Question

moment of inertia of a ring of mass m and radius r about any diameter is mr^2/2.what is its moment of inertia about any tangent of the ring?? ​

Answer 1

I= Icm+Mh^2

= MR^2/2+MR^2

=3MR^2/2

Answer 2

The moment of inertia about any tangent of the ring is $I=\dfrac{3mr^2}{2}$.

Explanation:

The moment of inertia of a ring of mass m and radius r about any diameter is, $I_{cm}=\dfrac{mr^2}{2}$

We need to find the moment of inertia about any tangent of the ring. It can be calculated using the theorem of parallel axis. Here,

$I=\dfrac{mr^2}{2}+mr^2$

$I=\dfrac{3mr^2}{2}$

So, the moment of inertia about any tangent of the ring is $I=\dfrac{3mr^2}{2}$. Hence, this is the required solution.

Learn more,

Theorem of parallel axis

https://brainly.in/question/5719860