System of Particles and Rotational Motion - Result Question 53

57. 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?

[2016]

(a) $15 MR^{2} / 32$

(c) $11 MR^{2} / 32$

(b) $13 MR^{2} / 32$

(d) $9 MR^{2} / 32$

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Answer:

Correct Answer: 57. (b)

Solution:

  1. (b) Moment of inertia of complete disc about point ’ $O$ ‘.

$I _{\text{Total disc }}=\frac{MR^{2}}{2}$

Mass of removed disc

$M _{\text{Removed }}=\frac{M}{4}$ (Mass $\propto$ area $)$

Moment of inertia of removed disc about point ‘O’.

$I _{\text{Removed }}$ (about same perpendicular axis)

$=I _{cm}+mx^{2}$

$=\frac{M}{4} \frac{(R / 2)^{2}}{2}+\frac{M}{4}(\frac{R}{2})^{2}=\frac{3 MR^{2}}{32}$

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

$I _{\text{Remaing disc }}=I _{\text{Total }}-I _{\text{Removed }}$

$=\frac{MR^{2}}{2}-\frac{3}{32} MR^{2}=\frac{13}{32} MR^{2}$

Moment of inertia of a part of a rigid body (Symmetrically cut from the whose mass) is the same as that of the whole body when whole mass is replaced by mass of that part.