Question

A horizontal disc is rotating about a transverse axis through its centre at 100 rpm. A 20 gram blob of wax falls on the disc and sticks to it at 5 cm from its axis. The moment of inertia of the disc about its axis passing through its centre is 2 × 10^-4 kg.m² . Calculate the new frequency of rotation of the disc.

Answer 1

Data : f₁ = 100 rpm, m = 20 g = 20 × 10^-3 kg, 

r = 5 cm = 5 × 10^-2 m, 

I₁ = I_disc = 2 × 10^-4 kg.m²

The MI of the disc and blob of wax is

I₂ = I₁ + mr²

\

By the principle of conversation of angular momentum, I₁\(\omega_1\) = I₂\(\omega_2\).

This is the new frequency of rotation.