Question

Small bead of mass 'm' moving with velocity 'v' gets threaded on a stationary semicircular ring of mass 'm' and radius 'r' kept on a horizontal table. The ring can freely rotate about its center o. The bead comes to rest relative to the ring. What would be the final angular velocity of the system?

Answer 1

The final angular velocity of the system will be $\frac{v}{2r}$

Explanation :

Given data,

mass of bead = m

mass of ring = m

velocity of bead = v

radius of ring = r

Let the angular velocity of the system be ω

using the conservation of angular momentum about point o we get,

mvr = ω(mr² + mr²)

=> mvr = 2mωr²

=> v = 2ωr

=> ω = $\frac{v}{2r}$

Hence the final angular velocity of the system will be $\frac{v}{2r}$