Question

An insect of mass 20 g crawls from the centre to the outside edge of a rotating disc of

mass 200g and radius 20 cm. The disk was initially rotating at 22.0 rads−1

. What will be

its final angular velocity? What is the change in the kinetic energy of the system?

Answer 1

Conservation of angular momentum principle is applicable here, as there is no external force /torque acting on the system or disc and insect.

m = 20gm = 0.020 kg
M = 200 gm = 0.200 kg,  R = 20cm = 0.20 m,   ω₁ = 22.0 rad/s
Final angular velocity of Disc & ant = ω₂

Moment of Inertia of the Disc: I₁ = M R² /2 = 0.200* 0.20²/2 = 0.004 units
Initial Moment of inertia of insect: I₂₁ = 0  (as it is at the center)
Final MOI of the insect:  I₂₂ = m R² = 0.020 * 0.20² = 0.0008 units

Angular momentum = L = I ω

L initial = I₁ ω₁ + I₂₁ ω₁ = 0.004 * 22.0 = 0.088 units
L final =  I₁ ω₂ + I₂₂ ω₂  = 0.0048 * ω₂  units
=>   ω₂ = 0.088/0.0048 = 55/3 = 18.333 rad/sec

PE remains same.
K.E. initial:    1/2 I₁ ω₁² + 0  (as ant is not rotating initially)
                   = 1/2 * 0.004 * 22.0² = 0.968 Joules

K.E. final :   1/2 I₁ ω₂²  + 1/2 I₂₂ ω₂² 
                 = 1/2 * [ 0.004 + 0.0008 ] (55/3)² Joules
                 = 0.807 Joules

Loss of KE = 0.968 - 0.807 = 0.161 Joules

% change in KE = - 161/968 * 100 = - 16.63%