Question

A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity ɷ. Two objects, each of mass m, are attached gently to the opposite ends of the diameter of the ring. The wheel now rotates with an angular velocity-
A. $\dfrac{\omega M}{M+m}$
B. $\dfrac{\omega (M-2m)}{M+2m}$
C. $\dfrac{\omega M}{M+2m}$
D. $\dfrac{\omega (M+2m)}{M}$

Answer 1

Hint:
If there is no external torque given to the system, then the angular momentum should remain the same. Calculate the angular momentum before and after attaching the objects. Equate those two expressions to find the result.

Formula Used:
Moment of Inertia (I) of a ring with respect to its axis is given by,
${{I}_{r}}=M{{R}^{2}}$
Where,
$M$ is the mass of the ring,
$R$ is the radius of the ring.

Moment of Inertia (I) of a point object is given by,
${{I}_{o}}=M{{R}^{2}}$
Where,
$M$ is the mass of the object,
$R$ is the distance of the object from the axis.
Angular momentum is given by,
$L=I\omega$
Where,
$I$ is the moment of inertia,
$\omega$ is the angular velocity.

Complete step-by-step answer:
The moment of inertia of the ring is,
${{I}_{r}}=M{{r}^{2}}$
Given, the angular velocity of the ring is ɷ.

So, the initial angular momentum is, ${{L}_{1}}=M{{r}^{2}}\omega$

We can consider the ring and objects separately to calculate the total angular momentum.

Let’s assume that the new angular velocity is,
${{\omega }^{'}}$

So, the angular momentum of the ring is,
$M{{r}^{2}}{{\omega }^{'}}$

The moment of inertia of the two point objects is, $m{{r}^{2}}$
Hence, the angular momentum of those two objects is, $2m{{r}^{2}}{{\omega }^{'}}$

As all the motion is happening around the same axis, we should not worry about the direction of angular momentum. We can simply add the two angular momentums.

Hence, the final angular momentum is,
${{L}_{2}}=M{{r}^{2}}{{\omega }^{'}}+2m{{r}^{2}}{{\omega }^{'}}$

Hence, we can write,
${{L}_{1}}={{L}_{2}}$ $\Rightarrow M{{r}^{2}}\omega =M{{r}^{2}}{{\omega }^{'}}+2m{{r}^{2}}{{\omega }^{'}}$
$\Rightarrow M{{r}^{2}}\omega =(M+2m){{r}^{2}}{{\omega }^{'}}$
$\Rightarrow \dfrac{M\omega }{(M+2m)}={{\omega }^{'}}$

So, the new angular velocity is, ${{\omega }^{'}}=\dfrac{M\omega }{(M+2m)}$

The correct option is- (C).

Note:
Angular momentum is a vector quantity like linear momentum. It depends on the angular velocity of the rotating object. In this question, we did not worry about the direction, because there is only one axis of rotation. If a problem has more than one axis of rotation, you need to apply the conservation of angular momentum theorem for each axis. You can only apply this technique in the absence of external torque.