26 + Interview Questions in ROTATIONAL MECHANICS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	A wheel of radius 20 cm is pushed to move it on a rough horizontal surface. It is found to move through a distance of 60 cm on they 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 Explanation: Perimeter of the wheel = 2π*0.20 =0.4π =1.26 m But it moves only 0.60 m. So the wheel slips on the road. At the point of contact with the road, the part of the wheel is sliding backward. Hence the force of friction acts forward i.e. along the velocity of the wheel.

1.

A wheel of radius 20 cm is pushed to move it on a rough horizontal surface. It is found to move through a distance of 60 cm on they 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

Explanation:

Perimeter of the wheel = 2π*0.20 =0.4π =1.26 m

But it moves only 0.60 m.

So the wheel slips on the road. At the point of contact with the road, the part of the wheel is sliding backward. Hence the force of friction acts forward i.e. along the velocity of the wheel.

Discussion

2.	A body is in pure rotation. The linear speed v of a particle, the distance r of the particle from the axis and the angular velocity ω of the body are related asω = v/r. thus(a) ω ∝1/r(b) ω ∝ r(c) ω = 0(d) ω is independent of r.
Answer» (d) ω is independent of r. Explanation: In fact, v ∝ r and ω is the constant of proportionality. Thus v = ω.r and since ω is a constant, so it is independent of r.

2.

A body is in pure rotation. The linear speed v of a particle, the distance r of the particle from the axis and the angular velocity ω of the body are related asω = v/r. thus(a) ω ∝1/r(b) ω ∝ r(c) ω = 0(d) ω is independent of r.

Answer»

(d) ω is independent of r.

Explanation:

In fact, v ∝ r and ω is the constant of proportionality. Thus v = ω.r and since ω is a constant, so it is independent of r.

Discussion

3.	A sphere is rolled on a rough horizontal surface. It 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 Explanation: When a sphere is pushed on a rough horizontal surface to roll, it is given some linear speed and it ties to slip on the surface which is opposed by the force of friction so it tries to decrease the linear velocity and due to its direction it tries to increase the angular velocity. Hence option (a) and (b) and not the options (c) and (d).

3.

A sphere is rolled on a rough horizontal surface. It 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

Explanation:

When a sphere is pushed on a rough horizontal surface to roll, it is given some linear speed and it ties to slip on the surface which is opposed by the force of friction so it tries to decrease the linear velocity and due to its direction it tries to increase the angular velocity. Hence option (a) and (b) and not the options (c) and (d).

Discussion

4.	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 a larger magnitude than the Explanation: When accelerating the torque on the rear wheel tries the part in contact with the road to slide backward so the friction on it is forward. Option (a). The body of the car tries to push the front wheel in the forward direction, hence the force of friction is in the backward direction. Option (b). The driving rear wheel has to produce a force in the forward direction (through the torque) to push the whole mass of the car while the torque on the front wheel has just to overcome the inertia of the wheel. Hence the friction on the rear wheel is more than the front wheel. Option (c). Net friction on the car is in forward direction because rear wheel friction is larger and in forward direction. Hence option (d) is not correct.

4.

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 a larger magnitude than the

Explanation:

When accelerating the torque on the rear wheel tries the part in contact with the road to slide backward so the friction on it is forward. Option (a).

The body of the car tries to push the front wheel in the forward direction, hence the force of friction is in the backward direction. Option (b).

The driving rear wheel has to produce a force in the forward direction (through the torque) to push the whole mass of the car while the torque on the front wheel has just to overcome the inertia of the wheel. Hence the friction on the rear wheel is more than the front wheel. Option (c). Net friction on the car is in forward direction because rear wheel friction is larger and in forward direction. Hence option (d) is not correct.

Discussion

5.	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.

Discussion

6.	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.

6.

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.

Discussion

7.	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 moment of inertia I increases.

7.

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 moment of inertia I increases.

Discussion

8.	A body rotating at 20 rad/s is acted upon by a constant torque providing it a deceleration of 2 rad/s 2 . At what time will the body have kinetic energy same as the initial value if the torque continues to act ?
Answer» Initial angular velocity = 20 rad/s Therefore α = 2 rad/s² t₁ = ω/α ₁ = 20/2 = 10 sec Therefore 10 sec it will come to rest. Since the same torque continues to act on the body it will produce same angular acceleration and since the initial kinetic energy = the kinetic energy at an instant. So initial angular velocity = angular velocity at that instant Therefore time require to come to that angular velocity, t₂ = ω₂/α ₂ = 20/2 = 10 sec therefore time required = t₁ + t₂ = 20 sec.

Discussion

9.	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» A disc of radius = 10 cm = 0.1 m Angular velocity = 20 rad/s Linear velocity on the rim = ωr = 20 × 0.1 = 2 m/s Linear velocity at the middle of radius = ωr/2 = 20 × (0.1)/2 = 1 m/s.

Discussion

10.	A sphere can roll on a surface inclined at an angle 0 if the friction coefficient is more than 2/7 g tanθ. Suppose the friction coefficient is 1/7 g 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 Explanation: Since there is some friction which will produce a torque on the sphere and there is no othe torque to balance it. Hnce the sphere will not stay at rest. Option (a) is not correct. Due to the torque produced by the friction it will have some rotational motion also. Hence option (b) is not correct. From the question, the friction coefficient is =(1/7)g.sinθ which is lessthan (2/7)g.sinθ {required for rolling only} so it will translate and rotate about the center. Option (c) is correct. As the sphere go down the inclined plane its angular velocity will get increasing. So the angular momentum of the sphere about its center will continue increasing. Option (d) is not correct.

10.

A sphere can roll on a surface inclined at an angle 0 if the friction coefficient is more than 2/7 g tanθ. Suppose the friction coefficient is 1/7 g 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

Explanation:

Since there is some friction which will produce a torque on the sphere and there is no othe torque to balance it. Hnce the sphere will not stay at rest. Option (a) is not correct. Due to the torque produced by the friction it will have some rotational motion also. Hence option (b) is not correct.

From the question, the friction coefficient is =(1/7)g.sinθ which is lessthan (2/7)g.sinθ {required for rolling only} so it will translate and rotate about the center. Option (c) is correct.

As the sphere go down the inclined plane its angular velocity will get increasing. So the angular momentum of the sphere about its center will continue increasing. Option (d) is not correct.

Discussion

11.	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» The Block is moving the rim of the pulley The pulley is moving at a ω = 10 rad/s Therefore the radius of the pulley = 20 cm Therefore linear velocity on the rim = tangential velocity = rω = 20 × 20 = 200 cm/s = 2 m/s.

Discussion

12.	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%. Explanation: With the increase of petrol input the angular velocity of the wheel increases but in the absence of friction on the road, no external force on the scooter is applied. Hence the linear velocity of the scooter is not increased.

12.

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%.

Explanation:

With the increase of petrol input the angular velocity of the wheel increases but in the absence of friction on the road, no external force on the scooter is applied. Hence the linear velocity of the scooter is not increased.

Discussion

13.	Find the angular velocity of a body rotating with an acceleration of 2 rev/s 2 as it completes the 5th revolution after the start.
Answer» θ = 5 rev, α = 2 rev/s², ω₀ = 0 ; ω = ? ω²= (2 α θ) ω = √(2 x 2x 5) = 2 √5 rev/s. or θ = 10π rad, α= 4π rad/s², ω₀ = 0, ω = ? ω = √2αθ = 2 × 4π × 10π = 4π 5 rad/s = 2 √5 rev/s.

Discussion

14.	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 0.20kg-m2 (b) is 0.40kg-m2 (c) is 0.20/2kg-m2 (d) cannot be deduced with this information.
Answer» (d) cannot be deduced with this information. Explanation: To know the moment of inertia we need to know the mass and shape of the body. None is available in the problem, hence the option (d).

14.

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 0.20kg-m2 (b) is 0.40kg-m2 (c) is 0.20/2kg-m2 (d) cannot be deduced with this information.

Answer»

(d) cannot be deduced with this information.

Explanation:

To know the moment of inertia we need to know the mass and shape of the body. None is available in the problem, hence the option (d).

Discussion

15.	A cubical block of mass M and edge a slides down a rough inclined plane of inclination 0 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) 1/2Mga sinθ.
Answer» (d) 1/2Mga sinθ.

Discussion

16.	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 Explanation: During the above rotation water will move away from the axis of rotation. And the moment of inertia is directly proportional to the square of the distance of the mass from the axis of rotation. Hence the option (a).

16.

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

Explanation:

During the above rotation water will move away from the axis of rotation. And the moment of inertia is directly proportional to the square of the distance of the mass from the axis of rotation. Hence the option (a).

Discussion

17.	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» Masses of 1 gm, 2 gm ……100 gm are kept at the marks 1 cm, 2 cm, ……1000 cm on he x axis respectively. A perpendicular axis is passed at the 50th particle. Therefore on the L.H.S. side of the axis there will be 49 particles and on the R.H.S. side there are 50 particles. Consider the two particles at the position 49 cm and 51 cm. Moment inertial due to these two particle will be = 49 × 1² + 51 + 1² = 100 gm-cm² Similarly if we consider 48th and 52nd term we will get 100 ×2²gm-cm² Therefore we will get 49 such set and one lone particle at 100 cm. Therefore total moment of inertia = 100 {1² + 2² + 3² + … + 49²} + 100(50)². = 100 × (50 × 51 × 101)/6 = 4292500 gm-cm² = 0.429 kg-m² = 0.43 kg-m².

Discussion

18.	Equal torques act on tlie discs 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 smother point on the rim of B are vA and vB respectively. We have (a) vA>vB (b) vA = vB (c) vA < vB (d) the relation depends on the actual magnitude of the torques.
Answer» The correct answer is (a) v_A > v_B Explanation: Let T be the torque and A and A' be the angular accelerations. A = T/IA and A' = T/IB. Clearly A >A'. In fact, the ratio of the moment of Inertias is 64. So at a later instant, the angular velocity of the first disc will be much greater than the second. Hence option (a).

18.

Equal torques act on tlie discs 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 smother point on the rim of B are vA and vB respectively. We have (a) vA>vB (b) vA = vB (c) vA < vB (d) the relation depends on the actual magnitude of the torques.

Answer»

The correct answer is

(a) v_A > v_B

Explanation:

Let T be the torque and A and A' be the angular accelerations. A = T/IA and A' = T/IB. Clearly A >A'. In fact, the ratio of the moment of Inertias is 64. So at a later instant, the angular velocity of the first disc will be much greater than the second. Hence option (a).

Discussion

19.	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 1/4. The relation between the moments of inertia IA and IB is (a) lA > IB (b) IA = IB (c)IA<IB (d) depends on the actual values of t and r.
Answer» (c) I_A <I_B Explanation: Let the density of iron plate be ρ. Mass of first disc m = πr²tρ M.I. = I_A = ½mr² == ½πr²tρr² = ½πr4tρ Mass of second disc = M = π(4r)²(t/4)ρ = 4πr²tρ M.I. = I_B = ½M(4r)² = ½4πr²tρ16r² = 32πr4tρ Clearly, I_A < I_B.

19.

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 1/4. The relation between the moments of inertia IA and IB is (a) lA > IB (b) IA = IB (c)IA<IB (d) depends on the actual values of t and r.

Answer»

(c) I_A <I_B

Explanation:

Let the density of iron plate be ρ.

Mass of first disc m = πr²tρ

M.I. = I_A = ½mr² == ½πr²tρr² = ½πr4tρ

Mass of second disc = M = π(4r)²(t/4)ρ = 4πr²tρ

M.I. = I_B = ½M(4r)² = ½4πr²tρ16r² = 32πr4tρ

Clearly, I_A < I_B.

Discussion

20.	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) I_A <I_B Explanation: Let the density of iron plate be ρ. Mass of first disc m = πr²tρ M.I. = I_A = ½mr² == ½πr²tρr² = ½πr4tρ Mass of second disc = M = π(4r)²(t/4)ρ = 4πr²tρ M.I. = I_B = ½M(4r)² = ½4πr²tρ16r² = 32πr4tρ Clearly, I_A < I_B.

20.

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) I_A <I_B

Explanation:

Let the density of iron plate be ρ.

Mass of first disc m = πr²tρ

M.I. = I_A = ½mr² == ½πr²tρr² = ½πr4tρ

Mass of second disc = M = π(4r)²(t/4)ρ = 4πr²tρ

M.I. = I_B = ½M(4r)² = ½4πr²tρ16r² = 32πr4tρ

Clearly, I_A < I_B.

Discussion

21.	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 el The force exerted by the clamp on the rod has a horizontal component(a) m w2l (b) zero(c) mg(d)1/2 mw2l
Answer» (d)1/2 mw²l. Explanation: Since the rod rotates on the horizontal surface, the horizontal component of the force applied by the clamp is the centripetal force = mω²r = mω²(l/2) = ½mω²l.

21.

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 el The force exerted by the clamp on the rod has a horizontal component(a) m w2l (b) zero(c) mg(d)1/2 mw2l

Answer»

(d)1/2 mw²l.

Explanation:

Since the rod rotates on the horizontal surface, the horizontal component of the force applied by the clamp is the centripetal force = mω²r = mω²(l/2) = ½mω²l.

Discussion

22.	A uniform rod is kept vertically on a horizontal smooth surface at a point 0. If it is rotated slightly and released, it falls down on the horizontal surface. The lower end will remain(a) at 0(b) at a distance less than l/2 from 0(c) at a distance l/2 from 0(d) at a distance larger than l/2 from 0.
Answer» (c) at a distance l/2 from 0 Explanation: Since the rod is uniform its Center of mass (CoM) will be in the middle i.e. at l/2 distance from the ends. When the rod is vertical the CoM will bel/2 distance vertically above the point O. Since there is no force in the horizontal direction the CoM will be on the point O after it falls down. Hence the lower end will remain at a distance l/2 from O.

22.

A uniform rod is kept vertically on a horizontal smooth surface at a point 0. If it is rotated slightly and released, it falls down on the horizontal surface. The lower end will remain(a) at 0(b) at a distance less than l/2 from 0(c) at a distance l/2 from 0(d) at a distance larger than l/2 from 0.

Answer»

(c) at a distance l/2 from 0

Explanation:

Since the rod is uniform its Center of mass (CoM) will be in the middle i.e. at l/2 distance from the ends. When the rod is vertical the CoM will bel/2 distance vertically above the point O. Since there is no force in the horizontal direction the CoM will be on the point O after it falls down. Hence the lower end will remain at a distance l/2 from O.

Discussion

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 Explanation: Since there is no friction on the incline, the only force on each of these objects along the incline is their weight component mg.sinθ. So each of them have same initial velocity=0 and same acceleration=mg.sinθ/m = g.sinθ So the time taken by each of them t is given by S=0t+½g.sinθt² t²=2S/g.sinθ t=√{2S/g.sinθ} where S is the length of the incline.

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

Explanation:

Since there is no friction on the incline, the only force on each of these objects along the incline is their weight component mg.sinθ.

So each of them have same initial velocity=0 and same acceleration=mg.sinθ/m = g.sinθ

So the time taken by each of them t is given by

S=0*t+½g.sinθ*t²

t²=2S/g.sinθ

t=√{2S/g.sinθ}

where S is the length of the incline.

Discussion

24.	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.
Answer» According to the question Mg – T₁= Ma …(1) (T₂ – T₁)R = Ia/R  (T₂ – T₁) = Ia/R² …(2) (T₂ – T₃)R = Ia/R² …(3) T₃ – mg = ma …(4) By adding equation (2) and (3) we will get, (T₁ – T₃) = 2 Ia/R² …(5) By adding equation (1) and (4) we will get – mg + Mg + (T₃ – T₁) = Ma + ma …(6) Substituting the value for T₃ – T₁ we will get Mg – mg = Ma + ma + 2Ia/R² a = (M -m)G /(M +m)+2l/R²)

24.

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.

Answer»

According to the question
Mg – T₁= Ma …(1)
(T₂ – T₁)R = Ia/R  (T₂ – T₁) = Ia/R² …(2)
(T₂ – T₃)R = Ia/R² …(3)
T₃ – mg = ma …(4)
By adding equation (2) and (3) we will get,
(T₁ – T₃) = 2 Ia/R² …(5)
By adding equation (1) and (4) we will get

– mg + Mg + (T₃ – T₁) = Ma + ma …(6)
Substituting the value for T₃ – T₁ we will get
Mg – mg = Ma + ma + 2Ia/R²
a = (M -m)G /(M +m)+2l/R²)

Discussion

25.	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 Explanation: Since the body is rotating uniformly, a particle not on the axis moves in a uniform circular motion. This circle will be horizontal with its center at the intersecting point with the axis. So it has a radial acceleration and hence a radial resultant force. Thus this force will be horizontal and intersecting the axis.

25.

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

Explanation:

Since the body is rotating uniformly, a particle not on the axis moves in a uniform circular motion. This circle will be horizontal with its center at the intersecting point with the axis. So it has a radial acceleration and hence a radial resultant force. Thus this force will be horizontal and intersecting the axis.

Discussion

26.	A thin circular ring of mass M and radius r is rotating about its axis with an angular speed co. Two particles having mass m each are now attached at diametrically opposite points. The angular speed ω of the ring will become(a) ωm/m+m(b)ωm/m+2m(c) ω m-2m)/m+2m(d) ω(m+2m)/m.
Answer» The correct answer is (b) Explanation: Moment of inertia of the ring I = Mr² Angular Momentum = I⍵ When the masses are attached, the moment of Inertia I'= Mr²+2mr² =(M+2m)r² Let the new angular speed be ⍵'. So the angular momentum =I'⍵'. Since the angular momentum is conserved. I'⍵'=I⍵ →⍵' = I⍵/I' =⍵Mr²/(M+2m)r² =⍵M/(M+2m)

26.

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed co. Two particles having mass m each are now attached at diametrically opposite points. The angular speed ω of the ring will become(a) ωm/m+m(b)ωm/m+2m(c) ω m-2m)/m+2m(d) ω(m+2m)/m.

Answer»

The correct answer is (b)

Explanation:

Moment of inertia of the ring I = Mr²

Angular Momentum = I⍵

When the masses are attached, the moment of Inertia I'= Mr²+2mr²

=(M+2m)r²

Let the new angular speed be ⍵'. So the angular momentum =I'⍵'.

Since the angular momentum is conserved. I'⍵'=I⍵

→⍵' = I⍵/I' =⍵Mr²/(M+2m)r² =⍵M/(M+2m)

Discussion

Explore topic-wise InterviewSolutions in .

A body is in pure rotation. The linear speed v of a particle, the distance r of the particle from the axis and the angular velocity ω of the body are related asω = v/r. thus(a) ω ∝1/r(b) ω ∝ r(c) ω = 0(d) ω is independent of r.

A sphere is rolled on a rough horizontal surface. It 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.

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

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?

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

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

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.

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 ?

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

A cubical block of mass M and edge a slides down a rough inclined plane of inclination 0 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) 1/2Mga sinθ.

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.

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 1/4. The relation between the moments of inertia IA and IB is (a) lA &gt; IB (b) IA = IB (c)IA&lt;IB (d) depends on the actual values of t and r.

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 &gt;IB(b) IA =IB(c) IA &lt;IB(d) depends on the actual values of t and r.

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 el The force exerted by the clamp on the rod has a horizontal component(a) m w2l (b) zero(c) mg(d)1/2 mw2l

A uniform rod is kept vertically on a horizontal smooth surface at a point 0. If it is rotated slightly and released, it falls down on the horizontal surface. The lower end will remain(a) at 0(b) at a distance less than l/2 from 0(c) at a distance l/2 from 0(d) at a distance larger than l/2 from 0.

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.

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.

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.

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed co. Two particles having mass m each are now attached at diametrically opposite points. The angular speed ω of the ring will become(a) ωm/m+m(b)ωm/m+2m(c) ω m-2m)/m+2m(d) ω(m+2m)/m.

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 1/4. The relation between the moments of inertia IA and IB is (a) lA > IB (b) IA = IB (c)IA<IB (d) depends on the actual values of t and r.

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.