## Saturday, August 09, 2008

### Multiple Choice Questions (MCQ) Involving Rotation

Questions on rotational motion have been discussed on many occasions on this site. You can access them by clicking on the label ‘rotation’ below this post. Now, see the following questions involving rotation.

(1) A light inextensible string is wound round a wheel of moment of inertia I and radius R. A mass m is attached to its hanging end as shown. The wheel and the mass are initially at rest. Assume that there are no frictional forces opposing the motion. When the mass is released, it moves down and rotates the wheel. When the mass has fallen down through a distance h, what is the angular velocity of the wheel?
(a) [2gh/(I + mR)]1/2

(b) [2mgh/ (I + 2m)]1/2

(c) [2mgh/ (I + 2mR2)]1/2

(d) [2mgh/ (I + mR2)]1/2

(e) [2mgh/ (I + m)]1/2

The mass loses gravitational potential energy and gains kinetic energy. The wheel also gains kinetic energy so that we have

mgh = ½ mv2 + ½ 2 where v is the velocity of the mass after falling through the distance h and ω is the angular velocity of the wheel.

Since v =ωR the above equation becomes

mgh = ½ 2R 2 + ½ 2

This yields ω = [2mgh/ (I + mR2)]1/2

(2) A simple pendulum has a spherical bob of mass 200 g. The string of the pendulum has negligible mass and it can withstand a maximum tension of 22 N. By holding the string in hand a student whirls the bob in a vertical circle of radius 1 m. The maximum possible angular velocity the bob can have throughout its motion (in radian per second) will be nearly

(a) 2

(b) 4

(c) 6

(d) 8

(e) 10

The tension in the string will be maximum when the bob is at the bottom point of the vertical circle. This can be seen by writing the expression for the resultant force on the bob and equating it to the centripetal force which makes the bob move along the circle:

T mg = mrω2, where m is the mass of the bob, T is the tension in the string, r is the radius of the circular path and ω is the angular velocity of revolution of the bob.

This gives ω = [(T mg)/mr]1/2

= [(22 – 0.2×10)/(0.2×1)]1/2 = 10 radian/sec

We have substituted the maximum tension the string can withstand and hence the above angular velocity is the maximum value possible.

You will find some useful posts on rotational motion at AP Physics Resources: AP Physics B and C– Multiple Choice Questions on Circular Motion and Rotation