## Sunday, May 11, 2008

### Kerala Engineering Entrance 2008 Questions on Rotational Motion

Here are the two questions on rotational motion which appeared in KEAM (Engineering) 2008 question paper:

(1) A thin circular ring of mass M and radius R rotates about an axis through the centre and perpendicular to its plane, with a constant angular velocity ω. Four small spheres each of mass m (negligible radius) are kept gently at the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

(a) 4ω

(b) Mω/4m

(c) [(M+4m)/M]ω

(d) [M/(M–4m)]ω

(e) [M/(M+4m)]ω

The angular miomentum of the system is conserved since there is no external torque. Therefore we have I1ω1 = I2ω2 where I1 and I2 are the initial and final moments of inertia and ω1 and ω2 are the initial and final angular velocities of the system.

We have I1= MR2, ω1 = ω and I2 = MR2+4mR2.

Therefore, MR2ω = (M+4m)R2ω2 from which ω2 = [M/(M+4m)]ω

[Note that the four masses can be placed anywhere on the ring and you will get the same answer as above].

(2) The angular velocity of a wheel increases from 100 rps to 300 rps in 10 seconds. The number of revolutions made during that time is

(a) 600

(b) 1500

(c) 1000

(d) 3000

(e) 2000

The angular acceleration (α) of the wheel is given by

α = Change of angular velocity/Time = (300–100)/10 = 20 revolutions per second2.

[We have retained the angular velocity in rps itself since the number of revolutions is to be found out]

Substituting the known values in the equation, ω2 = ω02 + 2αθ (which is similar to v2 = v02 +2as in linear motion), we have

3002 = 1002 + 2×20×θ

[The angular displacement θ will be in revolutions for obvious reasons]

From the above equation, θ = 2000 revolutions.

You can find all posts on rotational motion on this site by clicking on the label ‘rotation’ below this post.

You will find many useful multiple choice questions (with soluton) involving rotational motion at apphysicsresources here as well as here