Question

From a disc of radius r and mass m a circular hole of diameter r whose rim passes through the centre is cut what is i of the remaining part of the disc about a perpendicular axis passing through the centre

Answer 1

given, mass of complete disc is m

so, moment of inertia of disc, $I_{disc}$ = mr²/2

mass of removed part from the disc, m' = m/(πr²) × π(r/2)²

= m/πr² × πr²/4

= m/4

now, moment of inertia of removed part from the disc about an axis passing through its centre , I = m'(r/2)²/2

= (m/4)(r/2)²/2 = mr²/32

now, moment of inertia of removed part about same perpendicular axis, $I_{removed}$ = I + m(r/2)²

= mr²/32 + mr²/16 = 3mr²/32

now, $I_{remaining}=I_{disc}-I_{removed}$

= mr²/2 - 3mr²/32

= 13mr²/32

Answer 2

Answer:

Hope this helps

Explanation: