First slide

Moment of intertia

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? [NEET 2016]

Moderate

A

$13 M R^{2} / 32$
B

$11 M R^{2} / 32$
C

$9 M R^{2} / 32$
D

$15 M R^{2} / 32$

Solution

Considering the information given in the question, let us draw the figure.
If the above figure is considered, then moment of inertia of disc will be given as
$I = I_{remaining} + I_{(R / 2)}$
$\Rightarrow I_{remaining} = I - I_{(R / 2)}$
Putting the values, we get
$= \frac{M R^{2}}{2} - [\frac{\frac{M}{4} {(\frac{R}{2})}^{2}}{2} + \frac{M}{4} {(\frac{R}{2})}^{2}]$
$= \frac{M R^{2}}{2} - [\frac{M R^{2}}{32} + \frac{M R^{2}}{16}] = \frac{M R^{2}}{2} - [\frac{M R^{2} + 2 M R^{2}}{32}]$
$= \frac{M R^{2}}{2} - \frac{3 M R^{2}}{32} = \frac{16 M R^{2} - 3 M R^{2}}{32}$
$I_{remaining} = \frac{13 M R^{2}}{32}$

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