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 the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

विकल्प

• 11 MR²/32

• 9 MR²/32

• 15 MR²/32

• 13 MR²/32

Answer 1

13 MR²/32

Explanation:

The disc's moment of inertia is given by

I_disc = I_r + I_hole  ......(I_r = M.I of remaining part)

∴ I_r = I_disc – I_hole   ......(i)

I_disc = `(MR^2)/2`  ......(ii)

According to the parallel axes theorem,

I_hole = `[(M/4 (R/2)^2)/2 + M/4(R/2)^2]`  .......`((because M_"hole" = (M_"disc")/4),(because "the surface density is same"))`

∴ I_hole = `[(MR^2)/32 + (MR^2)/16]`  .....(iii)

We get by substituting equations (iii) and (iii) in equation (i)

I_r = `(MR^2)/2 - (MR^2)/32 - (MR^2)/16`

= `MR^2 [1/2 - 1/32 - 1/16]`

= `13/32` MR²