Question

The M.I. of a ring of mass 500 gm and radius 0.5 m about a tangent perpendicular to its plane is-------------- ​

Answer 1

Moment Inertia OF RING:

• Now, for the axis which is tangential and perpendicular to ring , we need to apply PARALLEL AXES THEOREM.

$\rm MI = m {r}^{2} + m {d}^{2}$

• Here d = radius (r)

$\rm \implies MI = m {r}^{2} + m {r}^{2}$

$\rm \implies MI = 2m {r}^{2}$

$\rm \implies MI = 2 \times \dfrac{500}{1000} \times {(0.5)}^{2}$

$\rm \implies MI = 2 \times 0.5 \times 0.25$

$\rm \implies MI = 0.25 \: kg {m}^{2}$

So, Moment of Inertia along given axis is 0.25 kgm².