Question

15. Two thin uniform circular rings each of radius 10cm and mass 0.1kg are arranged such that they have
a common centre and their planes are perpendicular to each other. The moment of inertal is
system about an axis passing through their common centre and perpendicular to the plane of one of
the rings in kgm2 is:​

Answer 1

Given:

• Radius of ring $(r) = 10 cm$

• Mass $= 0.1kg$

To Find:

• The moment of inertia of the system about an axis passing through their common centre and perpendicular to the plane of either of the ring

Solution:

The moment of inertia of the ring about the diameter is given as,

              $I_d = \frac{1}{2} mr^2$

The moment of inertia of the ring about the axis perpendicular to the plane is given by

               $I_p = mr^ 2$

The moment of inertia of the system about an axis passing through their common center and perpendicular to the plane of either of the ring is given as,

            $I = mr^ 2 + \frac{1}{2} mr^ 2$

Now, putting the values

        $I = \frac{3}{2} (0.1) (0.1)^ 2$

        $I = \frac{3}{2} (0.001)$

        $I = 0.0015$

        $I = 1.5 \times 10^{-3} kg /m$

Thus,  The moment of inertia is  system about an axis passing through their common centre and perpendicular to the plane of one of  the rings

 is $1.5 \times 10^{-3} kg /m$