AP Physics B & C - Multiple Choice Practice Questions on Circular Motion and Rotation

"I know not with what weapons World War III will be fought, but World War IV will be fought with sticks and stones."

– Albert Einstein

Questions on circular motion and rotation were discussed on many occasions on this site. You can access all those questions by clicking on the label ‘circular motion and rotation’ below this post. Or, you may try a search for circular motion and rotation using the search box provided on this page.

Today we will discuss a few more multiple choice practice questions in this section.

The following questions will be useful for AP Physics B as well as AP Physics C aspirants:

(1) A 7-metre long uniform iron girder of mass 300 kg is placed on the horizontal top of a building with 3 m extending over the edge of the building. A fireman weighing 75 kg starts walking towards the protruding end of the girder (Fig.). What is the maximum distance (from the edge of the building) upto which he can walk safely?

(a) 2 m

(b) 2.2 m

(c) 2.4 m

(d) 2.5 m

(e) 2.6 m

When the fireman moves just beyond the maximum safe distance, the girder starts rotating about the ege of the building. If the maximum safe istance is d, we have

75×d = 300×0.5

[We have equated the magnitude of the moment of the weight of the fireman (which is the deflecting torque due to the fireman’s weight) to the magnitude of the moment of the weight of the girder (which is the restoring torque due to the weight of the girder): 75g×d = 300g×0.5]

This gives d = 2 m.

(2) A uniform metre stick of mass 0.2 kg can be balanced horizontally on a knife edge at the 40 cm mark, when a stone is suspended (using a string of negligible mass) from the 20 cm mark. The mass of the stone is

(a) 0.5 kg

(b) 0.3 kg

(c) 0.2 kg

(d) 0.1 kg

(e) 0.05 kg

The weight of the metre stick acts through its centre of gravity located at its middle (the 50 cm mark). Since metre stick is in equilibrium, the moment of the weight of the metre stick about the knife edge is equal and opposite to the moment of the weight of the stone about the knife edge. Therefore we have (on equating the magnitudes of the moments)

0.2g×0.1 = Mg×0.2 where M is the mass of the stone.

[Note that the distance of the knife edge from the centre of gravity of the metre stick is 0.1 m where as the distance from the line of action of the weight of the stone is 0.2 m].

This gives M = 0.1 kg.

(3) A square plate ABCD of side ‘a’ and mass ‘m’ is suspended so that it can rotate about a horizontal axis passing through A and perpendicular to its plane. To rotate the plate, a horizontal force of magnitude F is applied to the plate at the corner C as shown in the figure. What is the magnitude of the torque produced by this force?

(a) aF

(b) maF

(c) (√2)aF

(d) aF/√2

(e) 2maF

The distance of the axis of rotation from the line of action of the force is √(a²+a²) = (√2)a. Therefore, the magnitude of the torque is (√2)aF.

[The mass of the plate (m) in the question just serves the purpose of distracting you!]

The following questions are meant for AP Physics C aspirants:

(4) A disc (of mass M and radius R) and a ring (of mass M and radius 2R) are released from rest from the top of an inclined plane. If they roll down the plane without slipping, which one of the following statements is correct?

(a) The ring will reach the bottom of the incline earlier

(b) They will reach the bottom of the incline with the same speed

(c) The ring will reach the bottom of the incline with twice the speed of the disc

(d) They will have the same acceleration down the incline

(e) They will reach the bottom of the incline with the same kinetic energy

Since the disc and the ring have the same mass and both have fallen down through the same height, they have lost the same amount of gravitational potential energy. Therefore, they have gained the same amount of kinetic energy so that the correct option is (e). Since the above question is a single correct answer type multiple choice question, you can easily arrive at the above answer.

[Let us consider the remaining options in the question (in view of the benefits you will have in certain situations).

The linear acceleration ‘a’ of a body rolling down an inclined plane of angle θ is given by

a = g sinθ/[1+(k²/R²)] where g is the gravitational acceleration, k is the radius of gyration about the axis of rolling and R is the radius of the rolling body.

In the case of a ring k²/R² = 1 where as in the case of a disc k²/R² = ½

(Remember that the moment of inertia of a body is k² times the mass of the body).

The acceleration of the ring is less than that of the disc. In fact a ring (and a pipe) has the least acceleration while rolling down an incline. Options (a), (b), (c) and (d) are therefore wrong].

(5) A fly wheel has radius R and its axle has radius R/10. The ends of the axle are mounted on ball bearings so that the wheel can rotate without appreciable friction at the bearings. Strings are wound round the rim of the wheel and the axle. A mass M is suspended vertically from the free end of the string wound round the rim of the wheel. The free end of the string wound round the axle is pulled in a direction perpendicular to the length of the axle and making an angle θ with the horizontal plane. If the mass M moves up with uniform velocity, what is the tension in the string wound round the axle?

(a) Mg

(b) Mg/10

(c) 10 Mg

(d) 10 Mg cosθ

(e) 10 Mgsinθ

The tension acting in the string wound round the rim of the wheel is Mg if the mass M is at rest or in uniform motion (moving with uniform velocity).

The torque produced by the weight of the suspended mas M is RMg. If the mass M is to be at rest or in uniform motion, the opposing torque to be applied by pulling on the string wound round the axle must be of the same magnitude RMg.

If T represents the tension in the string wound round the axle, we have

(R/10)×T = RMg

Therefore, T = 10Mg

[The angle θ in the question just serves the purpose of a distraction].

(6) A solid sphere of mass M and radius R rotates about its central axis with uniform angular acceleration α, on exerting a tangential force F₁. Another solid sphere of mass 2M and radius 2R rotates about its own central axis with the same angular acceleration α, on exerting a tangential force F₂. Then F₁ and F₂ are related as

(a) F₂ = 2F₁

(b) F₂ = 4F₁

(c) F₁ = 4F₂

(d) F₁ = 2F₂

(e) F₁ = F₂

The angular acceleration is the ratio of torque to the moment of inertia. Therefore, the angular accelerations of the spheres A and B are respectively τ₁/I₁ and τ₂/I₂ where τ₁ and τ₂ are the torques exerted on spheres A and B respectivly by the forces F₁ and F₂ and I₁ and I₂ are the moments of inertia of spheres A and B about their central axes.

Since the spheres have the same angular acceleration we have

τ₁/I₁ = τ₂/I₂

Or, RF₁/(2MR²/5) = 2RF₂/(2×2M×4R²/5)

This gives F₂ = 4F₁

[Even if you don’t remember the expression for the moment of inertia of a solid sphere, you can work out this question if you know that the moment of inertia is directly proportional to the mass and the square of the radius].

You will find a few more multiple choice questions in this section here.

AP Physics C – Circular Motion and Rotation – Multiple Choice Questions Based on Some Interesting Aspects of Rolling

“I object to violence because when it appears to do good, the good is only temporary; the evil it does is permanent.”

– Mahatma Gandhi

Rolling motion is very much related to our daily life since the wheels used in transportation move between places by rolling. All of you know that the invention of wheels for transportation was a turning point in the history of mankind. It is therefore important that in your AP Physics course you are required to understand the special features of rolling motion.

Even though you come across rolling bodies very commonly in your daily life, many among you might be unaware of some of the interesting aspects of rolling. For instance, some of you might not have noted that a smooth sphere cannot roll on a smooth surface, whether the surface is horizontal or inclined. For a body to roll along a surface, there has to be friction between the body and the surface. If friction is absent, the body will just slide along the surface. If the friction is insufficient, the body will roll as well as slide (slip) along the surface. In pure rolling there will be no slipping.

This post is meant for making you aware of similar aspects of rolling, by working out a few multiple choice practice questions. Here are the questions:

(1) The adjoining figure shows a disc of radius R rolling without slipping along a horizontal surface. Assume that the centre of mass C of the disc is displaced along the positive x-direction. If the angular velocity of the disc is ω, what is the velocity V_A of the point A of the disc at the instant it is in contact with the horizontal surface?

(a) ωR directed along the positive x-direction

(b) 2ωR directed along the positive x-direction

(c) ωR directed along the negative x-direction

(d) 2ωR directed along the negative x-direction

(e) Zero

Since the disc is rolling without slipping, the contact point A of the disc (to be more precise, the line of contact through A) must be at rest.

The translational velocity V_cm of the centre of mass C of the disc is ωR which is directed parallel to the horizontal surface. Since the disc is a rigid body all points of the disc must move forward (along the positive x-direction) with this translational velocity. But all points of the disc have an additional linear velocity V_r because of the rotation of the disc about its central axis passing through C.

We have V_r = ωr where r is the distance of the point from the centre of mass C. In the case of the contact point A of the disc the distance from the centre of mass is R so that V_r = ωR and is directed along the negative x-direction.

Thus the contact point has the common translational velocity V_cm = ωR directed along the positive x-direction and the linear velocity (due to rotation about C) V_r = ωR directed along the negative x-direction, with the result that it is at rest. The correct option is (e).

(2) What is the velocity V_B of the topmost point B of the disc in question No.1?

(a) ωR directed along the positive x-direction

(b) 2ωR directed along the positive x-direction

(c) Zero

(d) ωR directed along the negative x-direction

(e) 2ωR directed along the negative x-direction

The topmost point B of the disc has translational velocity V_cm = ωR directed along the positive x-direction as in the case of all other points of the disc. The additional linear velocity on account of the rotation is V_r = ωR. This too is directed along the positive x-direction in the case of the topmost point A. Therefore, the resultant velocity of the topmost point is 2ωR directed along the positive x-direction [Option (b)].

(3) What is the velocity V_D of the point D (at the left edge) of the disc in question No.1?

(a) ωR directed along the positive x-direction

(b) (√2)Rω directed along the positive x-direction

(c) Zero

(d) ωR directed vertically upwards

(e) (√2)ωR directed at 45º to the horizontal

The velocity of the point D has two parts (as explained in the solution of question No.1):

(i) The translational velocity V_cm = ωR directed horizontally (along the positive x-direction).

(ii) The additional linear velocity on account of the rotation given by V_r = ωR, directed vertically upwards.

The resultant velocity of the point D is (√2)ωR, directed at an angle of 45º as shown in the adjoining figure [Option (e)].

[Note that the two velocities of equal magnitude ωR at right angles give a resultant velocity of magnitude (√2)ωR inclined equally (at 45º in this case) to both].

(4) Consider any arbitrary point P on a thin disc rolling on any surface. If the centre of mass is C and the point of contact with the surface is A at any instant, the instantaneous velocity of the point P is directed

(a) along the radius through P

(b) parallel to the surface

(c) perpendicular to the line AP

(d) along the line CP

(e) perpendicular to the line CP

You can forget about the velocities V_cm and V_r we considered in solving the previous questions and imagine that the disc is rotating about a parallel axis through the point of contact A (or, in other words, rotating about the line of contact) with angular velocity ω. You can easily obtain the velocities of the points we discussed above. Try yourself!

You will easily arrive at the answer to the present problem: [Perpendicular to the line AP given in option (c)].

[The magnitude of the velocity of point P will be ωr’ where r’ is the distance AP].

(5) A solid sphere and a hollow sphere made of two different metals, but of the same mass and radius are given to you. Which one of the following methods will be suitable for identifying them?

(a) Arrange two simple pendulums of equal length with the hollow sphere and the solid sphere as bobs and compare their periods of oscillation.

(b) Supply equal charges to them and compare the electric potentials on them.

(c) Allow them to fall freely under gravity from the same height and compare their times of fall.

(d) Allow them to roll down from the top of the same inclined plane and compare the times taken to reach the bottom of the incline.

(e) Compare their apparent loss of weight when fully submerged under water.

In all cases except (d) the quantities measured will not distinguish the hollow sphere from the solid sphere. The time for rolling down the incline will be greater for the hollow sphere since its moment of inertia is greater. It has more laziness (inertia) to roll!

[The acceleration a of a body rolling down an incline of angle θ (with respect to the horizontal) is given by

a = (g sin θ)/[1 + (k²/R²)] where g is the acceleration due to gravity, R is the radius of the rolling body and k is the radius of gyration defined by I = Mk² where I is the moment of inertia of the body about its central axis (axis of rolling) and M is the mass of the body.

For a solid sphere k²/R² = 2/5 since I = (2/5)MR².

For a thin hollow sphere k²/R² = 2/3 since I = (2/3)MR²

For a thick hollow sphere of outer radius R and inner radius R₁ the moment of inertia about the central axis is (2/5)M[(R⁵ – R₁⁵)/(R³ – R₁³)].

The quantity k²/R² is certainly more than 2/5 in this case also. Since k²/R² appears in the denominator of the expression for acceleration, the solid sphere has greater acceleration and it reaches the bottom of the incline earlier.

In fact in the case of regularly shaped bodies such as ring, disc, cylinder, hollow sphere, solid sphere etc., made of material of uniform density, the solid sphere has the maximum acceleration while rolling down an incline and a thin ring has the least acceleration (since it has k²/R² = 1).

If you have a solid sphere with non uniform density such that there is more concentration of material near the centre, it will accelerate faster than a solid sphere made of a material of uniform density].

AP Physics C – Circular Motion and Rotation - Multiple Choice Practice Questions

“Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time”.

– Thomas A. Edison

Questions on circular motion and rotation were discussed earlier on this site. You can access them by clicking on the label ‘circular motion and rotation’ below this post’. Today we will discuss a few more interesting multiple choice practice questions in this section:

(1) A long, flexible ribbon of mass M rolled in the form of a cylinder (Fig.) of radius R is placed on horizontal floor. When a negligibly small push is applied, the ribbon unrolls without sliding along the surface (thereby reducing the radius of the cylinder). The horizontal velocity of the axis of the cylinder when the radius becomes (¾)R is

(a) √(2Rg)

(b) √(3Rg/2)

(c) √(27Rg/17)

(d) √(37Rg/27)

(e) √(47Rg/37)

When the roll of ribbon unrolls, the radius of the cylindrical roll decreases, thereby lowering the centre of gravity of the cylinder. Initially the mass of the cylinder is M and its centre of gravity is at a height R. When the radius of the cylinder becomes (¾)R, the mass of the cylinder is (9/16) M and its centre of gravity is at a height (¾)R

[When the radius is R, mass = M = πR²Lρ where L is the length and ρ is the density of the cylinder. When the radius is (¾)R, mass = π(9/16)R²Lρ = (9/16) M].

The initial gravitational potential energy of the cylinder is MgR and the final gravitational potential energy is (9/16)Mg×(¾)R = (27/64)MgR.

Therefore the loss of potential energy = MgR – (27/64)MgR = (37/64)MgR

The cylinder gains an equal amount of kinetic energy (partially rotational and partially translational) which is ½ Iω² + ½ mv² where I is the moment of inertia [of the cylinder of radius (¾)R] about its axis, ω is its angular velocity, m is its mass and v is its linear velocity. Thus we have

(37/64)MgR = (½)(mr²/2)(v²/r² )+ ½ mv² = (¾)mv²

[Remember that the moment of inertia of a cylinder of mas m and radius r about its own axis is mr²/2 and the angular velocity is v/r where v is the linear velocity].

Therefore, (37/64)MgR = (¾)(9/16)Mv²

This gives v = √(37Rg/27), as given in option (d).

[The above question is perhaps more suitable as part of a free response question].

(2) The angular momentum of a fly-wheel changes from 2L to 5L in 5 seconds under the action of a constant torque. The torque is

(a) 5L

(b) 3L

(c) 2L

(d) (3/5)L

(e) (2/5)L

The change in the angular momentum of the fly-wheel is 3L. The angular impulse received by the fly-wheel during the time t is τ t where τ is the torque. Since the angular impulse is equal to the change in angular momentum, we have

τ t = 3L

Since t = 5 seconds, τ = (3/5)L

(3) The adjoining figure shows a bar pendulum. It consists of a sphere of mass M attached to the end B of a light rigid rod AB suspended from a horizontal nail (at A) on the wall. Initially the pendulum is at rest in its equilibrium position, with the rod vertical. What is the magnitude of the minimum velocity to be imparted to the sphere so that it just completes its circular path?

(a) √(gL)

(b) √(2gL)

(c) √3gL)

(d) √(4gL)

(e) √(5gL)

Since the rod is light, this problem is quite simple. It is enough to equate the initial kinetic energy of the sphere to its potential energy (gravitational) at its topmost position. Therefore we have

½ Mv² = Mg×2L where ‘v’ is the magnitude of the minimum velocity to be imparted to the sphere.

[Note that the height through which the centre of gravity of the sphere rises is 2L].

Therefore, v = √(4gL)

[If the sphere is suspended by means of a string (as in the case of a simple pendulum), the minimum speed to be imparted will be greater {equal to √(5gL)}, since the sphere should possess kinetic energy also at its topmost position, in order to maintain its circular path. In that case the tension in the string is zero and the speed at the top of the path (v_top) is such that the centrifugal force on the sphere has the same magnitude as the weight of the sphere:

Mv_top²/L = Mg from which v_top = √(gL)

The kinetic energy at the top is ½ Mv_top² = ½ MgL

The minimum kinetic energy to be imparted to the sphere in its initial position is ½ MgL + (Mg×2L) = (5/2) MgL

Therfore, the minimum velocity ‘v’ to be imparted to the sphere is given by

½ Mv² = (5/2) MgL from which v = √(5gL)]

(4) A simple pendulum of length L has a bob of mass M. The pendulum is suspended from the point O and is held initially in the position OA, as shown in the adjoining figure, so that the string is horizontal. On releasing the pendulum, what will be the resultant acceleration of the bob when the pendulum is in the position OB? (Acceleration due to gravity = g)

(a) g [1 + 3 sin²θ]^1/2

(b) g [1 + 3 cos²θ]^1/2

(c) 3g sin²θ

(d) 3g (1+sinθ)

(e) 3g/2

The forces acting on the bob of the pendulum in the position OB are the weight Mg of the bob acting vertically downwards and the tension T (Fig.) in the string which supplies the centripetal force required for the circular motion of the bob.

The speed v of the bob in the position B is given by

½ Mv² = Mg Lsinθ

Therefore v = √(2gL sinθ) ……….(i)

[In position B the bob has fallen down by a distance L sinθ, losing gravitational potential energy mg Lsinθ and gaining kinetic energy ½ Mv²]

The centripetal force on the bob is the resultant of the tension T in the string and the component Mg sinθ of the weight of the bob. Therefore we have

Mv²/L = T – Mg sinθ

Substituting for v from Eq.(i) we have

2Mg sinθ = T – Mg sinθ

This gives T = 3Mg sinθ

The resultant force F acting on the bob is the resultant of the forces Mg and T (with an angle 90º+θ between them) and has magnitude equal to [(Mg)² + T² + (2×Mg× T) cos(90+θ)]^1/2.

Substituting for T, the resultant force is

F = [(Mg)² + (3Mg sinθ)² + (2× Mg× 3Mg sinθ) cos(90+ θ)]^1/2 = [(Mg)² + 3(Mg sinθ)²]^1/2 = Mg [1 + 3 sin²θ]^1/2

[Remember that cos(90+ θ) = – sinθ]

The resultant acceleration of the bob in position B is F/M = g [1 + 3 sin²θ]^1/2.

(5) A disc of mass M and radius R and a ring of mass M and radius 2R are released from rest from the top of an inclined plane. If they roll down the plane without slipping, which one of the following statements is correct?

(a) The ring will reach the bottom of the incline earlier

(b) They will reach the bottom of the incline with the same speed

(c) The ring will reach the bottom of the incline with twice the speed of the disc

(d) They will have the same acceleration down the incline

(e) They will reach the bottom of the incline with the same kinetic energy

The acceleration ‘a’ of a body of radius R rolling down an incline of angle θ is given by

a = g sinθ/[1 + k²/R²] where k is the radius of gyration of the body.

[The radius of gyration k is defined by I = Mk² where I is the moment of inertia (about the axis of rolling) and M is the mass].

Note that the acceleration is independent of the mass and size of the body. The ratio k²/R² also is independent of the mass and size of the body

For a given inclined plane, the acceleration depends only on the ratio k²/R².

For a given inclined plane, the acceleration depends only on the ratio k²/R².

[You may substitute the value of moment of inertia in the equation I = Mk² and find the value of k²/R². The maximum value in the case of regularly shaped bodies is for a thin ring and is equal to 1 since I = MR² = Mk². The least value is 2/5 which is the case for a solid sphere since I = (2/5)MR². Better, see this post ]

Moment of inertia of a disc (about the axis of rolling) is MR²/2 where as the moment pf inertia of a ring (about the axis of rolling) is MR².

The acceleration in the case of a disc is g sinθ/[1+ ½] = (2/3) g sinθ

The acceleration in the case of a ring is g sinθ/[1+ 1] = (½) g sinθ

All options except the last one (e) in the question are wrong. The kinetic energy with which the disc and the ring will reach the bottom of the incline will be the same since they have the same mass and they start from the same height.

[The kinetic energy at the bottom of the incline is the same as the potential energy at the top of the incline].