Question

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

Answer 1

In this question we are asked to calculate the angular velocity of the ring after the two objects with mass M are attached to it. To solve this question, we will use the concept of conservation of angular momentum. Therefore, we will calculate the angular momentum of the ring at the given two instances and equate it with the initial angular momentum. Thus, we can calculate the new angular velocity of the ring.

Complete step by step answer:
It is said that the two objects with mass M are gently placed on the ring. It means that no external torque will be applied on the ring due to this. Now, we know that this is the condition for conservation of angular momentum.
Therefore, we can say that
${{L}_{i}}={{L}_{f}}$ …………….. (1)
Now, the initial angular momentum ${{L}_{i}}$ will be given for just the mass of the ring as initially the objects are not added.
Therefore,
We know,
${{L}_{i}}=m{{r}^{2}}\omega$ …………. (2)
Now, for the final angular momentum
We know that,
${{L}_{f}}=({{I}_{i}}+{{I}_{f}}){{\omega }_{n}}$ ……………. (3)
Where, ${{I}_{i}}$ and ${{I}_{f}}$ is the moment of inertia before two objects are attached and after the two objects are attached. Also, ${{\omega }_{n}}$ is the new angular velocity.
We know,
${{I}_{i}}=m{{r}^{2}}$
Also,
${{I}_{f}}=2M{{r}^{2}}$
Now substituting the above values in equation (3)
We get,
${{L}_{f}}=(m{{r}^{2}}+2M{{r}^{2}}){{\omega }_{n}}$ ……….. (4)
Now, from (1), (2) and (4)
We can say that,
$m{{r}^{2}}\omega =(m{{r}^{2}}+2M{{r}^{2}}){{\omega }_{n}}$
On solving
We get,
${{\omega }_{n}}=\dfrac{m\omega }{m+2M}$

So, the correct answer is “Option A”.

Note: Just as linear momentum is conserved when there is no external force applied, the angular momentum is conserved when there is no external torque on the object. For an object in a circle of radius r, the angular momentum depends on the mass of the object, angular speed and the radius of the circle. Newton’s first law tells us that if there is no external torque the angular momentum will always be conserved.