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*’

*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

*M*

_{1 }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 a_{1}* *of the mass *M*_{1} in terms of the given parameters.

(c) In terms of the given parameters obtain the angular acceleration of the cylinder when the mass *M*_{1} 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 = I*α where *I* is the moment of inertia (*MR*^{2}*/*2)* *of the cylinder about its axis (axle). Therefore, *TR = *(*MR*^{2}*/*2)α

But α = *a/R* so that *TR = *(*MR*^{2}*/*2)(*a/R*)

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

*M *(*g* – *a*)*R = *(*MR*^{2}*/*2)(*a/R*) from which (*g* – *a*)* = a/*2 so that *a = *2*g*/3

(b) The force driving the mass *M*_{1} downwards is its weight *M*_{1}*g*. The tension *T*_{1}* *in the sring tries to oppose the downward motion (fig.). The downward acceleration a_{1} of the mass *M*_{1} is therefore given by* M*_{1}*g* – *T*_{1}* = M*_{1}a_{1} from which

*T*_{1}* = M*_{1}(*g* – a_{1})

Since the *torque* produced by the tension *T*_{1} is *T*_{1}*R *and this is responsible for the rotation of the cylinder, we have

*T*_{1}*R = I*α_{1} where α_{1} is the angular acceleration of the cylinder.

Substituting for *T*_{1} and the moment of inertia *I*, we have

*M*_{1}(*g* – a_{1})*R* *= *(*MR*^{2}*/*2)α_{1}

Substituting for α_{1} = a_{1}*/R*, the above equation becomes

*M*_{1}(*g* – a_{1})*R* *= *(*MR*^{2}*/*2)(a_{1}*/R*)

Or, 2*M*_{1}(*g* – a_{1}) = *M*a_{1}, from which **a _{1} = 2M_{1}g/(2M_{1}+M)**

(c) The angular acceleration α_{1 }of the cylinder when the mass *M*_{1} moves down, unwinding the string is given by

α_{1} = a_{1}/*R* = **2 M_{1}g/ [(2M_{1}+M)R]**

(d) The kinetic energy of the cylinder at time *t* is due to its rotational motion and is equal to ½ *I**ω*^{2} where *ω* is its angular velocity at time *t*.

But *ω* is related to the angular acceleration α_{1} of the cylinder as

α_{1} = *d**ω/dt*

Therefore, *d**ω/dt = *2*M*_{1}*g/ *[(2*M*_{1}+*M*)*R*]

Integrating, *ω =*2*M*_{1}*gt / *[(2*M*_{1}+*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 =*2*M*_{1}*gt / *[(2*M*_{1}+*M*)*R*]

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

*ω* = *ω*_{0} + α*t* where *ω*_{0 }= 0 and α = α_{1} = 2*M*_{1}*g/*{(2*M*_{1}+*M*)*R*}]

Therefore, kinetic energy of cylinder = ½ *I**ω*^{2} = ½ ×(*MR*^{2}/2) ×4*M*_{1}^{2}*g*^{2}*t*^{2}* / *[(2*M*_{1}+*M*)*R*]^{2}.

This yields the value *MM*_{1}^{2}*g*^{2}*t*^{2}* / *(2*M*_{1}+*M*)^{2}.

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

Suppose you were asked to determine the kinetic energy of the mass *M*_{1} at time *t*. You will then proceed as follows:

At time *t *the mass *M*_{1} has kinetic energy equal to ½ *M*_{1}*v*^{2} where *v* is the velocity acquired by it in time *t* while moving down with acceleration a_{1}. Since it started from rest we have

*v = v*_{0} + a_{1}*t ^{ }= *0 + [2

*M*

_{1}

*g/*(2

*M*

_{1}+

*M*)]

*t*

Therefore, kinetic energy of mass *M*_{1} = ½ *M*_{1}*v*^{2} = ½ ×*M*_{1}[2*M*_{1}*g*t* /*(2*M*_{1}+*M*)]^{2}

This yields the value 2*M*_{1}^{3}*g*^{2}*t*^{2}* /*(2*M*_{1}+*M*)^{2}

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