
Q1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body? Ans: Geometric centre; No
The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For the given geometric shapes having a uniform mass density, the C.M. lies at their respective geometric centres.
The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc., lies outside the body.
Q3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system? Ans: No change
The child is running arbitrarily on a trolley moving with velocity v. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy's motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the boytrolley system, the boy's motion will produce no change in the velocity of the centre of mass of the trolley.
Q11 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time? Ans: Let m and r be the respective masses of the hollow cylinder and the solid sphere.
The moment of inertia of the hollow cylinder about its standard axis,I_{I} = mr^{2}
The moment of inertia of the solid sphere about an axis passing through its centre, I_{II} = 2/5 mr^{2}
We have the relation:
τ = I α
Where, α = Angular acceleration
τ = Torque
I = Moment of inertia
For the hollow cylinder, τ_{I} = I_{I} α_{I}
For the solid sphere, τ_{II} = I_{II} α_{II}
As an equal torque is applied to both the bodies, τ_{I} = τ_{2}
∴ α_{II }/ α_{I }= I_{I }/ I_{II = }mr^{2 }/ 2/5 mr^{2 }= 2/5
α_{II }> αI .... (i)
Now, using the relation:
ω = ω_{0} + αt
Where, ω_{0} = Initial angular velocity
t = Time of rotation
ω = Final angular velocity
For equal ω_{0} and t, we have:
ω ∝ α … (ii)
From equations (i) and (ii), we can write:
ω_{II} > ω_{I}
Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.
Q12 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis? Ans: Mass of the cylinder, m= 20 kg
Angular speed, ω = 100 rad s^{1}
Radius of the cylinder, r= 0.25 m
The moment of inertia of the solid cylinder:
I = mr^{2} / 2
= 1/2 x 20 x (0.25)^{2}
= 0.625 kgm^{2}
∴Kinetic energy = 1/2 I ω^{2}
= 1/2 x 6.25 x 100^{2} = 3125 J
∴Angular momentum, L= Iω = 6.25 × 100 = 62.5 Js
Q13 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy? Ans: (a) 100 rev/min
Initial angular velocity, ω_{1}= 40 rev/min
Final angular velocity = ω_{2}
The moment of inertia of the boy with stretched hands = I_{1}
The moment of inertia of the boy with folded hands = I_{2}
The two moments of inertia are related as:
I_{2 }= 2/5 I_{1}
Since no external force acts on the boy, the angular momentum L is a constant.
Hence, for the two situations, we can write:
I_{2} ω_{2 = }I_{1} ω_{1}
ω_{2} = I_{1} ω_{1 }/ I_{2}
= I_{1} / 2/5 I_{2 } x 40 = 5/2 x 40_{ }
= 100 rev/min
(b) Final K.E. = 2.5 Initial K.E.
Final kinetic rotation,E_{F} = 1/2 I_{2} ω_{2}^{2}
Initial kinetic rotation, E_{I }= 1/2 I_{1} ω_{1}^{2}
E_{F }/ E_{I} = 1/2 I_{2} ω_{2}^{2 }/ 1/2 I_{1} ω_{1}^{2}
= 2/5 I_{1 }/ I_{1 }x 100^{2} / 40^{2}
= 5/2 = 2.5
∴ E_{F }= 2.5 E_{I}
The increase in the rotational kinetic energy is attributed to the internal energy of the boy.
Q14 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping. Ans: Mass of the hollow cylinder, m = 3 kg
Radius of the hollow cylinder, r = 40 cm = 0.4 m
Applied force, F = 30 N
The moment of inertia of the hollow cylinder about its geometric axis:
I = mr^{2}
= 3 × (0.4)^{2} = 0.48 kg m^{2}
Torque, τ = F x r
= 30 × 0.4 = 12 Nm
For angular acceleration α, torque is also given by the relation:
τ = Iα
α = τ / I = 12 / 0.48
= 25 rad s^{2}
Linear acceleration = rα = 0.4 × 25 = 10 m s^{–2}
Q15 To maintain a rotor at a uniform angular speed of 200 rad s–1, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100 % efficient. Ans: Angular speed of the rotor, ω = 200 rad/s
Torque required, τ = 180 Nm
The power of the rotor (P) is related to torque and angular speed by the relation:
P = τω
= 180 × 200 = 36 × 10^{3} = 36 kW
Hence, the power required by the engine is 36 kW.
Q19 A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it? Ans: Radius of the hoop, r = 2 m
Mass of the hoop, m = 100 kg
Velocity of the hoop, v = 20 cm/s = 0.2 m/s
Total energy of the hoop = Translational KE + Rotational KE
E_{r} = 1/2 mv^{2} + 1/2 I ω^{2}
Moment of inertia of the hoop about its centre, I = mr^{2}
E_{r} = 1/2 mv^{2} + 1/2 (mr^{2}) ω^{2}
But we have the relation, v = rω
∴ Er = 1/2 mv^{2} + 1/2 mr^{2} ω^{2}
∴ Er = 1/2 mv^{2} + 1/2 mv^{2}
∴ Er = mv^{2}
The work required to be done for stopping the hoop is equal to the total energy of the hoop.
∴Required work to be done, W = mv^{2} = 100 × (0.2)^{2} = 4 J
Q29 Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated. (a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins. (b) What is the force of friction after perfect rolling begins? Ans: A torque is required to roll the given disc. As per the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.
(a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.
(b) Since frictional force acts opposite to the direction of velocity at point B, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.
Q32 Read each statement below carefully, and state, with reasons, if it is true or false; (a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body. (b) The instantaneous speed of the point of contact during rolling is zero. (c) The instantaneous acceleration of the point of contact during rolling is zero. (d) For perfect rolling motion, work done against friction is zero. (e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion. Ans: (a) False Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.
(b) True Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero.
(c) False When a body is rolling, its instantaneous acceleration is not equal to zero. It has some value.
(d) True When perfect rolling begins, the frictional force acting at the lowermost point becomes zero. Hence, the work done against friction is also zero.
(e) True The rolling of a body occurs when a frictional force acts between the body and the surface. This frictional force provides the torque necessary for rolling. In the absence of a frictional force, the body slips from the inclined plane under the effect of its own weight.