From a disc of radius R and mass M , a circular hole of diameter R...

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? From a disc of radius R and mass M , a circular hole of diameter R , whose rim passes through the centre is cut. What

From a disc of radius R and mass M , a circular hole of diameter R , whose rim passes through the centre is cut. What

Options

A $15\dfrac{M{R}^{2}}{32}$
B $13\dfrac{M{R}^{2}}{32}$
C $11\dfrac{M{R}^{2}}{32}$
D $9\dfrac{M{R}^{2}}{32}$

Correct answer

B. $13\dfrac{M{R}^{2}}{32}$

Step-by-step solution

$I = \dfrac{M{R}^{2}}{2} - \dfrac{3\sigma}{2}\pi{\left(\dfrac{R}{2}\right)}^{2}{\left(\dfrac{R}{2}\right)}^{2}$

Where $\sigma = \dfrac{M}{\pi{R}^{2}}$

From a disc of radius R and mass M , a circular hole of diameter R , whose rim passes through the centre is cut. What

$I = \dfrac{M{R}^{2}}{2} - \dfrac{3}{32}M{R}^{2}$

$I = \dfrac{13}{32}M{R}^{2}$

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