Question

from a disc of radius r and mass m a circular hole of radius r/4 is removed from a distance 3r/4 from centre. the moment of inertia of remaining part of disc about an axis perpendicular to plane and passing through centre of original disc is​

Answer 1

Explanation:

I

Total disc

=

2

MR

2

As mass is proportional to area, M

Removed

=

4

M

Now, about the same perpendicular axis:

I

Removed

=

4

M

2

(R/2)

2

+

4

M

(

2

R

)

2

=

32

3MR

2

⇒I

Remaining Disc

=I

Total

−I

Removed

=

2

MR

2

−

32

3MR

2

=

32

13MR

2

Answer 2

Answer:

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