First slide

Moment of inertia

Question

A circular portion of diameter R is cut out from a uniform circular disc of mass M and radius R as shown in Fig. The moment of inertia of the remaining (shaded) portion of the disc about an axis passing through the centre O of the disc and perpendicular to its plane is

Moderate

A

$\frac{15}{32} {MR}^{2}$
B

$\frac{7}{16} {MR}^{2}$
C

$\frac{13}{32} {MR}^{2}$
D

$\frac{3}{8} {MR}^{2}$

Solution

Moment of inertia of complete disc about O is $I = \frac{1}{2} {MR}^{2}$ . Mass of the cut-out part is $m = (\frac{M}{4})$ .
The moment of inertia of the cut-out portion about its own centre $I_{0} = \frac{1}{2} {mr}^{2} = \frac{1}{2} (\frac{M}{4}) {(\frac{R}{2})}^{2} = \frac{1}{32} {MR}^{2}$ because r = R/2. From the parallel axes theorem, the moment of inertia of the cut out portion about O is
$\begin{array}{l} I_{c} = I_{0} + {mr}^{2} = \frac{1}{32} {MR}^{2} + (\frac{M}{4}) {(\frac{R}{2})}^{2} \\ = \frac{3}{32} {MR}^{2} \end{array}$
$∴$ Moment of inertia of the shaded portion about O is $I_{s} = I - I_{c} = \frac{1}{2} {MR}^{2} - \frac{3}{32} {MR}^{2} = \frac{13}{32} {MR}^{2}$ .

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App