Question

The M.I. of a disc about its diameter
units. Its M.I. about axis through a point
its rim and in the plane of the disc is?​

Answer 1

Question :

The Moment of Inertia about it's diameter is 2 units. It's MI about axis through a point on it's rim and in the plane of the disc is _______

Answer :

Let 'r' be the radius of the disc.

Given,

Moment of inertia of the disc about PQ,

$\implies$ $\boxed{\sf{I_{PQ} = 2\:units}}$

Moment of inertia of the disc about AB,

$\implies$ $\boxed{\sf{I_{AB} = \frac{1}{2}\:Mr^{2}}}$

We need the MI of the disc about RS or $\sf{I_{RS}}$.

Now, from theorem of perpendicular axes,

$\implies$ $\sf{I_{AB} = 2I_{PQ}}$

$\implies$ $\sf{\frac{1}{2}\:Mr^{2} = 2}$

$\implies$ $\sf{Mr^{2} = 4}$

Now, by theorem of parallel axes,

$\implies$ $\sf{I_{RS} = I_{PQ} + Mr^{2}}$

$\implies$ $\sf{I_{RS} = 2 + 4}$

$\implies$ $\sf{I_{RS} = 6\:units}$

Therefore,

The Moment of Inertia about it's diameter is 2 units. It's MI about axis through a point on it's rim and in the plane of the disc is 6 units.

_____________________

Answered by: Niki Swar, Goa❤️