Question

A horizontal disc is freely rotating about a transverse axis passing through its centre at the rate of 100 revolutions per minute. A 20 gram blob of wax falls on the disc and sticks to the disc at a distance of 5 cm from its axis. Moment of intertia of the disc about its axis passing through its centre of mass is 2 x 10^-4 kg m². Calculate the new frequency of rotation of the disc. 

Answer 1

I₁ = Moment of inertia of disc = 2 × 10⁻⁴ kgm²

I₂ = moment of inertia of the disc + moment of inertia of the bob of wax on the disc

= 2 × 10⁻⁴ + mr² = 2 × 10⁻⁴ + 20 × 10⁻³ (0.05)²

= 2 × 10⁻⁴ + 0.5 × 10⁻⁴ = 2.5 × 10⁻⁴ kgm²

By the principle of conservation of angular momentum

I₁n₁ = I₂n₂ ⇒ 2 × 10⁻⁴ × 100 = 2.5 × 10⁻⁴ n₂

n₂ = `(100 xx 2)/2.5` = 80r ±