Answer to Question #156136 in Quantum Mechanics for Anand

Question

A disc rotates with a period of 0.50s. Its moment of inertia about its axis of rotation is 0.08 Kg m²

. A small mass is dropped onto the disc and rotates with it. The moment of 

inertia of the mass about the axis of rotation is 0.02 kg m²

. Determine the final period 

of the rotating disc and mass system.

Answer 1

Let's apply the law of conservation of angular momentum:

$$L_i=L_f,$$ $$I_{disc}\omega_{disc,i}+I_{mass}\omega_{mass,i}=(I_{disc}+I_{mass})\omega_{f}$$

Since, initially small mass is at rest, we get:

$$I_{disc}\omega_{disc,i}=(I_{disc}+I_{mass})\omega_{f}.$$

Taking into account that $\omega=\dfrac{2\pi}{T},$ we get:

$$I_{disc}\dfrac{2\pi}{T_{disc,i}}=(I_{disc}+I_{mass})\dfrac{2\pi}{T_{disc, f}},$$ $$T_{disc,f}=(\dfrac{I_{disc}}{T_{disc,i}(I_{disc}+I_{mass})})^{-1},$$ $$T_{disc,f}=(\dfrac{0.08\ kg\cdot m^2}{0.5\ s\cdot(0.08\ kg\cdot m^2+0.02\ kg\cdot m^2)})^{-1}=0.625\ s.$$

Answer:

$T_{disc,f}=0.625\ s.$