.
No external torque is applied on the person or on the table; therefore, the angular momentum will be conserved.

Question 20:

The centre of a wheel rolling on a plane surface moves with a speed

ν0. A particle on the rim of the wheel at the same level as the centre will be moving at speed
(a) zero

(b)

ν0(c)

2ν0(d)

2ν0.

Answer:

(c)

2ν0For pure rolling,

ωr=v0

As shown in the figure, the velocity of the particle will be the resultant of v0 and ωr.
Therefore, we have:

vnet=v02+ωr2vnet=v02+v02vnet=2v0

Question 21:

A wheel of radius 20 cm is pushed to move it on a rough horizontal surface. If is found to move through a distance of 60 cm on the road during the time it completes one revolution about the centre. Assume that the linear and the angular accelerations are uniform. The frictional force acting on the wheel by the surface is
(a) along the velocity of the wheel
(b) opposite to the velocity of the wheel
(c) perpendicular to the velocity of the wheel
(d) zero.

Answer:

(a) along the velocity of the wheel

As the distance covered in one revolution about the centre is less than the perimeter of the wheel, it means that the direction of torque due to frictional force opposes the motion of wheel, i.e., the frictional force acting on the wheel by the surface is along the velocity of the wheel.
FIG.

Question 22:

The angular velocity of the engine (and hence of the wheel) of a scooter is proportional to the petrol input per second. The scooter is moving on a frictionless road with uniform velocity. If the petrol input is increased by 10%, the linear velocity of the scooter is increased by
(a) 50%
(b) 10%
(c) 20%
(d) 0%.

Answer:

(d) 0%.

On a frictionless road, we have:
Angular velocity of the engine = 0
Therefore, increase in petrol input will not affect the angular velocity and hence the linear velocity of the scooter will remain the same.

Question 23:

A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of a smooth incline and released. Least time will be taken in reaching the bottom by
(a) the solid sphere
(b) the hollow sphere
(c) the disc
(d) all will take same time.

Answer:

(d) all will take same time.

The incline is smooth; therefore, all bodies will slip on the incline. Also, as the mass of  bodies is same, they will reach the bottom in equal time.

Question 24:

A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top on an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. Least time will be taken in reaching the bottom by
(a) the solid sphere
(b) the hollow sphere
(c) the disc
(d) all will take same time.

Answer:

(d) all will take same time

Let θ be the inclination angle.

From the free body diagram, we have:

N=mgcosθ    …ima=mgsinθ-fr    …iiPutting fr=μN in ii we get,a=gsinθ-μcosθThe friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling; therefore, all the bodies come down with the same acceleration.

Question 25:

In the previous question, the smallest kinetic energy at the bottom of the incline will be achieved by
(a) the solid sphere
(b) the hollow sphere
(c) the disc
(d) all will achieve same kinetic energy.

Answer:

(b) the hollow sphere

Torque is same for all the bodies; therefore, the angular momentum will be conserved.
Now, total kinetic energy =

12mv2+L22ISo, the body with greater value of moment of inertia will have smallest kinetic energy at the bottom of the incline.

Order of the moment of inertia of the bodies:
hollow sphere > disc > solid sphere

Hence, the hollow sphere will have the smallest kinetic energy at the bottom.

Question 26:

A string of negligible thickness is wrapped several times around a cylinder kept on a rough horizontal surface. A man standing at a distance l from the cylinder holds one end of the string and pulls the cylinder towards him (figure 10-Q4). There is no slipping anywhere. The length of the string passed through the hand of the man while the cylinder reaches his hands is
(a) l
(b) 2l
(c) 3l
(d) 4l.
Figure

Answer:

(b) 2l

For pure rolling,

ωr=v0Figure:

As shown in the figure, the velocity of the string will be resultant of v0 and ωr.

vnet=v0+ωrvnet=2v0Let:
Linear distance travelled by the cylinder in time (t),

v0t=l
∴ Linear distance travelled by the string in same time,

2v0t=2l

Question 1:

The axis of rotation of a purely rotating body
(a) must pass through the centre of mass
(b) may pass through the centre of mass
(c) must pass through a particle of the body
(d) may pass through a particle of the body.

Answer:

(b) may pass through the centre of mass
(d) may pass through a particle of the body

It is not necessary that the axis of rotation of a purely rotating body should pass through the centre of mass or through a particle of the body. It can also lie outside the body.

Question 2:

Consider the following two equations
(A)

L=I ω(B)

dLdt=ΓIn noninertial frames
(a) both A and B are true
(b) A is true but B is false
(c) B is true but A is false
(d) both A and B are false.

Answer:

(b) A is true but B is false.

In non-inertial frames,

dLdt=ΓtotalHere,

Γtotalis is the total torque on the system due to all the external forces acting on the system. So, equation (B) is not true as in non-inertial frames, pseudo force must be applied to study the motion of the object.

Question 3:

A particle moves on a straight line with a uniform velocity. Its angular momentum
(a) is always zero
(b) is zero about a point on the straight line
(c) is not zero about a point away from the straight line
(d) about any given point remains constant.

Answer:

(b) is zero about a point on the straight line
(c) is not zero about a point away from the straight line
(d) about any given point remains constant

(b) Angular momentum =

mr→×v→If the point is on the straight line,

r→and v→will have the same direction and their cross product will be zero. Hence, angular momentum is zero.

(c) If the point is not on the straight line,

r→and v→will not have the same direction and their cross product will not be zero. Hence, angular momentum is non-zero.

(d) No external torque is applied on the body; therefore, its angular momentum about any given point remains constant.

Question 4:

If there is no external force acting on a nonrigid body, which of the following quantities must remain constant?
(a) angular momentum
(b) linear momentum
(c) kinetic energy
(d) moment of inertia.

Answer:

(a) angular momentum
(b) linear momentum

 

F→ext=0⇒τ→ext=0That is, the change in linear momentum and angular momentum is zero. This is because:

dP→dt=F→extAnd dL→dt=τ→ext

Question 5:

Let IA and IB be moments of inertia of a body about two axes A and B respectively. The axis A passes through the centre of mass of the body but B does not.
(a) IA < IB
(b) If IA < IB, the axes are parallel
(c) If the axes are parallel, IA < IB
(d) If the axes are not parallel, IAIB.

Answer:

(c) If the axes are parallel, IA < IB

If axes A and B are parallel, we get:

IB=IA+mr2
Here, r is the distance between two axes and m is the mass of the body.

IA < IB

Question 6:

A sphere is rotating about a diameter.
(a) The particles on the surface of the sphere do not have any linear acceleration.
(b) The particles on the diameter mentioned above do not have any linear acceleration.
(c) Different particles on the surface have different angular speeds.
(d) All the particles on the surface have same linear speed.

Answer:

(b) The particles on the diameter mentioned above do not have any linear acceleration.

Explanation:
Linear acceleration of a rotating particle is given as:

a→=r→×α→(b)
The sphere is rotating about a diameter; therefore, the position vector of the particles on the diameter is zero. Thus, linear acceleration of the particle is zero.

(c) and (d)
All the particles of the body have the same angular velocity. All the particle on the surface have different linear speeds that depend on the position of the particle from the axis of rotation.

Page No 195:

Question 1:

A wheel is making revolutions about its axis with uniform angular acceleration. Starting from rest, it reaches 100 rev/sec in 4 seconds. Find the angular acceleration. Find the angle rotated during these four seconds.

Answer:

Given:
t = 4 s
Initial angular velocity =

ω0=0Final angular velocity =

ω=100 rev/s

ω=ω0+αtα=ωtα=1004 rev/s2=25 rev/s2Now, we have:θ=ω0t+12αt2⇒θ=12×25×16      =200°⇒θ=200×2π radians     =400π radians

Question 2:

A wheel rotating with uniform angular acceleration covers 50 revolutions in the first five seconds after the start. Find the angular acceleration and the angular velocity at the end of five seconds.

Answer:

Given:
Angular displacement of the wheel =

θ=50×2π=100πInitial angular velocity of the wheel =

ω0=0After, t = 5 seconds

θ=ω0t+12αt2⇒100π=12×α× (5)2⇒100π=12×α× 25⇒α=8π rad/s2  or 4 rev/sω=ω0+2αt⇒ω=0+8π×5=40π rad/s⇒ω=20 rev/s

Question 3:

A wheel starting from rest is uniformly accelerated at 4 rad/s2 for 10 seconds. It is allowed to rotate uniformly for the next 10 seconds and is finally brought to rest in the next 10 seconds. Find the total angle rotated by the wheel.

Answer:

It is given that the area under the

ω-tcurve gives the total angular displacement.
∴ Maximum angular velocity =

ω=αt

ω=4×10=40 rad/s

Area under the curve

=12×10×40+40×10+12×40×10=800 rad∴ Total angle rotated in 30 s = 800 rad.

Question 7:

The density of a rod gradually decreases from one end to the other. It is pivoted at an end so that it can move about a vertical axis though the pivot. A horizontal force F is applied on the free end in a direction perpendicular to the rod. The quantities, that do not depend on which end of the rod is pivoted, are
(a) angular acceleration
(b) angular velocity when the rod completes one rotation
(c) angular momentum when the rod completes one rotation
(d) torque of the applied force.

Answer:

(d) torque of the applied force

The torque of the applied force does not depend on the density of a rod. It depend on the distance between the pivot and the point where F is applied. So, it does not depend on which end of the rod is pivoted.

Question 8:

Consider a wheel of a bicycle rolling on a level road at a linear speed

ν0(figure 10-Q5).
(a) the speed of the particle A is zero
(b) the speed of B, C and D are all equal to

v0(c) the speed of C is 2

v0(d) the speed of B is greater than the speed of O.
Figure

Answer:

(a) the speed of the particle A is zero
(c) the speed of C is 2

v0(d) the speed of B is greater than the speed of O

For pure rolling,

ωr=v0Velocity of the particle at A, B, C and D will be resultant of v0 and ωr.

At point B,

vnet=v02+ωr2vnet=v02+v02vnet=2v0At point C,

vnet=v0+ωrvnet=2v0At point A,

vnet=v0-ωrvnet=0At point O,
r = 0

∴ vnet=v0

Question 9:

Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping,
(a) the heavier sphere reaches the bottom first
(b) the bigger sphere reaches the bottom first
(c) the two spheres reach the bottom together
(d) the information given is not sufficient to tell which sphere will reach the bottom first.

Answer:

(c) the two spheres reach the bottom together

Acceleration of a sphere on the incline plane is given by:

a=gsinθ1+ICOMmr2ICOM for a solid sphere=25mr2So, a=gsinθ1+2mr25mr2=57gsinθa is independent of mass and radii; therefore, the two spheres reach the bottom together.

Question 10:

A hollow sphere and a solid sphere having same mss and same radii are rolled down a rough inclined plane.
(a) The hollow sphere reaches the bottom first.
(b) The solid sphere reaches the bottom with greater speed.
(c) The solid sphere reaches the bottom with greater kinetic energy.
(d) The two spheres will reach the bottom with same linear momentum.

Answer:

(b) The solid sphere reaches the bottom with greater speed.

Acceleration of a sphere on the incline plane is given by:

a=gsinθ1+ICOMmr2ICOM for a solid sphere=25mr2So, a=gsinθ1+2mr25mr2=57gsinθ

ICOM for a hollow sphere=23mr2So, a’=gsinθ1+2mr23mr2=35gsinθThe acceleration of the solid sphere is greater; therefore, it will reach the bottom with greater speed.

Question 11:

A sphere cannot roll on
(a) a smooth horizontal surface
(b) a smooth inclined surface
(c) a rough horizontal surface
(d) a rough inclined surface.

Answer:

(a) a smooth horizontal surface
(b) a smooth inclined surface

A sphere cannot roll on a smooth inclined surface and on a smooth horizontal surface because there is no backward force (force of friction) to prevent its slipping.

Question 12:

In rear-wheel drive cars, the engine rotates the rear wheels and the front wheels rotate only because the car moves. If such a car accelerates on a horizontal road the friction
(a) on the rear wheels is in the forward direction
(b) on the front wheels is in the backward direction
(c) on the rear wheels has larger magnitude than the friction on the front wheels
(d) on the car is in the backward direction.

Answer:

(a) on the rear wheels is in the forward direction
(b) on the front wheels is in the backward direction
(c) on the rear wheels has larger magnitude than the friction on the front wheels

Explanation:
(a) On the rear wheels, friction force is in the forward direction because it favours the motion and accelerates the car in forward direction.
(b) Because of the movement of the car in forward direction, front wheels push the road in forward direction and in reaction, the road applies friction force in the backward direction.
(c) As the car is moving in forward direction, the rear wheels have larger magnitude of friction force (in forward direction) than on the front wheels.

Question 13:

A sphere can roll on a surface inclined at an angle θ if the friction coefficient is more than

27g tanθ. Suppose the friction coefficient is

17g tanθ. If a sphere is released from rest on the incline,
(a) it will stay at rest
(b) it will make pure translational motion
(c) it will translate and rotate about the centre
(d) the angular momentum of the sphere about its centre will remain constant.

Answer:

(c) it will translate and rotate about the centre

The given coefficient of friction (

17g tanθ) is less than the coefficient friction (

27g tanθ) required for perfect rolling of the sphere on the inclined plane. Therefore, sphere may slip while rolling and it will translate and rotate about the centre.

Question 14:

A sphere is rolled on a rough horizontal surface. If gradually slows down and stops. The force of friction tries to
(a) decrease the linear velocity
(b) increase the angular velocity
(c) increase the linear momentum
(d) decrease the angular velocity.

Answer:

(a) decrease the linear velocity
(b) increase the angular velocity

If a sphere is rolled on a rough horizontal surface, the force of friction tries to oppose the linear motion and favours the angular motion.

Question 15:

Figure (10-Q6) shows a smooth inclined plane fixed in a car accelerating on a horizontal road. The angle of incline θ is related to the acceleration a of the car as a = g tanθ. If the sphere is set in pure rolling on the incline,
(a) it will continue pure rolling
(b) it will slip down the plane
(c) its linear velocity will increase
(d) its linear velocity will slowly decrease.
Figure

Answer:

(a) it will continue pure rolling

From the free body diagram of sphere, we have:
Net force on the sphere along the incline,
Fnet = mgsinθmacosθ    …(i)
On putting a = gtanθ in equation (i), we get:
Fnet = 0
Therefore, if the sphere is set in pure rolling on the incline, it will continue pure rolling.

Page No 196:

Question 4:

A body rotates about a fixed axis with an angular acceleration of one radian/second. Through what angle does it rotate during the time in which its angular velocity increases from 5 rad/s to 15 rad/s.

Answer:

Given:
Angular acceleration of the body =

α=2 rad/s2Initial angular velocity of the body =

ω0=5 rad/sFinal angular velocity of the body =

ω=15 rad/s

We know that:ω=ω0+αt⇒t=ω-ω0α=15-51=10 sAlso, θ=ω0t+12αt2⇒θ=5×10+12×1×100⇒θ=50+50=100 rad

Question 5:

Find the angular velocity of a body rotating with an acceleration of 2 rev/s2 as it completes the 5th revolution after the start.

Answer:

Angular displacement of the body,

θ=5 rev=10π radα=2 rev/s2=4π rad/s2Initial angular velocity, ω0=0Final angular velocity, ω=?ω2=2αθ⇒ω=2×4π×10π        =4π5 rad/s ⇒ω=25 rev/s

Question 6:

A disc of radius 10 cm is rotating about its axis at an angular speed of 20 rad/s. Find the linear speed of
(a) a point on the rim,
(b) the middle point of a radius.

Answer:

Radius of disc = r = 10 cm = 0.1 m
Angular velocity of the disc = ω = 20 rad/s
∴ Linear velocity of point on the rim =

v=ωr

⇒v=20×0.1=2 m/s∴ Linear velocity of point at the middle of radius

v=ωr2=20×0.12= 1 m/s(i)   (ii)

Question 7:

A disc rotates about its axis with a constant angular acceleration of 4 rad/s2. Find the radial and tangential accelerations of a particle at a distance of 1 cm from the axis at the end of the first second after the disc starts rotating.

Answer:

Angular acceleration of the disc,

α=4 rad/s2Distance of the particle from the axis of rotation,

r=1 cm=0.01 m

So, ω=αt=4 rad/s  (t= 1 s)Radial acceleration,

αr=ω2r=42×0.01   =0.16 m/s2=16 cm/s2Tangential acceleration,

αT=αr=0.04 m/s2   =4 cm/s2

Question 8:

A block hangs from a string wrapped on a disc of radius 20 cm free to rotate about its axis which is fixed in a horizontal position. If the angular speed of the disc is 10 rad/s at some instant, with what speed is the block going down at that instant?

Answer:

It is given that the string is moving on the rim of the disc and block is connected with the string.
Therefore, the speeds of the block going down and the rim will be same.
Angular speed of the disc, ω = 10 rad/s
Radius of the pulley, r = 20 cm

Linear velocity of the rim, v = Tangential velocity = rω
v = 10 × 20 = 200 cm/s = 2 m/s
Therefore, velocity of the block = 2 m/s

Question 9:

Three particles, each of mass 200 g, are kept at the corners of an equilateral triangle of side 10 cm. Find the moment of inertial of the system about an axis
(a) joining two of the particles and
(b) passing through one of the particles and perpendicular to the plane of the particles.

Answer:

(a) The distance of mass at A from the axis passing through side BC,

AD=32×10=53 cm

Therefore, we have:
Moment of inertia of mass about the axis BC,

l=mr2=200×532=200×25×3=15000 gm-cm2=1.5×10-3 kg-m2(b) Now, let the axis of rotation passes through A and is perpendicular to the plane of triangle.
Therefore, we have:
Net moment of inertia,

l=mr2+mr2 =2 mr2 =2×200×102 =400×100 =40000 gm-cm2 =4×10-3 kg-m2

Question 10:

Particles of masses 1 g, 2 g, 3 g, ………., 100 g are kept at the marks 1 cm, 2 cm, 3 cm, ……….., 100 cm respectively on a metre scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale.

Answer:

It is given that the perpendicular bisector of the metre scale is passed through the 50th particle.
Therefore, on the L.H.S. of the axis, there will be 49 particles and on the R.H.S., there will be 50 particles.
Consider the two particles positioned at 49 cm and 51 cm.
Moment of inertia due to these two particles = 49 × (1)2 + 51 × (1)2
I1 = 100 × 1 = 100 gm-cm2
Similarly, if we consider particles positioned at 48 cm and 52 cm, we get:
I2  = 100 × (2)2 gm-cm2.
Thus, we will get 49 such sets and one particle at 100 cm. Therefore, total moment of inertia,

I=I1+I2+I3…..+I49 +I’Here, I’ is the moment of inertia of particle at 100 cm.

So, I=100 12+22+32+…+492+100 502       =100 12+22+32+…+502       =100×50×51×1016        =100×25×17×101=4292599 gm-cm2Or, I = 0.429 kg-m2 ≃ 0.43 kg-m2

Question 11:

Find the moment of inertia of a pair of spheres, each having a mass mass m and radius r, kept in contact about the tangent passing through the point of contact.

Answer:

It is given that two bodies of mass m and radius r are moving along a common tangent.

Moment of inertia of the first body about XY tangent,

I’=Icom+mr2

So, I’=25mr2+mr2=75mr2Moment of inertia of the second body about XY tangent,

I”=25mr2+mr2=75mr2Therefore, net moment of inertia,

I=I’+I”I=75mr2+75mr2

=145mr2 units

Question 12:

The moment of inertia of a uniform rod of mass 0⋅50 kg and length 1 m is 0⋅10 kg-m2 about a line perpendicular to the rod. Find the distance of this line from the middle point of the rod.

Answer:

Given:
Length of the rod, l = 1 m
Mass of the rod, m = 0.5 kg
Let the rod moves at a distance d from the centre.
On applying parallel axis theorem, we get:
Moment of inertia about that axis,

I=ml212+md2=0.10

I=0.5×l212+0.5×d2=0.10⇒112+d2=0.2⇒d2=0.118⇒d=0.342 m from the centre

Question 13:

Find the radius of gyration of circular ring of radius r about a line perpendicular to the plane of the ring and passing through one of its particles.

Answer:

Moment of inertia of the ring about a point on the rim of the ring and the axis perpendicular to the plane of the ring = mR2 + mR2 = 2mR2 (from parallel axis theorem)

We know that:mK2=2mR2K=Radius of the gyration⇒K=2R2=2R

Question 14:

The radius of gyration of a uniform disc about a line perpendicular to the disc equals its radius. Find the distance of the line from the centre.

Answer:

Moment of inertia of the disc about the centre and perpendicular to the plane of the disc =

12mr2
Radius of gyration of the disc about a point = Radius of the disc

Therefore, mk2=12mr2+md2(k = Radius of gyration about acceleration point; d = Distance of that point from the centre)

⇒K2=r22+d2⇒r2=r22+d2⇒r22=d2⇒d=r2

Question 15:

Find the moment of inertia of a uniform square plate of mass m and edge a about one of its diagonals.

Answer:

Let there be a small sectional area of width dx at a distance x from the x-axis.

So,
Mass of element

=ma2×a×dx Moment of inertia about x-axis,

Ixx=2∫0a/2ma2×adx×x2

⇒Ixx=2×max330a/2=2maa33×8=ma212Similarly, Iyy=ma212Now, Izz=Ixx+Iyy Perpendicular axis theorem⇒Izz=2×ma212=ma26The two diagonals are perpendicular to each other; therefore, we have:

Izz=Ix’x’+Iy’y’Also,Ixx=Iyy⇒Izz=2Ix’x’ ⇒Ix’x’=ma212

Question 16:

The surface density (mass/area) of a circular disc of radius a depends on the distance from the centre as

ρr=A+Br. Find its moment of inertia about the line perpendicular to the plane of the disc thorough its centre.

Answer:

Consider a ring of thickness dx at a distance r from the centre of the disc.
Mass of the ring ,

dm=A+Br×2πr×dr 

Therefore, moment of inertia about the centre,

I= ∫0 ar2dm

=∫oadA+Br 2πr.dr×r2=∫oa2πAr3 dr+∫oa 2πBr4 dr=2πA r44+2πB r550a=2πAa44+Ba55

Question 17:

A particle of mass m is projected with a speed u at an angle θ with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.

Answer:

Range of the particle

=u2sin2θgAt the highest point, we have:
Total force acting on the particle = mg (downward)
Distance between the line of force and the point of projection,

total range2=u2sin2θ2g

So,τ→=F→×d⊥=mg×u2sin2θ2g τ→=mu2sin2θ2   =mu2sinθcosθTherefore, the direction of torque is perpendicular to the plane of the motion.

Question 18:

A simple pendulum of length l is pulled aside to make an angle θ with the vertical. Find the magnitude of the torque of the weight ω of the bob about the point of suspension. When is the torque zero?

Answer:


Distance between the line of force and point of suspension, r = lsinθ

Torque, τ→=F→×r→⇒τ=wrsinθ=wlsinθHere, w is the weight of the bob.
The torque will be zero when the force acting on the body passes through the point of suspension, i.e., at the lowest point of suspension.

Question 19:

When a force of 6⋅0 N is exerted at 30° to a wrench at a distance of 8 cm from the nut, it is just able to loosen the nut. What force F would be sufficient to loosen it if it acts perpendicularly to the wrench at 16 cm from the nut?

Figure

Answer:

In the first case,

τ1=6sin30°×8100In first case,

τ2=F×16100To loosen the nut, torque in both the cases should be the same.
Thus, we have:
τ1=τ2

⇒F×16100=6sin30°×8100

⇒F=8×316=1.5 N

Question 20:

Calculate the total torque acting on the body shown in figure (10-E2) about the point O.

Figure

Answer:

Torque about a point = Total force × Perpendicular distance
Let the anticlockwise torque and clockwise acting torque be positive and negative, respectively.
Torque at O due to 5 N force is zero as it is passing through O.
Torque at O due to 15 N force,

τ1=15×6×10-2×sin37°   =15×6×10-2×35   =0.54 N-m anticlockwiseTorque at O due to 10 N force,

τ2=10×4×10-2=0.4 N-m clockwiseTorque at O due to 20 N force,

τ3=20×4×10-2×sin30°=20×4×10-2×12=0.4 N-m AnticlockwiseResultant torque acting at O,

τ=0.54-0.4+0.4  =0.54 N-m anticlockwise

Question 21:

A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its centre.

Answer:

Let N be the normal reaction on the block.

From the free body diagram of the block, it is clear that forces N and mgcosθ pass through the same line. So, there will be no torque due to N and mgcosθ. The only torque will be produced by mgsinθ.

∴ τ→=F→×r→                        a is the edge of the cube. Therefore, we have: r=a2∴τ=mgsinθ×a2    =12mgasinθ

Question 22:

A rod of mass m and length L, lying horizontally, is free to rotate about a vertical axis through its centre. A horizontal force of constant magnitude F acts on the rod at a distance of L/4 from the centre. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time t after the motion starts.

Answer:

Torque about the centre due to force,

τ→=F→×r→=F×L4×sin90°τ=F×L4Let the torque produces an angular acceleration α.

τ=Iα⇒τ=mL212×α  I of a rod=mL212⇒FL4=mL212×α⇒α=3FmLNow, θ=12αt2 (initially at rest)⇒θ=3Ft22mL

Question 23:

A square plate of mass 120 g and edge 5⋅0 cm rotates about one of the edges. If it has a uniform angular acceleration of 0⋅2 rad/s2, what torque acts on the plate?

Answer:

Consider a small area element of width dx and length a at a distance x from the axis of rotation.
Mass of element, dm

=ma2×adx(m = mass of the square, a = side of the plate)

l=∫0ax2dml=∫0ama2×ax2 dx =max330a=ma23

Now, torque produced,

τ=ma23×α  =120×10-3×52×10-43×.02  =40×25×10-7×0.2  =2×10-5 N-m.

Question 24:

Calculate the torque on the square plate of the previous problem if it rotates about a diagonal with the same angular acceleration.

Answer:

Moment of inertia of a square plate about its diagonals,
I =

ma212Therefore, we have:
Torque produced,

τ=Iα ⇒τ=ma212×α      =120×10-3×52×10-412  ×0.2     =10×25×10-4× 0.2     =0.5×10-5 N-m

Question 25:

A flywheel of moment of inertia 5⋅0 kg-m2 is rotated at a speed of 60 rad/s. Because of the friction at the axle, it comes to rest in 5⋅0 minutes. Find (a) the average torque of the friction, (b) the total work done by the friction and (c) the angular momentum of the wheel 1 minute before it stops rotating.

Answer:

Let the angular deceleration produced due to frictional force be α.
Initial angular acceleration,

ω0=60 rad/sFinal angular velocity,

ω=0t = 5 min =300 s
We know that:

ω=ω0+αt

⇒α=-ω0t⇒α=-60300=-15 rad/s2(a) Torque produced by the frictional force (R),

τ=Iα=5×-15  =1 N-m opposite to the rotation of wheel(b) By conservation of energy,
Total work done in stopping the wheel by frictional force = Change in energy

W=12Iω2   =12×5×60×60   =9000 joule=9 kJ(c) Angular velocity after 4 minutes,

ω=ω0+αt    =60-4×605    =605=12 rad/sSo, angular momentum about the centre,

L=Iω  =5×12=60 kg-m2/s

Question 26:

Because of the friction between the water in oceans with the earth’s surface, the rotational kinetic energy of the earth is continuously decreasing. If the earth’s angular speed decreases by 0⋅0016 rad/day in 100 years, find the average torque of the friction on the earth. Radius of the earth is 6400 km and its mass is 6⋅0 × 1024 kg.

Answer:

Rate of change of angular velocity, i.e., angular acceleration,
α =

0.0016100 rad/day

⇒α=0.0016864002×100×365 1 year=365 days=365×86400 secTorque produced by the ocean water in decreasing the Earth’s angular velocity,

τ=Iα=25mr2α =25×6×1024×64×1052×0.0016864002×100×365 =5.8×1020 N-m

Page No 197:

Question 27:

A wheel rotating at a speed of 600 rpm (revolutions per minute) about its axis is brought to rest by applying a constant torque for 10 seconds. Find the angular deceleration and the angular velocity 5 seconds after the application of the torque.

Answer:

Initial angular velocity of the wheel,

ω0=600 rpm=60060=10 revolutions per secondAfter 10 seconds,
Final angular velocity of the wheel,

ω=0So, ω0=-at⇒α=-1010=-1 rev/s2Now, t=5 sWe know that:ω’=ω0+at⇒ ω’=10-1×5=5 rev/s

Question 28:

A wheel of mass 10 kg and radius 20 cm is rotating at an angular speed of 100 rev/min when the motor is turned off. Neglecting the friction at the axle, calculate the force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions.

Answer:

Initial angular velocity of the wheel,

ω=100 rev/min   =10060=53 rev/s=10π3 rad/sθ=10 rev=20π radr=0.2 mAngular deceleration produced by the tangential force in order to stop the wheel after 10 revolutions,

α=ω22θTorque by which the wheel will come to rest,

τ=Icmα

Or,⇒ τ=Fr=IcmαPutting the values of Icm and α, we get:Fr=12mr2×10π32×12×20π⇒F=12×10×0.2×100π29×2×20π⇒F=5π18=15.718=0.87 N

Question 29:

A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed?

Answer:

After t seconds, let the angular velocity of two cylinders be ω.
For the first cylinder,

ω0=50 rev/s, α=1 rev/s2

∴ ω=50-αt⇒ t=ω-50-1For the second cylinder,

ω0=0, α=1 rev/s2

∴ ω=αt⇒ t=ω1

On equating the value of t, we get:ω=ω-50-1⇒2ω=50⇒ω=25 rev/s∴ t=251 s=25 s

Question 30:

A body rotating at 20 rad/s is acted upon by a constant torque providing it a deceleration of 2 rad/s2. At what time will the body have kinetic energy same as the initial value if the torque continues to act?

Answer:

Given:
Initial angular acceleration, ω0 = 20 rad/s
Angular deceleration,

α=2 rad/s2

⇒t1=ω0α1=202=10 sSo, in 10 s the body will come to rest.
The same torque continues to act on the body, therefore, it will produce the same angular acceleration.
The kinetic energy of the body is same as the initial value; therefore, its angular velocity should be equal to the initial value ω0.
Therefore, time required to come to that angular velocity,

t2=ωα=202=10 sSo, total time required,

t=t1+t2=20 s

Question 31:

A light rod of length 1 m is pivoted at its centre and two masses of 5 kg and 2 kg are hung from the ends as shown in figure (10-E3). Find the initial angular acceleration of the rod assuming that it was horizontal in the beginning.

Figure

Answer:

Total moment of inertia of the system about the axis of rotation,

Inet=m1r12+m2r22τnet=F1r1-F2r2Also, τnet=Inet×αOn equating the value of

τnetand putting the value of Inet, we get:

F1r1-F2r2=m1r12+m2r22×α

-2×10×0.5+5×10×0.5=5122+2122 α⇒15=74 α⇒α=607=8.57 rad/s2

Question 32:

Suppose the rod in the previous problem has a mass of 1 kg distributed uniformly over its length.
(a) Find the initial angular acceleration of the rod.
(b) Find the tension in the supports to the blocks of mass 2 kg and 5 kg.

Answer:

Total moment of inertia of the system about the axis of rotation,

Inet=m1r12+m2r22+ml212m and l are the mass and length of the rod, respectively. τnet= F1r1-F2r2Also, τnet=Inet×αOn equating the value of

τnetand putting the value of Inet, we get:

F1r1-F2r2=m1r12+m2r22+ml212×α

-2×10×0.5+5×10×0.5=5122+2122+1212 α

⇒ 15=1.75+0.084 α⇒α=1500175+8.4=1500183.4       =8.1 rad/s2 g=10       =8.01 rad/s2 if g=9.8

(b) From the free body diagram of the block of mass 2 kg,

T1-m1g=m1a⇒T1=2 a+g       =2αr+g   using, a=αr       =28×0.5+9.8⇒T1=27.6 NFrom the free body diagram of the block of mass 5 kg,

m2g-T2=m2a⇒T2=m2g-a       =5 g-a=5 9.8-8×0.5    a=αr       =5×5.8=29 N

Question 33:

Figure (10-E4) shows two blocks of mass m and M connected by a string passing over a pulley. The horizontal table over which the mass m slides is smooth. The pulley has a radius r and moment of inertia I about its axis and it can freely rotate about this axis. Find the acceleration of the mass M assuming that the string does not slip on the pulley.

Figure

Answer:


According to the equation, we have:

Mg-T1=Ma           … (i)          T2=ma           … (ii)⇒T1-T2=Iar2    …(iii)    Because, a=raIf we add the equations (i) and (ii), we get:

Mg+T2-T1=Ma+ma         … (iv)⇒Mg-Iar2=Ma+ma⇒M+m+Ir2 a=Mg⇒a=MgM+m+Ir2

Question 34:

A string is wrapped on a wheel of moment of inertia 0⋅20 kg-m2 and radius 10 cm and goes through a light pulley to support a block of mass 2⋅0 kg as shown in figure (10-E5). Find the acceleration of the block.

Figure

Answer:

Moment of inertia of the bigger pulley, I = 0.20 kg-m2
r = 10 cm = 0.1 m,
Smaller pulley is light. Therefore, on neglecting its moment of inertia, we have:
Mass of the block, m = 2 kg


From the free body diagram, we get:

mg-T=ma    … (i)

Tr=IαAnd

a=αr

⇒T=Iar2    … (ii)Using equations (i) and (ii), we get:

mg=m+Ir2 a⇒ a=mgm+Ir2       =2×9.82+0.20.01       =19.622=0.89 m/s2

Question 35:

Suppose the smaller pulley of the previous problem has its radius 5⋅0 cm and moment of inertia 0⋅10 kg-m2. Find the tension in the part of the string joining the pulleys.

Answer:

Given:
m = 2 kg, I1 = 0.10 kg-m2
r1 = 5 cm = 0.05 m
I2 = 2.20 kg-m2
r2 = 10 cm = 0.1 m

From the free body diagram, we have:

mg-T1=ma    … (i)T1-T2 r1=I1 α    … (ii)T2r2=I2α    … (iii)Substituting the value of T2 in the equation (ii), we get:

⇒T1-I2αr2 r1=I1α⇒T1-I2ar22=I1ar12⇒T1=I1r12+I2r22aSubstituting the value of T1 in the equation (i), we get:

mg-I1r12+I2r22a=ma

⇒mgI1r12+I2r22+m=a⇒a=2×9.80.10.0025+0.20.01+2       =0.316 m/s2⇒T2=I2ar22=0.20×0.3160.01=6.32 N

Question 36:

The pulleys in figure (10-E6) are identical, each having a radius R and moment of inertia I. Find the acceleration of the block M.

Figure

Answer:

Free the body diagram of the system,

For block of mass M,

Mg-T1=Ma    … (i)

T1-T2 R=Iαusing, a=αr⇒T1-T2=IaR2    … (ii) For pully 1Similarly, T2-T3=IaR2    … (iii) For pully 2For block of mass m,

T3-mg=ma    …(iv) For block mAdding equations (ii) and (iii), we get:

T1-T3=2IaR2    …(v) Adding equations (i) and (iv), we get:

-mg+Mg+T3-T1=Ma+ma    …(vi)Using equations (v) and (vi), we get:

Mg-mg = Ma+ma+2IaR2⇒a=M-mgM+m+2IR2

Question 37:

The descending pulley shown in figure (10-E7) has a radius 20 cm and moment of inertia 0⋅20 kg-m2. The fixed pulley is light and the horizontal plane frictionless. Find the acceleration of the block if its mass is 1⋅0 kg.

Figure

Answer:

Let the mass of block be m1 and mass of pulley be m2.
Acceleration of the massive pulley will be half of that of the block.

From the free body diagram, we have:

T1=m1a    … (i)T2-T1 r=IαT2-T1=Ia2r2=5a2    … (ii)m2g-m2a2=T1+T2    … (iii)Putting the value of mass in equation (i) and using equation (i) in equation (ii), we get:

T1=a and T2=72a

m2g=m2a2+72a+aOn replacing the value of m2 using 12mr2=I, we get:2Ir2g=2Ir2a2+92a   ⇒98=5a+4.5a⇒a=989.5=10.3 m/s2

Question 38:

The pulley shown in figure (10-E8) has a radius 10 cm and moment of inertia 0⋅5 kg-m2 about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4⋅0 kg block.

Figure

Answer:

From the figure, we have:

T1-m1gsinθ=m1a     … (i)T2-T1=Iar2    … (ii)m2gsinθ-T2=m2a    … (iii)Adding equations (i) and (iii), we get:

m2gsinθ+-T2+T1-m1gsinθ= m1+m2a⇒ m2-m1 gsinθ+-Ir2a=(m1+m2)a⇒ m2-m1 gsinθ=(m1+m2)a+Ir2a⇒ a=m2-m1gsinθm2+m1+Ir2        =4-2×10×124+2+0.5/0.01        =2×10×126+50        =0.248=0.25 m/s2

Question 39:

Solve the previous problem if the friction coefficient between the 2⋅0 kg block and the plane below it is 0⋅5 and the plane below the 4⋅0 kg block is frictionless.

Answer:


From the figure, we have:

m2gsinθ-T2=m1a    …iT1-m1gsinθ+μm1gcos θ=m1a    …iiT2-T1=Iar2    …iiiAdding equations (i) and (ii), we get:

m2gsinθ-m1gsinθ-μm1gcos θ+T1-T2=m1a+m2a

m2gsinθ-m1gsinθ+μm1gcos θ-Iar2=m1a+m2am2gsinθ-m1gsinθ+μm1gcos θ=m1a+m2a+Iar2

4×9.8×12-2×9.8×12+0.5×2×9.8×12=4+2+0.50.01a⇒27.80-13.90+6.95=56a⇒27.8-20.8=56a⇒a=756=0.125 m/s2

Question 40:

A uniform metre stick of mass 200 g is suspended from the ceiling thorough two vertical strings of equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tensions in the two strings.

Answer:

Given:
Mass of the stick =

m1=200 gMass of the small object =

m2=20Length of the string =

l=1 mAs the system is in equilibrium, we have:

τtotal=0  about O

T1×r1-T2×r2-m1g×r3=0 ⇒T1×0.7-T2×0.3-2×0.2×g=0⇒7T1-3T2=3.92    … (i)Now, we have:
Total upward force = Total downward force

T1+T2=m1g+m2g            =0.2×9.8+0.02×9.8⇒T1+T2=2.156    …ii

Solving equations (i) and (ii), we get:

T1=1.038 N≈1.04 N;T2=1.18≈1.12 N

Question 41:

A uniform ladder of length 10⋅0 m and mass 16⋅0 kg is resting against a vertical wall making an angle of 37° with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60⋅0 kg climbs up the ladder. If he stays on the ladder at a point 8⋅00 m from the lower end, what will be the normal force and the force of friction on the ladder by the ground? What should be the minimum coefficient of friction for the electrician to work safely?

Answer:

Let N2 be the normal force on the ladder by the ground.
Let N1 be the normal force on the ladder by the wall.
Let fr be the force of friction on the ladder by the ground.
As system is in translation equilibrium, we have:

N1=fr=μN2;N2=16 g+60 g=745 NApplying condition of rotational equilibrium at point O,
i.e., about point O, we have:

Γtotal=0

N1×10 cos37°=16g×5sin37°+60g×8sin37°

⇒8N1=48g+288g⇒N1=336×9.88=412 N∴ μ=N1N2=412745=0.553

Page No 198:

Question 42:

Suppose the friction coefficient between the ground and the ladder of the previous problem is 0⋅540. Find the maximum weight of a mechanic who could go up and do the work form the same position of the ladder.

Answer:

Let the maximum mass of a mechanic who could go up be m.
The system is in translation and rotational equilibrium; therefore, we have:

N2=16 g+mg    …iN1=μN2    …iiR1×10cos37°=16g×5sin37°-mg×8 sin37°    …iii ⇒8N1=48g+245mgFrom eq. (ii), we have:N2=48 g+245 mg8×0.54

Putting the value of N2 in eq. (i), we have:

16g+mg=24.0g+24mg5×8×0.54⇒16+m=24.0+24m40×0.54⇒m=44 kgTherefore, weight of the mechanic, who can go up and do the work, should be less than 44 kg.

Question 43:

A 6⋅5 m long ladder rests against a vertical wall reaching a height of 6⋅0 m. A 60 kg man stands half way up the ladder. (a) Find the torque of the force exerted by the man on the ladder about the upper end of the ladder. (b) Assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth, find the force exerted by the ground on the ladder.

Answer:

Given:
Mass of the man = m = 60 kg
Ladder length = 6.5 m
Height of the wall = 6 m

(a) We have to find the torque due to the weight of the body about the upper end of the ladder.

τ=60×10×6.52sinθ⇒τ=600×6.52×1-cos2θ⇒τ=600×6.52×1-66.52⇒τ=740 N-m(b) Let us find the vertical force exerted by the ground on the ladder.

R2=mg=60×9.8=588 NVertical force exerted by the ground on the ladder =

μR2=R1As system is in rotational equilibrium, we have:

τnet=0 about O

⇒6.5N1cosθ=60g×6.52sinθ

⇒N1=1260gtanθ       =1260g×2.56  using, tanθ=2.56⇒N1=252g⇒N1=122.5 N≈120 N

Question 44:

The door of an almirah is 6 ft high, 1⋅5 ft wide and weighs 8 kg. The door is supported by two hinges situated at a distance of 1 ft from the ends. If the magnitudes of the forces exerted by the hinges on the door are equal, find this magnitude.

Answer:

It is given that the magnitudes of the forces exerted by the hinges on the door are equal.

Therefore, we have:
Resultant of N1 and F1 at point A = Resultant of N2 and F2 at point B

⇒N12+F12=N22+F22    …(i)
System is in translation equilibrium. Therefore, we have:
N1 = N2;

8g=F1+F2Putting the value in eq. (i), we get:
F1 = F2

⇒2F1=8g⇒F1=40Let us take the torque about point B. We get:

N1×4=8g×0.75⇒N1=80×34×4=15 NNow, the forces exerted by the hinges A on the door,

F12+N12=402+152=42.72=43 N

Question 45:

A uniform rod of length L rests against a smooth roller as shown in figure (10-E9). Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make the horizontal is θ.

Figure

Answer:

Given:
Length of rod = L
It makes an angle θ with the floor.
Height of the wall is h.

The system is in translational equilibrium; therefore, we get:

N2=mg-N1cosθ    … iN1sin θ=μN2                 …  iiSystem is in rotational equilibrium; therefore, we have:N1cosθ×htanθ+N1sinθ×h=mg×L2 cosθ⇒N1=mg×L2cosθcos2θsinθh+hsinθ⇒N1cosθ=12mgLcos2θcos2θsinθh+hsinθFrom equation (ii), we have:

μ=N1sinθN2  =N1sinθmg-N1cosθ⇒μ=L/2cosθsinθ×2sinθ2 cos2θ +sin2θh-Lcos2θsinθ⇒μ=Lcosθsin2θ2h-Lcos2θsinθ

Question 46:

A uniform rod of mass 300 g and length 50 cm rotates at a uniform angular speed of 2 rad/s about an axis perpendicular to the rod through an end. Calculate (a) the angular momentum of the rod about the axis of rotation, (b) the speed of the centre of the rod and (c) its kinetic energy.

Answer:

Given:
Mass of the rod = m = 300 gm
Length of the rod = l = 50 cm
Angular velocity of the rod =

ω=2 rad/s

(a) Moment of inertia about one end of the rod,

I=mL23=0.3×0.5023⇒I=0.2510=0.025 kg-m2Angular momentum about that point,

L=IωL = 0.025 × 2 = 0.05 kg-m2/s

(b) Speed of the centre of the rod,

v=ωr=2×0.502=0.5 m/s(c) Kinetic energy,

K=12Iω2

⇒K=12×0.025×22=12×0.025×4=0.05 joule

Question 47:

A uniform square plate of mass 2⋅0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0⋅10 N-m. Calculate the angular momentum and the kinetic energy of the plate at the end of the fifth second after the start.

Answer:

Given:
Torque acting on the plate =

τ=0.10 N-mMass of the plate =

m=2 kg

On applying τ=Iα, we get:mr212×α=0.10 N-m⇒α=60 rad/sLet ω be the angular velocity after time t (t = 5 s).

Therefore, we have: ω=ω0+at⇒ω=60×5=300 rad/sAngular momentum, Iω=0.1060×300    =0.50 kg-m2/sKinetic energy,12Iω2=12×0.1060×3002         =75 joule

Question 48:

Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth = 6400 km and radius of the orbit of the earth about the sun = 1⋅5 × 108 km.

Answer:

Given:
r = 6400 km = 6.4 × 106 m;
R = 1⋅5 × 108 km = 1⋅5 × 1011 m
About its axis, we have:
T = 1 day = 86400 s;

ω=2πTAngular momentum of the Earth about its axis,

L=Iω   =25mr2×2π86400About the Sun’s axis,
T = 365 day = 365 × 86400 s
Angular momentum of the Earth about the Sun’s axis,

L’=mR2×2π86400×365Ratio of angular momentums,

LL’=2/5mr2×2π/86400mR2×2π/86400×365     =2r2×3655R2=2×6.4×1062×3655×1.5×10112     =2.65×10-7

Question 49:

Two particles of masses m1 and m2 are joined by a light rigid rod of length r. The system rotates at an angular speed ω about an axis through the centre of mass of the system and perpendicular to the rod. Show that the angular momentum of the system is

L=μ r2ωwhere

μis the reduced mass of the system defined as

μ=m1+m2m1+m2.

Answer:

Distance of centre of mass from mass m1,

d1=m2rm1+m2Angular momentum due to the mass m1 at the centre of system,

L1=I1ω=m1d12ω=m1m2rm1+m22ω ⇒L1=m1m22r2m1+m22ω         … 1Similarly, the angular momentum due to the mass m2 at the centre of system,

L2=m2m1rm1m22ω   =m2m12r2m1+m22ω           … 2Therefore, net angular momentum,

L=L1+L2  =m1m22r2ωm1+m22+m2m12r2ωm1+m22  =m1m2 m1+m2 r2 ωm1+m22  =m1m2m1+m22r2ω  =μr2ω

Question 50:

A dumb-bell consists of two identical small balls of mass 1/2 kg each connected to the two ends of a 50 cm long light rod. The dumb-bell is rotating about a fixed axis thorough the centre of the rod and perpendicular to it at an angular speed of 10 rad/s. An impulsive force of average magnitude 5⋅0 N acts on one of the masses in the direction of its velocity for 0⋅10 s. Find the new angular velocity of the system.

Answer:

Moment of inertia of the dumb-bell,

I=mr2+mr2=2mr2Torque,

τ=Iα

⇒ F×r=mr2+mr2 α⇒5×0.25 = 2mr2×α⇒α=1.252×0.5×0.252=20 rad/s2Given:

ω0=10 rad/s and t=0.10 s

Using ω=ω0+αt, we get:ω=10+20×0.10=10+2=12 rad/s

Question 51:

A wheel of moment of inertia 0⋅500 kg-m2 and radius 20⋅0 cm is rotating about its axis at an angular speed of 20⋅0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel.

Answer:

Given:
Initial moment of inertia of the system,
I1 = 0.500 kg-m2;
r = 0.2 m;
ω = 20 rad/s
Mass of the stationary particle, m = 0.2 kg
Final moment of inertia of the system,
I2 = I1 + mr2
It is given:
External torque = 0
Angular momentum is conserved; therefore, we have:

I1ω1=I2ω2

⇒0.5×20=0.5+0.2×0.22 ω2⇒ω2=100.508≈19.7 rad/s

Question 52:

A diver having a moment of inertia of 6⋅0 kg-m2 about an axis thorough its centre of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to decrease the moment of inertia to 5⋅0 kg-m2, what will be the new angular speed?

Answer:

Initial moment of inertia of diver, I1 = 6 kg-m2
Initial angular velocity of diver, ω1 = 2 rad/s
Final moment of inertia of diver, I2 = 5 kg-m2
Let ω2 be the final angular velocity of the diver.
We have:
External torque = 0

∴ I1ω1 = I2ω2

⇒ω2=6×25=2.4 rad/s

Question 53:

A boy is seated in a revolving chair revolving at an angular speed of 120 revolutions per minute. Two heavy balls form part of the revolving system and the boy can pull the balls closer to himself or may push them apart. If by pulling the balls closer, the boy decreases the moment of inertia of the system from 6 kg-m2 to 2 kg-m2, what will be the new angular speed?

Answer:

Given:
Initial angular speed of the system,

ω1=120 rpm=120×2π60=4π rad/sInitial moment of inertia of the system,

I1=6 kg-m2Final moment of inertia of the system,

I2=2 kg-m2Two balls are inside the system; therefore, we get:
Total external torque = 0

∴I1ω1=I2ω2⇒6×4π=2ω2⇒ω2=12π rad/sOr, ω2=6 rev/s=360 rev/minute

Question 54:

A boy is standing on a platform which is free to rotate about its axis. The boy holds an open umbrella in his hands. The axis of the umbrella coincides with that of the platform. The moment of inertia of “the platform plus the boy system” is 3⋅0 × 10−3 kg-m2 and that of the umbrella is 2⋅0 × 10−3 kg-m2. The boy starts spinning the umbrella about the axis at an angular speed of 2⋅0 rev/s with respect to himself. Find the angular velocity imparted to the platform.

Answer:

Given:
Moment of inertia of umbrella = I1 = 2 × 10−3 kg-m2
Moment of inertia of the system = I2 = 3 × 10−3 kg-m2
Angular speed of the umbrella with respect to the boy = ω1 = 2 rev/s
Let the angular velocity imparted to the platform be ω2.
Taking the Earth as the reference, we have:
Angular velocity of the umbrella = (ω1ω2)
Applying conservation of angular momentum, we get:

I1ω1-ω2=I2ω2⇒2×10-3 2-ω2=3×10-3ω2⇒5ω2=4⇒ω2=0.8 rev/s

Question 55:

A wheel of moment of inertia 0⋅10 kg-m2 is rotating about a shaft at an angular speed of 160 rev/minute. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with a common angular speed of 200 rev/minute. Find the moment of inertia of the second wheel.

Answer:

Given:
For the first wheel,
I1 = 10 kg-m2 and ω1 = 160 rev/min
For the second wheel, ω2 = 300 rev/min
Let I2 be the moment of inertia of the second wheel.
After they are coupled, we have:
ω = 200 rev/min
If we take the two wheels to be an isolated system, we get:
Total external torque = 0
Therefore, we have:

I1ω1+I2ω2=I1+I2 ω

⇒0.10 ×160+I2×300=0.10+I2 × 200⇒16+300I2=20+200I2⇒100I2=4⇒I2=4100=0.04 kg-m2

Question 56:

A kid of mass M stands at the edge of a platform of radius R which can be freely rotated about its axis. The moment of inertia of the platform is I. The system is at rest when a friend throws a ball of mass m and the kid catches it. If the velocity of the ball is

νhorizontally along the tangent to the edge of the platform when it was caught by the kid, find the angular speed of the platform after the event.

Answer:

On considering two bodies as a system, we get:
Moment of inertia of kid and ball about the axis

=M+m R2Applying the law of conservation of angular momentum, we have:

mνR=I+M+m R2 ω

⇒ω=mνRI+M+m R2

Question 57:

Suppose the platform of the previous problem is brought to rest with the ball in the hand of the kid standing on the rim. The kid throws the ball horizontally to his friend in a direction tangential to the rim with a speed

νas seen by his friend. Find the angular velocity with which the platform will start rotating.

Answer:

Initial angular momentum of the system is zero (L1 = 0).
Let the platform starts rotating with an angular velocity ω.
Therefore, we have:
Final angular momentum of the system = L2

=Iω+MR2ω-mvR(m = mass of the ball, v = velocity of the ball)
Total external torque = 0
Also,
Initial angular momentum = Final angular momentum

⇒Iω+MR2 ω-mvR=0⇒ω=mvRI+MR2

Question 58:

Suppose the platform with the kid in the previous problem is rotating in anticlockwise direction at an angular speed ω. The kid starts walking along the rim with a speed

νrelative to the platform also in the anticlockwise direction. Find the new angular speed of the platform.

Answer:

Let ω’ be the angular velocity of platform after the kid starts walking.
Let ω be the angular velocity of the wheel before walking.
When we see the (kid-wheel) system from the initial frame of reference, we can find that the wheel moves with a speed of

ωand the kid with a speed of

ω’+νR, after the kid has started walking.
Initial angular momentum of the system = (I + MR2)ω
Angular velocity of the kid after it starts walking =

ω’+vRExternal torque is zero; therefore, angular momentum is conserved.

⇒I+MR2ω=Iω’+MR2 ω’+vR⇒I+MR2ω=Iω’+MR2 ω’+vR⇒I+MR2ω-MvR= I+MR2 ω’⇒ω’=I+MR2ω -MvRI+MR2⇒ω’=ω- MvRI+MR2

Page No 199:

Question 59:

A uniform rod of mass m and length l is struck at an end by a force F perpendicular to the rod for a short time interval t. Calculate
(a) the speed of the centre of mass, (b) the angular speed of the rod about the centre of mass, (c) the kinetic energy of the rod and (d) the angular momentum of the rod about the centre of mass after the force has stopped to act. Assume that t is so small that the rod does not appreciably change its direction while the force acts.

Answer:

Given:
Mass of the rod = m
Length of the rod = l

(a) For the centre of mass,
Acceleration,

a=Fm

Velocity,  v=at⇒v=Ftm(b) Let the angular speed about the centre of mass be ω.
Moment of inertia of rod about centre of mass =

I=ml212

Iω=mvr

⇒ml212×ω=mνl2⇒ml212×ω=m×Ftm×l2⇒ω=6Ftml(c) Kinetic energy,

K.E.=12mν2+12Iω2      =12×mFtm2+12×Iω2      =12×mF2t2m2+12×ml212×36F2t2m2l2      =2F2t2m(d) Angular momentum about the centre of mass,

L=mνr  =m×Ftm×l2=Flt2

Question 60:

A uniform rod of length L lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Find the distance travelled by the centre of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic.

Answer:

Let initial velocity of the particle = u1
Let final velocity of the particle = v1
Let time taken to move π/2 angle = t
Given v1 = 0
Let initial velocity of CM of the rod =  u2
Given u2 = 0
Let final velocity of the CM of rod = v2
By conservation of linear momentum
mu1 + Mu2 = mv1 + Mv2
=> v2 = (m/M)u1 —(1)

Now by we consider the angular momentum imparted by the particle to the rod.
mu1(L/2)  = Iω
For rod about its CM, I = ML2/12
mu1(L/2)  = ωML2/12
=> ω = 6mu1/ML  —2.

Now we know
ω = θ/t
θ = π/2
=> t = θ/ω = (π/2)/(6mu1/ML)
=> t = πML/(12mu1)

Linear distance moved by the CM of the rod will be
s = v2t
By equation 1.
s = [πML/(12mu1)]× (m/M)u1
= πL/12

Initial KE of the particle is ½m(u1)2
KE of the CM of the rod =  ½M(v1)2 = ½M{(m/M)u1}2  [by equation 1.]
= ½(u1)2m2/M
KErot of the rod = ½Iω2
KErot = ½ (ML2/12)(6mu1/ML)2
=  (3/2) m2(u1)2/M

To be elastic collision KE must be conserved
½m(u1)2 = ½(u1)2m2/M + (3/2) m2(u1)2/M
=> M = 4m

Hence Proved.

Question 61:

Suppose the particle of the previous problem has a mass m and a speed

νbefore the collision and it sticks to the rod after the collision. The rod has a mass M. (a) Find the velocity of the centre of mass C of the system constituting “the rod plus the particle”. (b) Find the velocity of the particle with respect to C before the collision. (c) Find the velocity of the rod with respect to C before the collision. (d) Find the angular momentum of the particle and of the rod about the centre of mass C before the collision. (e) Find the moment of inertia of the system about the vertical axis through the centre of mass C after the collision. (f) Find the velocity of the centre of mass C and the angular velocity of the system about the centre of mass after the collision.

Answer:

(a) It is given that no external torque and force is applied on the system.
Applying the law of conservation of momentum, we get:

mν=M+m ν’

⇒ν’=mνM+m(b) Velocity of the particle w.r.t. centre of mass (COM) C before the collision =

vc=v-v’

⇒vc=v-mvM+m=MvM+v(c) Velocity of the particle w.r.t. COM C before collision

=-MνM+m(d) Distance of the COM from the particle,

xcm=m1x1+m2x2m1+m2⇒r=M×L2+m×0M+m⇒r=ML2M+m∴ Angular momentum of body about COM

=mvr=m×MvM+m×ML2M+m=M2 mvL2M+m2∴ Angular momentum of rod about COM

=M×mvM+m×12mLM+m= Mm2vL2 M+m2(e) Moment of inertia about COM = I

=I1+I2 I=mML2M+M2+ML212+MML2m+M2  =mM2L24m+M2+ML212+Mm2L24M+m2 =3mM2L2+Mm+M2 L2+3Mm2L212m+M2 =MM+4mL212M+m(f) About COM,

Vcm=mνM+m∴ Iω=mvr=mv×ML2 M+m⇒ ω=mνML2 M+m×12 M+mM M+4mL2       =6mνM+4mL

Question 62:

Two small balls A and B, each of mass m, are joined rigidly by a light horizontal rod of length L. The rod is clamped at the centre in such a way that it can rotate freely about a vertical axis through its centre. The system is rotated with an angular speed ω about the axis. A particle P of mass m kept at rest sticks to the ball A as the ball collides with it. Find the new angular speed of the rod.

Answer:

Let the new angular speed of the rod be ω‘.
Here, net torque on the system is zero.
So, the angular momentum is conserved.
Therefore, we get:

Iω=I’ω’

mL22+mL22ω=2mL22+mL22ω’

2mL2ω=3mL2ω’⇒ω’ =23ω

Question 63:

Two small balls A and B, each of mass m, are joined rigidly to the ends of a light rod of length L (figure 10-E10). The system translates on a frictionless horizontal surface with a velocity

ν0in a direction perpendicular to the rod. A particle P of mass m kept at rest on the surface sticks to the ball A as the ball collides with it. Find
(a) the linear speeds of the balls A and B after the collision, (b) the velocity of the centre of mass C of the system A + B + P and (c) the angular speed of the system about C after the collision.

Figure

Answer:

(a) Collision will not affect the velocity of ball B because the light rod will exert a force on it only along its length.
Therefore, we have:
Velocity of B = v0

For ball A,

On applying the law of conservation of linear momentum, we get:
mν0=2m×ν

⇒v=v02

∴ Velocity of A

=ν02(b) If we consider the three bodies to be a system, we have:
Net external force = 0

vCM=m×v0+2m×v0/2n+2m     =mv0+mv03m     =2v03(Direction will be same as the initial velocity before collision.)

(c) Velocity of (A + P) w.r.t. the centre of mass

=2v03-v02=v06Velocity of B w.r.t. the centre of mass

=v0-2v03=v03Distance of the (A + P) from the centre of mass

=l3Distance of the B from the centre of mass =

2l3

Applying mvcomr=lcm×ω, we have:2m×vo6×l3+m×v03×2l3=2ml32+2l32m×ω⇒6mv0l18=6ml9ω⇒ω=v02l

Question 64:

Suppose the rod with the balls A and B of the previous problem is clamped at the centre in such a way that it can rotate freely about a horizontal axis through the clamp. The system is kept at rest in the horizontal position. A particle P of the same mass m is dropped from a height h on the ball B. The particle collides with B and sticks to it. (a) Find the angular momentum and the angular speed of the system just after the collision. (b) What should be the minimum value of h so that the system makes a full rotation after the collision.

Answer:

(a) Angular momentum

=mνrLet the particle P collides the ball B with a speed u and system moves with speed v just after the collision.
Applying the law of conservation of linear momentum, we get:

mu=2mν-mν=mν

∴u=ν

Velocity, u=2ghand r=L2Initial angular momentum of system about COM of rod,

mur=m×2gh×L2           =mLgh2Angular momentum of system about COM of the rod just after the collision

=lω

I=2mL24+mL24=3mL24Applying the law of conservation of angular momentum and obtaining the value of ω, we get:

ω=mLgh23mL24=8gh3L(b) When the mass 2m and m are at the top most position and at the lowest point, respectively, they will automatically rotate. In this position, we have:
Total gain in potential energy

=2mgL2-mgL2Kinetic energy =

12Iω2Therefore, by the law of conservation of energy, we have:

mgL2=12×34mL2×8gh9gL2⇒h=3L2

Question 65:

Two blocks of masses 400 g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia

1·6×10-4 kg-m2and a radius 2⋅0 cm, Find (a) the kinetic energy of the system as the 400 g block falls through 50 cm, (b) the speed of the blocks at this instant.

Answer:

From the free body diagram, we have:

0.4g-T1=0.4a    …(i)T2-0.2g=0.2a    …(ii)T1-T2 r=lar    …(iii)On solving the above equations, we get:

a=0.4-0.2 g0.4+0.2+1.60.4=g5On solving the (b) part of the question first, we have:
Speed of the blocks =

v=2ah=2×g5×0.5=9.85=1.4 m/s(a) Total kinetic energy of the system

=12m1v2+12m2v2+12Iω2=12m1v2+12m2v2+12Ivr2=12×0.4×1.42+12×0.2×1.42+12×1.6×10-4×1.42×10-22=0.2+0.1+0.21.42=0.5×1.96=0.98 joule

Question 66:

The pulley shown in figure (10-E11) has a radius of 20 cm and moment of inertia 0⋅2 kg-m2. The string going over it is attached at one end to a vertical spring of spring constant 50 N/m fixed from below, and supports a 1 kg mass at the other end. The system is released from rest with the spring at its natural length. Find the speed of the block when it has descended through 10 cm. Take g = 10 m/s2.

Answer:

Given:
Moment of inertia of the pully = I = 0.2 kg-m2
Radius of the pully = r = 0.2 m
Spring constant of the spring = k = 50 N/m
Mass of the block = m = 1 kg
g = 10 ms2 and h = 0.1 m
On applying the law of conservation of energy, we get:

mgh=12mv2+12kx2+12Iωr2On putting x = h = 0.1 m, we get:

1=12×1×v2+12×0.2×v20.04+12×50×0.01⇒1=0.5v2+2.5v2+14⇒3v2=34 ⇒v2=14⇒v=12=0.5 m/s

Question 67:

A metre stick is held vertically with one end on a rough horizontal floor. It is gently allowed to fall on the floor. Assuming that the end at the floor does not slip, find the angular speed of the rod when it hits the floor.

Answer:

Let the mass of the rod and its angular velocity be m and ​ω (when hits the ground), respectively.
It is given that the rod has rotational motion only.

On applying the law of conservation of energy, we get:

12Iω2=mgl2⇒ml23×ω2=mgl⇒ω2=3gl⇒ω= 3gl= 3×9.81⇒ω= 5.42 rad/s

Question 68:

A metre stick weighing 240 g is pivoted at its upper end in such a way that it can freely rotate in a vertical place through this end (figure 10-E12). A particle of mass 100 g is attached to the upper end of the stick through a light string of length 1 m. Initially, the rod is kept vertical and the string horizontal when the system is released from rest. The particle collides with the lower end of the stick and sticks there. Find the maximum angle through which the stick will rise.

Figure

Answer:

Let the angular velocity of stick after the collision be ω.
Given:
Mass of the stick = m = 240 g
Mass of the particle = m‘ = 100 g
Length of the string (or rod) = r = l = 1 m
Moment of inertia of the particle about the pivoted end =

I’=m’r2

⇒I’=0.1×12⇒I’=0.1 kg-m2Moment of inertia of the rod about the pivoted end =

I=ml23

⇒I=0.243×12⇒I=0.08 kg-m2Applying the law of conservation of energy, we get:
Final energy of the particle = Initial energy of the rod

mgh=12Iω2⇒12Iω2=0.1×10×1⇒ω=20For collision, we have:
0.1×12×20+0

=0.243×12+0.12 12 ω⇒ω=2010×0.18⇒0-12Iω2=-m1gl1-cos θ -m2g12 1-cos θ⇒12 Iω2=-mgl 1-cos θ -m2g12 1-cos θ

⇒12Iω2= 0.1×10 1-cos θ -0.24×10×0.5 1-cos θ⇒12×0.18×20324=2.2×9 1-cos θ⇒1-cos θ=12.2×1.8⇒1-cos θ=0.252⇒cos θ=1-0.252=0.748⇒θ=cos-1 0.748=41°

Question 69:

Find the magnitude of the force acting on a particle of mass dm at the tip of the rod when the rod makes and angle of 37° with the vertical.A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60° and then released.

Answer:

Let the length of the rod be l.
Mass of the rod be m.
Let the angular velocity of the rod be ω when it makes an angle of 37° with the vertical.

On applying the law of conservation of energy, we get:

12Iω2-0=mgl2cos37°-cos60°⇒12×ml2ω23=mgl245-12⇒ω2=9g10l

Let the angular acceleration of the rod be α when it makes an angle of 37° with the vertical.

Using τ=Iα, we get:Iα=mgl2sin37°⇒ml23α=mgl2×35⇒α=0.9glForce on the particle of mass dm at the tip of the rod

Fc=centrifugal force    =dmω2l=dm9g10ll⇒Fc=0.9gdmFt=tangential force   =dmαl⇒Ft=0.9gdmSo, total force on the particle of mass dm at the tip of the rod will be the resultant of Fc and Ft.

∴ F=Fc2+Ft2        =0.92gdm

Question 70:

A cylinder rolls on a horizontal place surface. If the speed of the centre is 25 m/s, what is the speed of the highest point?

Answer:

Let vc be the translational velocity of the cylinder.
Let ω be the rotational velocity of the cylinder.
Let r be the radius of the cylinder.
For rolling, we have:
vc =   

Speed of the highest point = vc +    = 2vc

2 × 25 m/s= 50 m/s

Page No 200:

Question 71:

A sphere of mass m rolls on a plane surface. Find its kinetic energy at an instant when its centre moves with speed

ν.

Answer:

Let radius of the sphere be R and its angular speed be ω.
Moment of inertia of sphere,

I=25mR2Total kinetic energy,

K=12Iω2+12mv2K=12×25mR2 ×v2R2+12mv2K=210mv2+12mv2K=2+5mv210=710mv2

Question 72:

A string is wrapped over the edge of a uniform disc and the free end is fixed with the ceiling. The disc moves down, unwinding the string. Find the downward acceleration of the disc.

Answer:

Let the radius of the disc be R.
Let the tension in the string be T.
Let the acceleration of the disc be a.

From the free body diagram, we have:

mg-T=ma    …(i)
Torque about the centre of disc,

T×R=I×α

⇒ T×R=12mR2×aR⇒T=12ma    … (ii)Putting this value in equation (i), we get:

mg-12ma=ma⇒mg=32ma⇒a=2g3

Question 73:

A small spherical ball is released from a point at a height h on a rough track shown in figure (10-E13). Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track.

Figure

Answer:

Let r be the radius of the ball.
Let v be the linear speed of the ball when it rolls on the horizontal part of the track.
Let ω be the angular speed of the ball when it rolls on the horizontal part of the track.
Potential energy the ball has gained w.r.t. the surface will be converted to angular kinetic energy about the centre and linear kinetic energy.
Therefore, we have:

mgh=12Iω2+mv2

⇒mgh=12×25mR2×vR2+12mv2⇒gh=15v2+12v2⇒v2=107gh⇒v=10gh7

Question 74:

A small disc is set rolling with a speed

νon the horizontal part of the track of the previous problem from right to left. To what height will it climb up the curved part?

Answer:

Let the radius of the disc be R.
Let the angular velocity of the disc ω.
Let the height attained by the disc be h.

On applying the law of conservation of energy, we get:

12mv2+12Iω2=mgh⇒12mv2+12×12mR2×vR2=mgh⇒12v2+14v2=gh⇒34v2=gh⇒h=3v24g

Question 75:

A sphere starts rolling down an incline of inclination θ. Find the speed of its centre when it has covered a distance l.

Answer:

Let radius of the sphere be r. Let r be negligible w.r.t. l.

Potential energy of the sphere, P.E. =

mglsinθTotal kinetic energy of the sphere of mass m rolling with speed v =

710mv2On applying the law of conservation of energy, we get:

mglsinθ=710mv2⇒glsinθ=710ν2⇒ν=107glsinθ

Question 76:

A hollow sphere is released from the top of an inclined plane of inclination θ. (a) What should be the minimum coefficient of friction between the sphere and the plane to prevent sliding? (b) Find the kinetic energy of the ball as it moves down a length l on the incline if the friction coefficient is half the value calculated in part (a).

Answer:

It is given that a hollow sphere is released from the top of an inclined plane of inclination θ.
(a) To prevent sliding, the body will make only perfect rolling. In this condition, we have:

mgl sin θ-f=ma                          … (1)f×R=23 mR2×aR⇒f=23 ma                                  … (2)On putting this value in the equation (1), we get:

mg sin θ-23 ma=ma

⇒    a=35 g sin θFrom equation (1), we have:

mg sin θ-f=35 mg sin θ⇒ f=25 mg sin θ⇒ μmg cos θ=25 mg sinθ⇒ μ=25 tan θ(b)

15 tan θ mg cos θ R=23 mR2α⇒          α=310 g sin θR

ac=g sin θ-g5 sin θ=45 g sin θ⇒t2=2lac       = 2l 4gsin θ5 52g sin θ ∴  ω=at     K.E.=12 mν2+12 Iω2=12 m 2al+12 l a2t2=12 m 4g sin θ5×2×l+12×23 mR2×9100=sin2 θR×5L2g sin θ=4 mgl sin θ5+3 mgl sin θ40=78 mgl sin θ

Question 77:

A solid sphere of mass m is released from rest from the rim of a hemispherical cup so that it rolls along the surface. If the rim of the hemisphere is kept horizontal, find the normal force exerted by the cup on the ball when the ball reaches the bottom of the cup.

Answer:

The radius of the solid sphere (r) is negligible w.r.t. the radius of the hemispherical cup (R).
Let the ball reaches the bottom of the cup with a velocity v and angular velocity ω.
On applying the law of conservation of energy, we get:

mgR=12Iω2+12mv2

⇒mgR=12×25 mν2+12 mν2⇒710 mν2=mgR⇒ν2=107gRAs shown in the free body diagram, we have:
Total normal force on the ball

=mg+mν2R

=mg+m107gRR=mg+mg 107=177mg

Question 78:

Figure (10-E14) shows a rough track, a portion of which is in the form of a cylinder of radius R. With what minimum linear speed should a sphere of radius r be set rolling on the horizontal part so that it completely goes round the circle on the cylindrical part.

Figure

Answer:

Let the sphere be thrown with velocity

v’ and its velocity becomes v at the top-most point.
From the free body diagram of the sphere, at the topmost point, we have:

mv2R-r=mg⇒v2=gR-rOn applying the law of conservation of energy, we have:

12mν’2+12Iω’2=2mgR-r+12mν2+12Iω2⇒710mν’2=2mgR-r+710mν2⇒710mν’2=2mgR-r+710mg(R-r)⇒710ν’2=27gR-r10⇒ν’=277 gR-r

Question 79:

Figure (10-E15) shows a small spherical ball of mass m rolling down the loop track. The ball is released on the linear portion at a vertical height H from the lowest point. The circular part shown has a radius R.
(a) Find the kinetic energy of the ball when it is at a point A where the radius makes an angle θ with the horizontal.
(b) Find the radial and the tangential accelerations of the centre when the ball is at A.
(c) Find the normal force and the frictional force acting on the if ball if H = 60 cm, R = 10 cm, θ = 0 and m = 70 g.

Figure

Answer:

(a) Let the velocity and angular velocity of the ball at point A be v and ω, respectively.
Total kinetic energy at point A

=12mv2+12Iω2

Total potential energy at point A

=mgR+RsinθOn applying the law of conservation of energy, we have:
Total energy at initial point = Total energy at A
Therefore, we get:

mgH=12mv2+12Iω2+mgR1+sinθ⇒mgH-mgR1+sinθ=12mν2+12Iω2⇒12mv2+12Iω2=mgH-R-Rsinθ    …1Total K.E. at A=mgH-R-Rsinθ(b) Let us now find the acceleration components.
Putting

I=25mR2 and ω=vRin equation (1), we get:

710mv2=mgH-R-Rsinθ⇒v2=107gH-R-Rsinθ    …2Radial acceleration,

ar=v2R=107gH-R-RsinθRFor tangential acceleration,

Differentiating equation 2 w.r.t. ‘t’ ,2vdvdt=-107gRcosθdθdt⇒ωRdvdt=-57 gRcosθdθdt⇒dvdt=-57 gcosθ⇒at=-57 gcosθ(c) At

θ=0, from the free body diagram, we have:

Normal force =

N=mar

N=m×107gH-R-RsinθR   =701000×107×10 0.6-0.10.1   = 5 NAt

θ=0, from the free body diagram, we get:

fr=mg-mat  (

fr = Force of friction)

⇒fr=mg-at       =m10-57×10       =0.0710-57×10       =1100 70-50=0.2 N

Question 80:

A thin spherical shell of radius R lying on a rough horizontal surface is hit sharply and horizontally by a cue. Where should it be hit so that the shell does not slip on the surface?

Answer:

If the shell does not slip on the surface, its motion should be pure rolling.
Let the cue hits at a height ‘h‘ above the centre.
Let the centre of shell moves with velocity vc and shell rotates with angular velocity ω after hitting.
For pure rolling,

vc=RωOn applying the law of conservation of angular momentum at point O, we get:

mvch=Iωmvch=23mR2 vcRh=2R3

Question 81:

A uniform wheel of radius R is set into rotation about its axis at an angular speed ω. This rotating wheel is now placed on a rough horizontal surface with its axis horizontal. Because of friction at the contact, the wheel accelerates forward and its rotation decelerates till the wheel starts pure rolling on the surface. Find the linear speed of the wheel after it starts pure rolling.

Answer:

Initial angular momentum,

L=Iω  =12mR2ωAngular momentum after it starts pure rolling,

L’=Iω’+mv×R   =12mR2vR+mvR   =32mVRAs no external torque is applied, angular momentum will be conserved.
Therefore, we have:
L = L’

⇒12mR2ω=32mvR⇒v=ωR3

Question 82:

A thin spherical shell lying on a rough horizontal surface is hits by a cue in such a way that the line of action passes through the centre of the shell. As a result, the shell starts moving with a linear speed

νwithout any initial angular velocity. Find the linear speed of the shell after it starts pure rolling on the surface.

Answer:

Initial angular momentum about point A,

L=mvRAngular momentum about point A’ after it starts pure rolling,

L’=Iω+mv×R   =23mR2v’R+mv’R   =53mv’RAs no external torque is applied, angular momentum will be conserved.
Therefore, we have:
L = L

⇒mvR=53mv’R⇒v’=3v5

Question 83:

A hollow sphere of radius R lies on a smooth horizontal surface. It is pulled by a horizontal force acting tangentially from the highest point. Find the distance travelled by the sphere during the time it makes one full rotation.

Answer:

Let M be the mass of the hollow sphere and α be the angular acceleration produced in the sphere by the tangential force F.

Torque due to this force,

τ=F×RAlso,

τ=Iα

So, F×R=23MR2α⇒α=3F2MRApplying

θ=ω0t+12αt2, we get:

2π=12αt2⇒t2=8πMR3FLet d be the distance travelled in this time t.
Acceleration,

a=FM

∴ d=12at2        = 12×FM×8πMR3F        =4πR3

Question 84:

A solid sphere of mass 0⋅50 kg is kept on a horizontal surface. The coefficient of static friction between the surfaces in contact is 2/7. What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface?

Answer:

Let α be the angular acceleration produced in the sphere.

Rotational equation of motion,

F×R-fr×R=Iα

⇒F=25mRα+μmg    … (1)Translational equation of motion,

F=ma-μmg⇒a=F+μmgmFor pure rolling, we have:

α=aR

⇒α=F+μmgmRPutting the value of

αin equation (1), we get:

F=25mRF+μmgmR+μmg⇒F=25F+μmg μmg⇒F=25F+25×27×0.5×10+27×0.5×10⇒3F5=47+107=2⇒F=5×23=103=3.3 N

Question 85:

A solid sphere is set into motion on a rough horizontal surface with a linear speed

νin the forward direction and an angular speed

ν/R in the anticlockwise directions as shown in figure (10-E16). Find the linear speed of the sphere (a) when it stops rotating and (b) when slipping finally ceases and pure rolling starts.

Figure

Answer:

(a)

Initial angular momentum about point A,

L=mv×R-Iω   =mvR-25mR2vR   =35mvRAngular momentum about point A’ after it stops rotating,

L’=mv’RAs no external torque is applied, angular momentum will be conserved.
Therefore, we have:
L = L’

⇒mvR=53mv’R⇒v’=3v5(b)

Angular momentum about point A after it stops rotating,

L’=mv’RAngular momentum about point A’ after it starts pure rolling.

L”=Iω”+mv”×R   =25mR2v”R+mv”R   =75mVRAs no external torque is applied, angular momentum will be conserved.
Therefore, we have:
L’ = L”

⇒m3V5R=75mv”R⇒v”=3v7

Question 86:

A solid sphere rolling on a rough horizontal surface with a linear speed

νcollides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in the backward direction.

Answer:

Consider two cases:

(a) Just after the collision with the wall, the sphere rebounds with velocity

νtowards left but it continues to rotate in the clockwise direction (as shown in figure)

Angular momentum of the sphere about centre,

L=mvR-Iω  =mvR-25mR2×νR  =35mvR(b) When pure rolling starts:
Let the velocity be

ν’ and the corresponding angular velocity be

v’R.

L’=mv’R+Iω’  =mv’R+25mR2×v’R  =75mv’RAngular momentum is constant.
Therefore, we have:

L=L’⇒v’=3v7So, the sphere will move with velocity

3v7.

HC Verma Solutions for Class 11 Physics – Part 1

HC Verma Solutions for Class 12 Physics – Part 2

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