Question

A disk of mass 3.0 kg and radius 65 cm with a small mass of 0.07 kg attached at the edge is rotating at 2.2 rev/s. The small mass suddenly flies off of the disk. What is the disk's final rotation rate (in rev/s)?

Answer 1

Answer:

$\omega_2=2.304\ rev/s$

Explanation:

Given that

Mass of disc,M=3 kg

radius r= 65 cm

Mass of small mass ,m=0.07 kg

Initial speed = 2.2 rev/s

If external torque is zero then angular momentum of system will remain conserve.

Moment of inertia of disc at initial condition

$I_1=\dfrac{Mr^2}{2}+mr^2$

$I_1=\dfrac{3\times 0.65^2}{2}+0.07\times 0.65^2$

$I_1=0.66\ kg.m^2$

Moment of inertia of disc at final condition

$I_2=\dfrac{Mr^2}{2}$

$I_2=\dfrac{3\times 0.65^2}{2}$

$I_2=0.63\ kg.m^2$

So from conservation of angular momentum

$I_1\omega_1=I_2\omega_2$

$0.66\times 2.2=0.63\times \omega_2$

$\omega_2=2.304\ rev/s$

This is final speed of disc after small mass flies off from the disc.