Question

Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be $\frac{MR^2}{4}$, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Answer 1

Given, moment of inertia of a disc of mass M and radius R about any of its diameters = $\frac{MR^2}{4}$

we have to find moment of inertia about an axis normal to the disc and passing through a point on its edge.

first , find M.I about an axis normal to the plane and passing through its centre.

From perpendicular axis theorem,
M.I about centre of disc , I = 2 × moment of inertia of disc about diameter.
= 2 × MR²/4 = MR²/2

Now, use parallel axis theorem,
Moment of inertia about an axis normal to the plane and passing through its centre = MR² + MR²/2
= 3/2 MR²

Answer 2

Answer:

Given, moment of inertia of a disc of mass M and radius R about any of its diameters = \frac{MR^2}{4}

4

MR

2

we have to find moment of inertia about an axis normal to the disc and passing through a point on its edge.

first , find M.I about an axis normal to the plane and passing through its centre.

From perpendicular axis theorem,

M.I about centre of disc , I = 2 × moment of inertia of disc about diameter.

= 2 × MR²/4 = MR²/2

Now, use parallel axis theorem,

Moment of inertia about an axis normal to the plane and passing through its centre = MR² + MR²/2

= 3/2 MR²