Answer to Question #253048 in Mechanics | Relativity for Mimi

Question

A horizontal disc of radius 9.00 cm rotates about an axis which passes vertically through the center of the disc with angular velocity of 100 r.p.m. A small particle of mass 10g is dropped onto the disc and sticks to the edge of the disc. If the angular velocity is reduced to 90 r.p.m, determine the moment of inertia of the disc about the axis.

Answer 1

We can find the moment of inertia of the disc from the law of conservation of angular momentum:

$$L_i=L_f,$$ $$I_{disc}\omega_i=(I_{disc}+I_{particle})\omega_f,$$ $$I_{disc}\omega_i=(I_{disc}+mr^2)\omega_f,$$ $$I_{disc}=\dfrac{mr^2\omega_f}{\omega_i-\omega_f},$$ $$I_{disc}=\dfrac{0.01\ kg\times(0.09\ m)^2\times90\ rpm}{100\ rpm-90\ rpm}=7.29\times10^{-4}\ kg\times m^2.$$