Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Hard2 minPYQ_2017

PhysicsHardsingle choice

A circular hole of radius $\frac{R}{4}$ is made in a thin uniform disc having mass and radius $R$ , as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point $O$ and perpendicular to the plane of the disc is-

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Answer:

Solution:

$I_{D} = \frac{m r^{2}}{2}$

Moment of inertia of removed portion about the axis.

$I_{r e m o v e d} = \frac{1}{2} (\frac{m}{16}) (\frac{r^{2}}{16}) + (\frac{m}{16}) (\frac{9 r^{2}}{16})$ (by parallel axis theorem)

$= \frac{m r^{2} + 18 m r^{2}}{512}$

$= \frac{19 m r^{2}}{512}$

$I_{r e m a i n i n g} = \frac{m r^{2}}{2} - \frac{19}{512} m r^{2}$

$= \frac{237}{512} m r^{2}$

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Calculation of M.I.

⏱ 2mℹ️ Source: PYQ_2017

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