HC Verma Questions and Solutions: Chapter 10: Rotational Mechanics- 1 | HC Verma Solutions - JEE PDF Download

Q.1. Can an object be in pure translation as well as in pure rotation?

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

Q.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?

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.
 

Q.3. In a rotating body,  Thus  Can you use the theorems of ration and proportion studied in algebra so as to write

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 + ω and/or α + a are not possible.

Q.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?

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 = 2π/T
After 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 = 2π
Time period = T
So, angular velocity is again 2π/T.
Thus, in both the cases, angular velocities are the same.

Q.5. The moon rotates about the earth in such a way that only one hemisphere of the moon faces the earth (see the following figure). 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?

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. 

Q.6. The torque of the weight of any body about any vertical axis is zero. If it always correct?

No, its not always correct.

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

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

Q.7. The torque of a force  about a point is defined as  Suppose  are all nonzero. Is  always true? Is it ever true?

No,  is not true.
In fact, it is never true.This is because:

Applying vector triple product, we get:

We know that r² is never negative and
This implies that both vectors may be antiparallel to each other but not parallel.

Q.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?

We know that

Given:-

The torque becomes

No, there will be no angular acceleration on the particle due to the torque.
Angular acceleration is given by  As the particle here moves in a straight line, the centre of rotation lies at a distance infinity  so, moment of inertia  of the particle is infinity.

Q.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?

Let  be the forces acting on a point P.
Let O be the point along which torques \(moments) will be taken.
Let:-

Moments of force (torque)  about O will be

The sum of the torques about O will be

Thus, we see that the torque of the resultant force  of the forces  gives the sum of the  moments of the torques.

Q.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?

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 figure). In the above case, although the sum of the forces acting on the body is zero  Still, the body will rotate along  So, it won't remain in equilibrium.
 

Q.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?

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.

Q.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?

No, it is not necessary that the torque about any other point be zero if it is zero about one point.
Let  be the resultant force due to all the forces acting on the plane of the body. Therefore, torque due to force \vec{F} at any point will be the resultant torque . Now, we see that the torque due to  at point Q will be zero because Q lies on the line of support of the force F but the torque due to force  will not be zero along the point P.

Q.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?

Yes , if the torque due to forces in translation equillibrium is 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 by forces  etc.
Let a force  act on a i^th particle and torque due to  be zero at a point Q.
Since the body is in translation equillibrium, we have

Again, torque about P is zero . Therefore, we have

Now, torque about point Q will be

Thus,  is zero about any other point Q.

Q.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.

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.

Q.15. When a fat person tries to touch his toes, keeping the legs straight, he generally falls. Explain with reference to the following figure.

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

Q.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?

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 + M)g 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.

Q.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?

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.

Q.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?

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.
F_A = Force required to rotated the rod clamped at A
R_A= Distance of CM from pivot A
M = Mass of the rod
F_B = Force required to rotate the rod clamped at B
R_B= Distance of CM from pivot B
We have R_A>R_B.
We have to find the torque required to rotate rod clamped at A to produce angular acceleration a.
T_A = MR_A²a = R_AF_A
⇒ F_A = MR_Aa
We have to find torque required to rotate rod clamped at B to produce angular acceleration a.
T_B = MR_B²a = R_BF_B
⇒ F_B = MR_Ba
On comparing, since R_A>R_B, we get
F_A>F_B

Q.19. When tall building are constructed on earth, the duration of day-night slightly increases. Is it true?

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.

Q.20. If the ice at the poles melts and flows towards the equator, how will it affect the duration of day-night?

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.

Q.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?

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.

Q.22. A sphere rolls on a horizontal surface. If there any point of the sphere which has a vertical velocity?

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

Question for HC Verma Questions and Solutions: Chapter 10: Rotational Mechanics- 1

Try yourself:Let  be a unit vector along the axis of rotation of a purely rotating body and  be a unit vector along the velocity of a particle P of the body away from the axis. The value of  is ____________ .

• A.

  1
• B.

  -1
• C.

  0
• D.

  None of these

Explanation

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

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Question for HC Verma Questions and Solutions: Chapter 10: Rotational Mechanics- 1

Try yourself:A body is uniformly rotating about an axis fixed in an inertial frame of reference. Let  be a unit vector along the axis of rotation and  be the unit vector along the resultant force on a particle P of the body away from the axis. The value of  is _________.

• A.

  1
• B.

  -1
• C.

  0
• D.

  None of these

Explanation

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

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Question for HC Verma Questions and Solutions: Chapter 10: Rotational Mechanics- 1

Try yourself:The following figure 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 = 2y
• B.

  x = y
• C.

  y = 2x
• D.

  None of these

Explanation

It is given that angular velocity is same for both the wheels.
Therefore, we have
v_A = ωR
v_B = ω2R
x = v_At = ωRT ...(1)
y = v_Bt = ω(2R)t ...(2)
From equations (1) and (2), we get
y = 2x

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Question for HC Verma Questions and Solutions: Chapter 10: Rotational Mechanics- 1

Try yourself:Let  be a force acting on a particle having position vector  be the torque of this force about the origin, then __________ .

• A.

  
• B.

  
• C.

  
• D.

  

Explanation

We have

Thus,
 is perpendicular to 
Therefore, we have

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Question for HC Verma Questions and Solutions: Chapter 10: Rotational Mechanics- 1

Try yourself:Consider the following two equations

In 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

Explanation

In non-inertial frames,
Here,
 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.

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*Multiple options can be correct

Question for HC Verma Questions and Solutions: Chapter 10: Rotational Mechanics- 1

Try yourself:Consider a wheel of a bicycle rolling on a level road at a linear speed v₀ (see the following figure)

• 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.

Explanation

For pure rolling,

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

At point B,

At point C,

At point A,

At point O,
r = 0

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Question for HC Verma Questions and Solutions: Chapter 10: Rotational Mechanics- 1

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

Explanation

The given coefficient of friction   is less than the coefficient friction  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.

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Question for HC Verma Questions and Solutions: Chapter 10: Rotational Mechanics- 1

Try yourself:The following figure 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

Explanation

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

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