Question

A small hole is made in a disc (mass M, radius R) at a distance R/4 from centre.
The disc is supported on horizontal peg through this hole. The moment of inertia
of disc about horizontal peg is :
(a) 1/2MR^2
(b) 5/16MR^2
(c)9/16MR^2
(d)5/4MR^2​

Answer 1

Answer:

Moment of Inertia of disc about its center is

2

1

MR

2

Therefore by parallel axis theorem,

I=I

CM

+Md

2

=

2

1

MR

2

+M(

4

R

)

2

=

2

1

MR

2

+

16

1

MR

2

=

16

9

MR

2

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