Multiple Choice Practice Questions on Rotation for AP Physics C

Some multiple choice practice questions (with solution) on rotational motion were posted earlier on this site. You can access those posts by clicking on the label ‘rotational motion’ or ‘rotation’ below this post. We will discuss a few more multiple choice questions (MCQ) for practice in this section today:

(1) A car moves forward on level road with uniform velocity of 4 ms^–1. If there is no slipping of the tyres, the velocities (with respect to the road) of the points at the bottom, centre and the top of a tyre are respectively

(a) 4 ms^–1, 4 ms^–1 and 4 ms^–1

(b) zero, 4 ms^–1and 8 ms^–1

(c) 4 ms^–1, zero and 8 ms^–1

(d) 8 ms^–1, 4 ms^–1and 8 ms^–1

(e) 8 ms^–1, 4 ms^–1 and zero

The wheel has translational motion as well as rotational motion. Because of the rotation, points at the bottom of the wheel move backward with speed 4 ms^–1 where as points at the top move forward with speed 4 ms^–1. The resultant speed of points at the bottom is therefore zero (4 ms^–1 – 4 ms^–1) where as that of points at the top is 8 ms^–1 (4 ms^–1 + 4 ms^–1). The linear velocity of the centre of the wheel because of its rotation is zero and hence the velocity of 4 ms^–1 due to the translational motion is preserved at the central point of the wheel. The correct option is (b).

(2) Three solid spheres of masses 10 kg, 8 kg and 2 kg are connected by two springs of negligible mass and force constants 10 Nm^–1 and 12 Nm^–1 as shown. The spheres are at rest on a horizontal frictionless surface so that their centres are along the x-axis. An impulsive force is applied on the 10 kg mass so that it starts moving with a velocity of 2 ms^–1 along the positive x-direction. The velocity of the centre of mass of the system of spheres is

(a) 0.5 ms^–1

(b) 0.8 ms^–1

(c) 1 ms^–1

(d) 1.6 ms^–1

(e) 2 ms^–1

Some of you will be confused by this simple problem. You should remember that the forces developed in the springs are internal forces in the system and hence they will not affect the velocity of the centre of mass. The law of conservation of momentum is obeyed and we can equate the initial momentum to the final momentum:

10×2 + 8×0 + 2×0 = (10+8+2)v where v is the velocity of the centre of mass.

This gives v = 1 ms^–1

(3) A horizontal turn table of mass M in the form of a uniform circular disc of radius R is rotating about a central vertical axis with uniform angular velocity ω. The friction at the axis is negligible. When two equal masses m and m are gently placed symmetrically on either side of the centre at distance R/2 (Fig), the angular velocity of the turn table is reduced to ω/2. Then each mass m is equal to

(a) M/4

(b) M/2

(c) M

(d) 3M/2

(e) 2M

This question is meant for checking your understanding of the law of conservation of angular momentum and for testing whether you can apply it in situations where it is needed.

The angular momentum is conserved in the absence of external torque. The angular momentum of the system before and after placing the two equal masses is the same so that we have

(MR²/2) ω = [(MR²/2) + 2m (R/2)²] ω/2

[Note that MR²/2 is the moment of inertia of the disc about the central axis and 2m (R/2)² is the total moment of inertia of the two masses m and m about the same axis which is distant R/2 from them].

From the above equation MR²/4 = mR²/4 so that m = M

(4) A solid sphere S₁ spinning about its own central axis with angular velocity ω moves along a frictionless horizontal surface and undergoes a head-on elastic collision with an identical solid sphere S₂ at rest on the surface. After the collision the spin angular velocities of S₁ and S₂ will be respectively

(a) ω/2 and ω/2

(b) ω and zero

(c) zero and ω

(d) zero and 2ω

(e) 2ω and zero

Since the line of action of the force exerted by S₁ on S₂ passes through the centre of S₂, there is no torque on S₂. Its spin angular momentum after the collision is therefore zero (same as that before the collision). Since the angular momentum of the system is to be conserved, the angular momentum of S₁ also is unchanged by the collision. Hence its spin angular velocity after the collision is ω. The correct option is (b).

(5) A car is decelerating uniformly from velocity v₁ to velocity v₂ while getting displaced by s. If the diameter of the wheel is d, the angular acceleration of the wheel is

(a) (v₁² – v₂²)/ds

(b) (v₂ – v₁)/ds

(c) (v₁ – v₂)/ds

(d) (v₂² – v₁²)/ds

(e) 2(v₂² – v₁²)/ds

The acceleration ‘a’ of the car is given by

v₂² = v₁² + 2as

[We have applied the equation of linear motion, v² = v₀² + 2as (or, v² = u² + 2as)].

Therefore, a = (v₂² – v₁²)/2s.

The angular acceleration ‘α’ of the wheel is given by

α = a/r where r is the radius of the wheel (which is equal to d/2).

Therefore, α = [(v₂² – v₁²)/2s]/(d/2) = (v₂² – v₁²)/ds

Answer to AP Physics C Free Response Practice Question on Rotation

In the post dated 30^th September 2009 A free response question (for practice) on rotational motion for AP Physics C aspirants was given. As promised I give below a model answer for the same. The question also is given:

A solid cylinder of mass M and radius R has a light inextensible string wound round it. The free end of the string is tied to a rigid support (Fig.) and the cylinder is released from its state of rest.

(a) Determine the acceleration ‘a’ of the cylinder as it moves down, unwinding the string.
The cylinder itself is now suspended from the support so that it is free to rotate about its axle. The friction at the axle of the cylinder is negligible. A mass M₁ is now attached to the free end of the string wound round the cylinder as shown in the adjoining figure and the system is released from rest at time t = 0. Now answer the following questions (b), (c) and (d):

(b) Determine the acceleration a₁ of the mass M₁ in terms of the given parameters.

(c) In terms of the given parameters obtain the angular acceleration of the cylinder when the mass M₁ moves down, unwinding the string.

(d) Determine the kinetic energy of the cylinder at time t.

(a) The weight Mg of the cylinder tries to drive it downwards while the tension T in the sring tries to oppose the downward motion (fig.). The net force on the cylinder is Mg – T. If the downward acceleration is a we have

Mg – T = Ma from which T = M (g – a)

The torque produced by the tension T rotates the cylinder and produces an angular acceleration α (let us say) so that we have

TR = Iα where I is the moment of inertia (MR²/2) of the cylinder about its axis (axle). Therefore, TR = (MR²/2)α

But α = a/R so that TR = (MR²/2)(a/R)

Since T = M (g – a), the above equation becomes

M (g – a)R = (MR²/2)(a/R) from which (g – a) = a/2 so that a = 2g/3

(b) The force driving the mass M₁ downwards is its weight M₁g. The tension T₁ in the sring tries to oppose the downward motion (fig.). The downward acceleration a₁ of the mass M₁ is therefore given by M₁g – T₁ = M₁a₁ from which

T₁ = M₁(g – a₁)

Since the torque produced by the tension T₁ is T₁R and this is responsible for the rotation of the cylinder, we have

T₁R = Iα₁ where α₁ is the angular acceleration of the cylinder.

Substituting for T₁ and the moment of inertia I, we have

M₁(g – a₁)R = (MR²/2)α₁

Substituting for α₁ = a₁/R, the above equation becomes

M₁(g – a₁)R = (MR²/2)(a₁/R)

Or, 2M₁(g – a₁) = Ma₁, from which a₁ = 2M₁g/(2M₁+M)

(c) The angular acceleration α₁ of the cylinder when the mass M₁ moves down, unwinding the string is given by

α₁ = a₁/R = 2M₁g/ [(2M₁+M)R]

(d) The kinetic energy of the cylinder at time t is due to its rotational motion and is equal to ½ Iω² where ω is its angular velocity at time t.

But ω is related to the angular acceleration α₁ of the cylinder as

α₁ = dω/dt

Therefore, dω/dt = 2M₁g/ [(2M₁+M)R]

Integrating, ω =2M₁gt / [(2M₁+M)R] + C where C is the constant of integration which can be found out from the initial condition.

Initially (when t = 0) the cylinder is at rest and hence ω = 0. Substituting in the above equation we have

0 = 0 + C. Therefore C = 0 and the expression for ω becomes

ω =2M₁gt / [(2M₁+M)R]

[Since the angular acceleration is constant, you can easily obtain ω from the equation,

ω = ω₀ + αt where ω₀ = 0 and α = α₁ = 2M₁g/{(2M₁+M)R}]

Therefore, kinetic energy of cylinder = ½ Iω² = ½ ×(MR²/2) ×4M₁²g²t² / [(2M₁+M)R]².

This yields the value MM₁²g²t² / (2M₁+M)².

* * * * * * * * * * * * * * * * * *

Suppose you were asked to determine the kinetic energy of the mass M₁ at time t. You will then proceed as follows:

At time t the mass M₁ has kinetic energy equal to ½ M₁v² where v is the velocity acquired by it in time t while moving down with acceleration a₁. Since it started from rest we have

v = v₀ + a₁t = 0 + [2M₁g/(2M₁+M)]t

Therefore, kinetic energy of mass M₁ = ½ M₁v² = ½ ×M₁[2M₁gt /(2M₁+M)]²

This yields the value 2M₁³g²t² /(2M₁+M)²

AP Physics C – A Free Response Practice Question on Rotation

Essential points and some practice questions on rotational motion have been discussed on this site earlier. You can access all posts related to rotational motion on this site by clicking on the label ‘rotation’ below this post.

Today we will discuss a free response question (for practice) in this section. This

question is meant for AP Physics C aspirants:

A solid cylinder of mass M and radius R has a light inextensible string wound round it. The free end of the string is tied to a rigid support (Fig.) and the cylinder is released from its state of rest.

(a) Determine the acceleration ‘a’ of the cylinder as it moves down, unwinding the string.

The cylinder itself is now suspended from the support so that it is free to rotate about its axle. The friction at the axle of the cylinder is negligible. A mass M₁ is now attached to the free end of the string wound round the cylinder as shown in the adjoining figure and the system is released from rest at time t = 0. Now answer the following questions (b), (c) and (d):

(b) Determine the acceleration a₁ of the mass M₁ in terms of the given parameters.

(c) In terms of the given parameters obtain the angular acceleration of the cylinder when the mass M₁ moves down, unwinding the string.

(d) Determine the kinetic energy of the cylinder at time t.

Try to answer this question which carries 15 points. You have 15 minutes at your disposal. I’ll be back with a model answer for you shortly.

Meanwhile find some useful multiple choice questions with solution in this section at physicsplus.

AP Physics B & C – Two Multiple Choice Questions (for practice) on Newton’s Laws of Motion

"Only two things are infinite, the universe and human stupidity, and I'm not sure about the former."

– Albert Einstein

Two practice questions (MCQ) on Newton’s laws of motion are given below with solution. Even though these are meant for AP Physics C aspirants, they should not be difficult for those who are preparing for AP Physics B Exam.

(1) In the system shown in the adjoining figure, the horizontal surface on which block A is placed is smooth and the pulleys are light and frictionless. If g = 10 ms^–2, the tension T in the string (which may be assumed to be weightless and inextensible) is

(a) 1 N

(b) 2N

(c) 5 N

The velocity of the iron block at the moment of hitting the nail is 6 ms^–1.

(d) 10 N

(e) 20 N

Net force driving the system = Weight of B = mg = 2×10 = 20 N.

Acceleration (a) of the system is given by

a = Driving force /Total mass moved = ²⁰/₄ = 5 ms^–2.

Tension, T = Weight of B moving down (as in a lift) = m(g – a) = 2(10 – 5) =10 N.

(2) An iron block of mass 8 kg falling vertically down hits a nail on a block of wood (fig.). The velocity of the iron block at the moment of hitting the nail is 6 ms^–1. If the nail penetrates through 2 cm into the wooden block due to the impact, what is the force exerted by the iron block on the nail?

(a) 1600 N

(b) 2200 N

(c) 4800 N

(d) 7200 N

(e) 9600 N

The acceleration (in fact, retardation), a of the iron block during the impact is given by

0 = 6² + 2a×(0.02), from the equation v² = u² + 2as

This gives a = – 900 ms^–2

The negative sign shows that the iron block is retarded.

The retarding force exerted by the nail on the iron block is of magnitude F given by

F = ma = 8×900 = 7200 N

The force exerted by the iron block on the nail is equal and opposite to that exerted by the nail on the iron block. The answer therefore is 7200 N.

[The above question can be modified as follows:

An iron block of mass 8 kg falling freely from rest under gravity from a height of 1.8 m hits a nail on a block of wood (fig.). If the nail penetrates through 2 cm into the wooden block due to the impact, what is the force exerted by the iron block on the nail? (g = 10 ms^–2)

(a) 1600 N

(b) 2200 N

(c) 4800 N

(d) 7200 N

(e) 9600 N

The answer will be unchanged since the velocity at the moment of hitting the nail will be 6 ms^–1 itself (v = √(2gh) = √(2×10×1.8) = √(36) = 6)].

AP Physics C – Multiple Choice Questions (for Practice) on Maxwell’s Equations and Electromagnetic Waves

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As promised in the post dated 26^th August 2009, I give below a few typical multiple choice practice questions for AP Physics C students:

(1) The displacement current in the dielectric between the plates of a 0.1 μF capacitor is 1 mA. The potential difference between the plates of this capacitor must be changing at the rate of

(a) 10⁴ volt/second

(b) 10³ volt/second

(c) 10² volt/second

(d) 10 volt/second

(e) 1 volt/second

This is a simple question contrary to the impression of some of you at the first glance! Whether it is conduction current or displacement current, we have the usual expression for electric current (i) as

i = dQ/dt

Since the charge Q = CV, we have i = CdV/dt

This gives dV/dt = i/C = 1 mA /0.1 μF = 10^–3 ampere /10^–7 farad = 10⁴ volt/second [Option (a)].

(2) A 10 μF electrolytic capacitor connected to a 6 V power supply is fully charged. The displacement current flowing through the dielectric is

(a) 100 mA

(b) 10 mA

(c) 1mA

(d) 0.1 mA

(e) zero

The correct option is (e) since there is no charging current in the case of the fully charged capacitor.

(3) The electric and magnetic fields in a radio wave propagating along the z-direction are given respectively by

E_x= E₀ sin (kz–ωt ) and

B_y= B₀ sin (kz–ωt )

where E₀ and B₀ are the amplitudes, k is the magnitude of the wave vector (propagation vector) and ω is the angular frequency. How is E₀ related to B₀?

(a) E₀ = B₀ kω

(b) E₀ = B₀k/ω

(c) E₀ = B₀√(kω)

(d) E₀ = B₀

(e) E₀ = B₀ω/k

The electric and magnetic fields in the wave are varying simple harmonically and the speed of the wave is given by c = ω/k. [Note that the speed of a wave propagating in the z-direction is given by the ratio of coefficient of t to the coefficient of z. [See the post dated 9^th April 2009 ‘AP Physics B- Wave Motion (including sound)- Points to be Noted’ which you can access by clicking on the label ‘waves’ below this post.

Since E = Bc we have E₀ = B₀c = B₀ω/k

(4) The charge on the plates of a capacitor of plate area 0.1 m² is 0.05 coulomb and is decreasing at the rate of 0.5 coulomb per second at an instant t during its discharge. What is the displacement current at the instant t?

(a) 0.005 A

(b) 0.05 A

(c) 0.5 A

(d) 5 A

(e) Data insufficient for arriving at an answer

Data is more than sufficient. The plate area given in the question is not required and it serves the purpose of a distraction. The displacement current is the time rate of change of charge (dQ/dt) on the plates and is equal to 0.5 ampere [Option (c)].

[If the capacitor is an ideal parallel plate capacitor and you are required to find the magnitude of the displacement current density, the answer is 0.5 A/0.1 m² = 5 Am^–2]

(5) At a particular point in space and time, the magnetic field in a plane electromagnetic wave traveling in free space along the positive x-direction is given by B = 3.1×10^–8 k T (where i, j, k are unit vectors along the x, y, z directions as usual). The electric field at the same point at the same instant is

(a) – 1.033 j Vm^–1

(b) 9.3 i Vm^–1

(c) – 9.3 j Vm^–1

(d) 1.033 k Vm^–1

(e) 9.3 j Vm^–1

The magnitude of the electric field is given by E = Bc = 3.1×10^–8 ×3×10⁸ Vm^–1 = 9.3 Vm^–1. Since the wave is proceeding along the positive x-direction and the magnetic field vector is along the positive z-direction (as indicated by the unit vector k), the electric field must be along the positive y-direction. [Note that the cross product vector E×B should point along the direction of propagation].

Therefore, E = 9.3 j Vm^–1.

(6) A microwave signal traveling through the earth’s atmosphere has an electric field of amplitude 0.02 Vm^–1. The intensity (power flow per unit area) of this signal is (Permittivity of free space, ε₀ = 8.85×10^–12 C² N^–1 m^–2)

(a) 5.31×10^–2 Wm^–2

(b) 5.31×10^–3 Wm^–2

(c) 5.31×10^–6 Wm^–2

(d) 5.31×10^–7 Wm^–2

(e) 5.31×10^–8 Wm^–2

The intensity (I) which is the power flow through unit area (with the plane of the area perpendicular to the direction of propagation) is given by

I = ε₀ c E_m²/2 where c is the speed of the signal and E_m is the amplitude of the electric field.

Therefore, I =(8.85×10^–12 ×3×10⁸×0.0004) / 2 = 5.31×10^–7 Wm^–2