KKMishra's Tutorials: Solutions to Problems on "ROTATIONAL MECHANICS" - H C Verma's Concepts of Physics, Part-I, Chapter-10, Questions for Short Answer

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ROTATIONAL MECHANICS:--
QUESTIONS FOR SHORT ANSWER

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

ANSWER: No. In pure translation, each particle of the body moves in parallel lines while in a pure rotation the particles of a body move in concentric circles, Both cannot be true at the same time.

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: Since in a pure rotation all particles of a body move in concentric circles, hence this motion of a simple pendulum is a pure rotation, only the direction of motion reverses periodically. The axis of rotation is the line perpendicular to the plane of rotation and passing through the fixed point where the thread of the pendulum is tied.

Since the oscillations are small, for all practical purposes the movement of the ball can be assumed on a straight line and also the motion can be taken as periodic and Simple Harmonic motion can be assumed.

But these are the differences in theory and practice.

3. In a rotating body, a=ɑr, v=ɷr. Thus a/ɑ=v/ɷ. Can you use the theorems of ratio and proportion studied in algebra so as to write
(a+ɑ)/(a-ɑ) = (v+ɷ)/(v-ɷ)?

ANSWER: In the given relations there are five unknowns. When the first one is divided by the second, in fact, one unknown "r" is being eliminated. But if we use the theorems of ratio and proportion to write it as (a+ɑ)/(a-ɑ) = (v+ɷ)/(v-ɷ), we are not eliminating any unknown but complicating the relation a/ɑ=v/ɷ. Also, this equation is not dimensionally correct. So we can not write it like that.

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 center? If yes, is this angular velocity equal to the angular velocity of the ball about the fixed point?

ANSWER: Yes the ball has an angular rotation about its center. If we notice the position of the point on the ball where it is tied with the string, we can see that it is moving about its center as its position changes. This point again comes to the same position when the ball makes a complete revolution around the fixed point. Thus the ball rotates about its center through 360° in the same time in which the ball revolves through 360° around the fixed point. Hence the angular velocity of the ball about its center is equal to the angular velocity of the ball about the fixed point.

5. The moon rotates about the earth in such a way that only one hemisphere of the moon faces the earth (figure-10Q1). 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 for Question-5

ANSWER: Since only one hemisphere of the moon faces the earth, we can never see the "other" face of the moon from the earth.

But a person on the moon can see all the faces of the earth. The reason is that the time period and the direction of rotation of the moon is exactly the same as the revolution around the earth (about 28 days)-just like a ball being rotated in a circle around a fixed point. But the time period of the earth's rotation is not the same which is about 24 hours. So during one revolution of the moon, the earth rotates about 28 times. So a person on the facing side of the moon can see all faces of the earth.

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

ANSWER: A force can only produce a torque around an axis if it has a component in the plane perpendicular to the axis and this component does not pass through the point of intersection of this plane and the axis. In the given case, the force of weight is parallel to any given vertical axis, hence it can not have a component in a plane perpendicular to this axis (Horizontal Plane). So it can not produce a torque about any vertical axis i.e. torque is zero. It is always correct.

7. The torque of a force F about a point is defined as Γ=rxF. Suppose r, F and Γ are all nonzero. Is rxΓ||F always true? Is it ever true?
ANSWER: The direction of the vector Γ (torque) is perpendicular to the plane in which r and F lie. The direction of the vector rxΓ will be perpendicular to the plane in which r and Γ lie. So it is not always true that rxΓ||F. It can only be true if r丄F but their directions will be opposite.

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: The torque acting on the particle about 50 cm away from the line of motion = mg*(0.50) N-m. Where m = mass of the particle in kg, g=acceleration due to gravity in m/s² and r = 50 cm = 0.50 m.

This torque cannot produce any angular acceleration in the particle because there is no force in the opposite direction of the weight at 50 cm east of the line of motion. For a torque to be effective through a force, there must be equal and opposite non-linear force acting on the particle.

9. If several forces act on a particle, the torque on the particle may be obtained by first finding the resultant of the 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 a common point?

ANSWER: Let F_1, F_2, F₃ .... be the forces acting on the particle and position vector of the particle from the point O (About which torque is being calculated) is r. Then the torque on the particle is equal to,
Γ= rxF₁ +rxF₂ + rxF₃ ....
= r x (F₁ +F₂ + F₃ ....) {Because vector cross product is distributive over addition}
= rxF
So, The Torque of the resultant force is equal to the sum of torques of individual forces on a particle.
Hence Proved.

Diagram for Q - 9

Second Part

If the forces act on different particles of a body such that their lines of actions intersect a common point then also this result is valid. Because we can shift the forces along their lines of action till their ends come to the point of intersection (Equivalent forces). This will not change the torque effect of a force (Since the magnitude of the torque of a force = Magnitude of the force x Shortest distance of the line of action of the force from the point about which torque is being calculated). Having done this we have a condition like first part where all the forces act at a point. So again
Γ= rxF₁ +rxF₂ + rxF₃ ....
= r x (F₁ +F₂ + F₃ ....) {Because vector cross product is distributive over addition}
= rxF

So, The Torque of the resultant force is equal to the sum of torques of individual forces on a body.

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: If the sum of all the forces acting on a body is zero then it is not necessarily in equilibrium. In fact, it will be in translatory equilibrium but may not be in rotational equilibrium. Suppose two equal and opposite forces having their line of actions at a certain distance apart act on a body, then the body will remain stationary at the point but will have rotation due to the torque produced.

If the sum of all the forces acting on a particle is zero then it is necessarily in equilibrium because there will not be a distance between antiparallel forces.

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.

We can express the angular momentum of a body about a point as mvr. Since it depends on 'r', it will be different about different points.

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: Let us understand it with a simple example as in the figure below:-

Diagram for Q-12

There are two forces having each a magnitude F, one horizontal and other vertical. Point O is equidistant from each force = d. The torque of each force is Fd about O but equal and opposite. So total torque is zero about O.

Now consider a point P which is in the line of the horizontal force. The total torque of the forces about the point P =Fr (anticlockwise). So it is not zero.

So if the resultant torque of all the forces acting on a body is zero about a point, it is not necessary that it will be zero about any other point.

13. A body is in transitional equilibrium under the actions 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. In an inertial or equivalent frame of reference, It will be zero about any other point.

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 slightly half the total length but it falls down if the length projected is slightly more than half the total length. Give reason.

ANSWER: The weight of the projected part produces a torque that will tend to overturn the brick about the edge of the table while the weight of the part on the table produces a torque that will tend to restore the brick from overturning. Assuming a uniform density of the brick, in the first case- the weight of projected part will be less than the weight of the part on the table. Also, the distance of the line of action of the weight of the projected part from the edge of the table is less in comparison to the other part. Hence the overturning torque, in this case, will be less than the restoring torque. So the brick remains at rest.

The situation is just reverse in the second case where the overturning torque is greater than the restoring torque, so the brick overturns and falls.

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

Figure for Q-15

ANSWER: We can understand the reason considering the body of the fat person in two parts - Legs up to waist as one part and rest of the body as the second part. (See diagram below).

Diagram for Q - 15

Center of the mass of the first part is at A (Somewhere near the knees) and the center of the mass of the second part is at B (Somewhere near the chest). So the center of mass of the whole body is at C on the line AB. The total weight of the fat person acts at C and the Normal force by the ground on his body (Which is also equal to W) acts opposite to the weight at his toes T. Though both weight and the Normal force are equal and opposite, their lines of action are not the same. Their lines of action are parallel having some distance d between them. So these antiparallel forces produce a torque equal to W*d on the fat person (Which is clockwise in the figure), and he generally falls.

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

ANSWER: The ladder tends to slip more when the man stands near the wall.

Let us first draw a diagram to understand the problem. AB is the ladder with end A resting on the ground and end B against the wall. OA = d, OB = h, and W is the weight of the man on the ladder at distance x from end A horizontally. R and R' are normal forces at the ends A and B also F and F' are friction forces at A and B.

Diagram for Q - 16

Assuming µ same at both contact surfaces,
F =µR, F' = µR'
Equating forces vertically,
R+F' =W
R+µR' =W
Equating forces horizontally
F = R'
Taking Moments of all forces about B, we have
W(d-x)+Fh-Rd = 0
→W(d-x)+µRh-Rd = 0
→W(d-x) = R(d-µh)
→R=W(d-x)/(d-µh)
Overturning Torque (anticlockwise) about A
= Wx
Restoring torque (Clockwise)
= R'h+F'd
So net overturning torque T
=Wx-(R'h+F'd)
=Wx-(Fh+µR'd)
=Wx-(Fh+µFd)
=Wx-F(h+µd)
=Wx-µR(h+µd)
=Wx-µW(d-x)(h+µd)/(d-µh)
={Wxd-Wxµh-µW(dh+µd²-xh-µxd)}/(d-µh)
={Wxd-~~Wxµh~~-µWdh+µ²d²W+~~µxhW~~+µ²Wxd)}/(d-µh)
={Wxd(1+µ²)+Wµd(µd-h)}/(d-µh)
In this expression all except x are constant and we can see that as x increases T also increases.
So, the ladder tends to slip more when the man stands near the wall.

17. When a body is weighed on an ordinary balance we demand that the arm 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: In fact, middle point (Point of support) is not in a straight line which joins the hanging point of the panes. This point of support (Middle point, pivot) is slightly above the center of the arms. Thus these three points make a triangle with a wide base. See diagram below,

Diagram for Q- 17

When the arms are horizontal, the resultant weight of the pans 2W (Downard) acts in the middle and the balancing normal force 2W also acts upwards a bit above but in the same line. So net torque is zero.
When the arm is kept at an angle with the horizontal, the lines of actions of resultant weight and the normal force are not the same but are at some distance d. (See the second diagram). These two equal and opposite forces each equal to 2W but at distance d produce a net restoring torque =2Wd which acts till the arms are horizontal because only then d=0 i.e. the torque is zero. Now it is in stable equilibrium.

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

ANSWER: By clamping at B and applying the force at A.

It is due to the fact that in this position the center of mass is comparatively closer to the clamp. Hence moment of inertia I of the rod about the clamp will be less. So for the same angular acceleration, the torque needed (=Iα) will be less.

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

ANSWER: Yes, True.

When tall buildings are constructed, the effective radius of the earth slightly increases. Though the mass remains constant, the moment of inertia of the earth I slightly increase. To conserve the angular momentum L of the earth the angular velocity ω (=L/I) slightly decrease. This means that the earth rotates slower now and takes more time to complete one rotation. So the duration of day-night slightly increases.

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

ANSWER: The duration of day-night slightly increases.

The reason is same as above problem. The moment of inertia I increases.

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: Since each object has the same radius, the total torque on each of them will be same. So angular acceleration α =Γ/I. Thus the object with minimum I will have maximum angular acceleration. Let us see the value of I for given objects:-

Hollow sphere = (2/3)MR²
Solid Sphere = (2/5)MR²
Disc = ½MR²
Ring = MR²
Obviously, minimum I of them is for the solid sphere. So the solid sphere will take the smallest time to cover a given length.

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

ANSWER: Theoretically, no point on the sphere has velocity vector pointing in the vertical direction.
Practically, at the points very near to the point which is in contact at the instant has a nearly vertical velocity with a very small magnitude. Just at the contact point, when the velocity becomes true vertical, its magnitude becomes zero. So we can not say that a vertical velocity.

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