Question

Two rings of the same radius and mass are placed such that their centers are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass of the ring = m and radius = r):
A. $\dfrac{1}{2}m{{r}^{2}}$
B. $m{{r}^{2}}$
C. $\dfrac{3}{2}m{{r}^{2}}$
D. $2m{{r}^{2}}$

Answer 1

Hint: The moment of inertia of a rigid composite system about an axis is the sum of the moment of inertia of its rigid subsystems about the same axis. We can solve this problem by finding the moments of inertia of each ring about the required axis and then adding both the moments of inertia to get the total moment of inertia of the system about the axis.
Complete step by step answer:
One of the additive properties of moment of inertia is that the moment of inertia of a rigid composite system about an axis is the sum of the moment of inertia of its rigid subsystems about the same axis. ---(1)
Here the given composite system is the two rings having the same radius r and mass m, with their centers coinciding and one of the rings placed perpendicular to the other. The axis about which the moment of inertia of the system has to be found is an axis that is perpendicular only to the plane of one ring and passing through its centre. Thus, the axis must lie in the plane of the second ring and pass through the common centers of the two rings.
Let the ring perpendicular to the axis be 1 and that parallel to the axis be 2.
Since, we know that the moment of inertia of a ring about an axis perpendicular to its plane and passing through its center is mr2, where m is mass of ring and r its radius,
Moment of Inertia of the ring 1 about the axis (I1) =$m{{r}^{2}}$ --(2)
Since, we know that the moment of inertia of a ring about an axis lying in its plane and passing through its center is$\dfrac{m{{r}^{2}}}{2}$ , where m is mass of ring and r its radius,
Moment of Inertia of the ring 2 about the axis (I2) =$\dfrac{m{{r}^{2}}}{2}$ --(3)
Now, using the property stated in (1)
Moment of inertia of system (${{I}_{system}}$ ) = ${{I}_{1}}+{{I}_{2}}$
Therefore, now using (2) and (3)
${{I}_{system}}=m{{r}^{2}}+\dfrac{m{{r}^{2}}}{2}=\dfrac{3}{2}m{{r}^{2}}$
Therefore the required moment of inertia of the system is $\dfrac{3}{2}m{{r}^{2}}$ .
So, the correct answer is C) $\dfrac{3}{2}m{{r}^{2}}$ .

Note: While writing the moment of inertia of different bodies in such questions, they must keep in mind the axis about which the moment of inertia has to be found out, since generally, the moment of inertia of a body is different about different axes.