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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Easy2 minPYQ_2020

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From a circular disc of radius $R$ and mass $9 M$ , a small disc of radius $\frac{R}{3}$ is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through $O$ is

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Answer:

Solution:

$I_{r e m a i n i n g} = I_{w h o l e} - I_{r e m o v e d}$

or $I = \frac{1}{2} (9 M) (R^{2}) - [\frac{1}{2} m {(\frac{R}{3})}^{2} + \frac{1}{2} m {(\frac{2 R}{3})}^{2}]$ ...(i)

Here, $m = \frac{9 M}{π R^{2}} \times π {(\frac{R}{3})}^{2} = M$

Substituting in Eq. (i), we have

$I = 4 M R^{2}$

Subject:PhysicsTopic:RotationSubtopic:Calculation of M.I.

⏱ 2mℹ️ Source: PYQ_2020

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