Question 644: Centre of Mass and Rotational Motion

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?

Accepted Answer

Moment of inertia of complete disc about
point 'O'.
$${I_{Total\,disc}} = {{M{R^2}} \over 2}$$
Mass of removed disc
$${M_{{\mathop{m Removed}
olimits} }} = {M \over 4}\left( {Mass \propto area} ight)$$
Moment of inertia of removed disc about point
'O'.
I
Removed
(about same perpendicular axis)
$$ = {I_{cm}} + m{x^2}$$
$$ = {M \over 4}{{{{\left( {R/2} ight)}^2}} \over 2} + {M \over 4}{\left( {{R \over 2}} ight)^2} = {{3M{R^2}} \over {32}}$$
Therefore the moment of inertia of the remaining
part of the disc about a perpendicular axis
passing through the centre,
I
Remaing disc
= I
Total
Ã¢ÂÂ I
Removed
$$ = {{M{R^2}} \over 2} - {3 \over {32}}M{R^2} = {{13} \over {32}}M{R^2}$$

Answer

11 MR
2
/32

Answer

9 MR
2
/32

Answer

15 MR
2
/32

Answer

13 MR
2
/32

Centre of Mass and Rotational Motion

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?

Choose the correct answer: