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