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 B & C - Multiple Choice Practice Questions Involving Rotational Motion

Example isn't another way to teach, it is the only way to teach.

– Albert Einstein

Questions (Multiple Choice and Free Response) involving rotational motion were discussed on various occasions on this site. Essential formulae to be remembered in this section were discussed in the post dated 20^th January 2008. You can access all the posts related to this section by clicking on the label ‘rotational motion’ below this post or by trying a search using the search box provided on this page. In both cases you need to use ‘older posts’ links to obtain all related posts.

Some of the questions on rotational motion may prove to be some what difficult and time consuming for many among you. Once you master the basic principles, you will become more confident. Today we will discuss a few more multiple choice practice questions involving rotational motion:

(1) A metre stick of mass 0.2 kg can be balanced on a knife edge at the 60 cm mark (see figure) when a piece of rock (specific gravity = 3) fully immersed in water is suspended from the 70 cm mark. If the piece of rock were in air, suspended from the 70 cm mark itself, what would be the position of the knife edge for the balance?

(a) 56 cm

(b) 58 cm

(c) 62 cm

(d) 64 cm

(e) 66 cm

The weight of the metre stick acts through its centre of gravity, which is at the 50 cm mark. The torque due to the weight of the metre stick tries to rotate it in the anticlockwise sense where as the torque due to the apparent weight of the piece of rock in water tries to rotate the metre stick in the clockwise sense. These torques have equal magnitudes since the metre stick is horizontal. Since the lever arms are equal (0.1 m each), the apparent weight of the piece of rock must be equal to the weight of the metre stick which is equal to 0.2 kgwt.

[Note that we got the above result by equating the magnitudes of the torques: 0.2×0.1 = w₂×0.1 where w₂ is the apparent weight (or, the weight inside water) of the piece of rock].

If the weight of the piece of rock in air is w₁, the loss of weight in water is w₁ – w₂ and the specific gravity, which is equal to 3 is given by

w₁/( w₁ – w₂) = 3

Or, w₁/( w₁ – 0.2) = 3

Therefore, w₁ = 0.3 kgwt.

When the piece of rock is in air, the knife edge should be at distance x from the centre of gravity such that

0.2×x = 0.3×(0.2 – x)

This gives x = 0.06/0.5 = 0.12 m = 12 cm

The position of the knife edge is therefore 50 cm + 12 cm = 62 cm.

(2) A sphere rolls without slipping along a horizontal surface. If the velocity of the centre of mass of the sphere is v, what is the velocity of the point of contact (with the horizontal surface) of the sphere?

(a) 2 v

(b) v

(c) v/2

(d) v/4

(e) zero

When the sphere rolls, it has an angular motion (rotation) and a linear motion (translation). If the the horizontal surface is perfectly smooth, the sphere will slip fully and will just spin about its central axis. In other words, it will have angular motion alone and the velocity of the contact point will be v, directed backwards.

When the surface is rough, the sphere rolls forward and the centre of mass moves forward with velocity v. Since the sphere is rigid, the contact point (lowest point) of the sphere too has to move forward with the velocity v. Since the contact point has an additional backward velocity v due to the spin of the sphere, the resultant velocity of the contact point on the sphere is zero [Option (e)].

[In the above question what is the velocity of the top point of the sphere? You can easily show that it is 2 v, forward].

(3) For opening a door 1 m wide, the minimum force required is 20 N. If the door is pushed at a distance of 0.4 m from the line of the hinges, what is the minimum force required for opening it?

(a) 8 N

(b) 10 N

(c) 20 N

(d) 40 N

(e) 50 N

The force required for opening the door will be minimum if it is applied perpendicular to the plane of the door, with the point of application of the force farthest away from the line of the hinges (at the edge of the door, opposite to the line of the hinges). This ensures that the torque produced by the force is maximum. Evidently the minimum torque required for opening the door is 20×1 Nm = 20 Nm.

[Remember that torque = force×lever arm. It is the torque (and not the force) that matters for opening the door].

If the force is applied at a distance of 0.4 m from the line of the hinges, the minimum force F required for opening the door is given by

F×0.4 = 20 Nm

Therefore, F = 50 N.

The following questions are specifically for the AP Physics C aspirants:

(4) The angular displacement θ of a fly wheel at the instant t is given by

θ =2t³ – 4 t² + 8

The angular acceleration of the fly wheel after 2s in is

(a) 20 rad/s

(b) 16 rad/s

(c) 12 rad/s

(d) 8 rad/s

(e) 6 rad/s

The angular acceleration α is given by

α = d²θ/dt²

Since dθ/dt = 6 t² – 8 t, we have

α = 12 t – 8

The angular acceleration after 2 s = (12×2) – 8 = 16 rad/s

(5) A cylinder of moment of inertia 2 kgm² has angular displacement θ given by

θ =3t² – 4 t

The torque acting on the cylinder

(a) decreases linearly with time

(b) increases linearly with time

(c) decreases non-linearly with time

(d) increases non-linearly with time

(e) remains constant

The angular acceleration of the cylinder is α = d²θ/dt² = 6 rad/s².

The torque τ is given by

τ = I α where I is the moment of inertia.

Therefore, τ =2×6 = 12 Nm

Since this is constant (independent of time), option (e) is correct.

[The expression for the angular displacement θ is similar to the expression s = ut + ½ at² for the linear displacement in uniformly accelerated motion. Therefore from the form of the expression itself you should be able to understand that the angular acceleration and hence the torque, is constant].

(6) Two identical solid hemispheres, each of mass m and radius r are welded together as shown. What is the moment of inertia of this system about the axis AB which is perpendicular to the plane surfaces of the hemispheres? [Moment of inertia of a solid sphere of mass M and radius R about any diameter is (²/₅)MR²]

(a) (¹⁴/₅)mr²

(b) (⁷/₅)mr²

(c) (⁶/₅)mr²

(d) (¹⁴/₅)mr²

(e) (¹⁴/₅)mr²

Since the moment of inertia of a solid sphere of mass M and radius R about any diameter is (²/₅)MR², its moment of inertia I about a tangent is (by the theorem of parallel axes) given by

I = (²/₅)MR² + MR² = (⁷/₅)MR²

The moment of inertia of a hemisphere about a tangent such as AB must be half of this. Since the axis AB is the common tangent to the two hemispheres, the total moment of inertia of the system containing the two hemispheres must be (⁷/₅)MR².

In the above expression M is the mass of the entire sphere and we have

M = 2m

Also R = r as given in the question.

Therefore, the answer is (⁷/₅)(2m)r² = (¹⁴/₅)mr²

You will find few more practice questions (with solution) in this section here.

AP Physics B & C- Multiple Choice Questions (for practice) on Conservation of Energy

Many of life's failures are people who did not realize how close they were to success when they gave up.

– Thomas A.Edison

The following two questions will be beneficial for AP Physics B as well as C:

(1) A rain drop of mass m at a height h starts from rest and gets accelerated under gravity initially. Because of the viscosity of air, its motion is resisted and it finally falls down with a constant terminal velocity v_T and hits the ground. You may neglect the force of buoyancy. Assume that the acceleration due to gravity, g is constant and the rain drop moves vertically down over the entire path. The magnitude of the average viscous force on the rain drop during the entire downward trip is

(a) mg

(b) mg – (mv_T²/h)

(c) mg + (mv_T²/h)

(d) mg + (mv_T²/2h)

(e) mg – (mv_T²/2h)

The initial gravitational potential energy of the drop at height h is mgh. On reaching the ground, it has lost its gravitational potential energy but has gained kinetic energy equal to ½ mv_T².

The loss of energy because of air resistance is mgh – ½ mv_T².

If the magnitude of the average viscous force on the rain drop is F, we have

Fh = mgh – ½ mv_T²

Therefore, F = mg – (mv_T²/2h)

(2) In the above question what is the viscous force when the rain drop moves down with its terminal velocity?

(a) mg upwards

(b) zero

(c) mg downwards

(d) mv_T²/h) upwards

(e) mv_T²/h) downwards

When the rain drop moves down with its terminal velocity, the net force acting on it is zero. This means that the weight mg of the rain drop and the viscous force offered by air are equal and opposite. Therefore, the viscous force has magnitude mg and is directed upwards.

The following question involves resistive torques in rotational motion.

(3) One end of a light inextensible string is tied to a solid cylinder of mass M and radius R. The string is wound round the cylinder many times and a mass m is suspended from the free end of the string so that on releasing the mass, the cylinder rotates about its central axis. There exists an appreciable amount of friction between the cylinder and the axle about which the cylinder rotates. When the mass m which is initially at a height h from the ground (Fig.) is released from rest, it strikes the ground with speed v. How much energy is lost from the system for doing work against friction during the fall of the mass m through the height h?

(a) mgh – (½ mv²+ ¼ Mv²)

(b) mgh – ½ mv²

(c) mgh – ½ (m + M)v²

(d) mgh – (½ mv²+ ½ MR²v²)

(e) mgh – (½ mv²+ ¼ MR²v²)

The system acquires kinetic energy at the cost of the gravitational potential energy of the suspended mass m.

The initial energy E₁ of the system is given by

E₁ = mgh.

The final energy E₂ of the system at the moment the mass m hits the ground is the sum of the translational kinetic energy of the mass m and the rotational kinetic energy of the cylinder:

E₂ = ½ mv² + ½ (MR²/2)×(v²/R²)

[The second term on the right hand side is ½ Iω² where I is the moment of inertia of the cylinder (MR²/2) and ω is its angular velocity (v/R) when the mass m strikes the ground].

Therefore, E₂ = ½ mv² + ¼ Mv²

The work (W) done by the system against frictional forces is given by

W = E₁ – E₂ = mgh – (½ mv²+ ¼ Mv²)