Question

A thin circular ring of mass ${\text{M}}$ and radius ${\text{R}}$ is rotating about its axis with an angular speed ${\omega _0}$. Two particles each of mass $m$ are now attached at diametrically opposite points. The new angular speed of the ring is $\dfrac{{{\omega _0}{\text{M}}}}{{{\text{M + xm}}}}$. Find ${\text{x}}$.

Answer 1

Hint: We should have such a relation which can relate the initial angular velocity and final angular velocity of the ring, so that we can get the new angular velocity after adding two masses diagrammatically in the term of mass of the ring ${\text{M}}$, initial angular velocity ${\omega _0}$ and masses of the added particles on the ring.

Complete step-by-step solution -
We can see in this problem that when we are not applying any external torque,
So, ${\tau _{ext}} = 0$.
And we know that when there is no external torque acting on the system then the angular momentum will be conserved. That means the angular momentum of the system before adding the two particles and after adding the two particles will not be different.
So to proceed with the problem now we have got the relation as follows,
Initial angular momentum= final angular momentum
${{\text{L}}_{i = }}{{\text{L}}_f}$-----equation (1)
Where ${{\text{L}}_i}$=initial angular momentum
And ${{\text{L}}_f}$=final angular momentum
Now ${{\text{L}}_i} = {{\text{I}}_{ring}}{\omega _0}$
Where ${{\text{I}}_{ring}}$=inertia of the ring=$M{R^2}$
${{\text{L}}_i} = M{R^2}{\omega _0}$ -----equation (2)
And ${\omega _0}$=angular velocity of the ring
Similarly, ${{\text{L}}_i} = {{\text{I}}_{ring + masses}}{\omega _0}$
Where ${{\text{I}}_{ring + masses}} = M{R^2} + (m{R^2} \times 2)$
So, ${{\text{L}}_f} = \{ M{R^2} + (m{R^2} \times 2)\} \times \omega $ -----equation (3)
Now putting the values in equation (1) from equation (2) and equation (3)
$M{R^2}{\omega _0} = {R^2}(M + 2m) \times \omega $
Now ${R^2}$ will cancel each other and we will be having the value of new angular velocity as following
$\omega = \dfrac{{{\omega _0}{\text{M}}}}{{{\text{M + 2m}}}}$
Now on comparing this value with the value given in the problem we will get the value of ${\text{x}}$.
Hence ${\text{x = 2}}$.

Note: If there is no external agent applying force or torque on a system then there will not be any momentum change in the system that means momentum is conserved. Adding masses does not add any external effect of force or torque on the system. Additional masses will show their effects internally not externally.