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