Question

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become
(a) ωMM+m
(b) ωMM+2 m
(c) ω(M-2 m)M+2 m
(d) ω(M+2 m)M.

Answer 1

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become $\omega M M+2 m$

Explanation:

No existing torque is applied to the circle; thus the angular momentum is preserved.

Step 1:

$I \omega$ is equal to $I^{\prime} \omega^{\prime}$

So that  

$I \omega=I^{\prime} \omega^{\prime}$

$\omega^{\prime}=\frac{I \omega}{I^{\prime}} \ldots \ldots e q^{n}(i)$

Step 2:

We know that  

$I=M r^{2}$

and

$I^{\prime}=M r^{2}+2 m r^{2}$

Step 3:

When these values are set in equation I we get:  

$\omega^{\prime}=\frac{\omega M}{(M+2 m)}$

angular momentum:- The quantity of body rotation produced by its inertia moment and angular velocity.