First slide

Moment of intertia

Question

A disc of mass ‘m’ and radius R has a concentric hole of radius ‘r’. Its M.I. about an axis through centre and normal to its plane is

Difficult

A

$\large \frac{m}{2}{(R- r)^2}$
B

$\large \frac{m}{2}{(R^2 - r^2)}$
C

$\large \frac{m}{2}{(R+ r)^2}$
D

$\large \frac{m}{2}{(R^2 + r^2)}$

Solution

Mass of ring of radius x and thickness dx } = dm
$\large dm = \frac{m}{{\pi \left( {{R^2} - {r^2}} \right)}}2\pi xdx = \frac{{2m}}{{{R^2} - {r^2}}}xdx$
MI of ring (DI) = dmx²of radius 'x'
$\large = \left( {\frac{{2mxdx}}{{{R^2} - {r^2}}}} \right){x^2}$
MI of disc = $\large \left( I \right) = \int\limits_r^R {dI} = \frac{{2m}}{{{R^2} - {r^2}}}\left| {\frac{{{X^4}}}{4}} \right|_r^R$
$\large = \frac{m}{{2\left( {{R^2} - {r^2}} \right)}}\left[ {{R^4} - {r^4}} \right]$
$\large I = \frac{m}{2}\left[ {{R^2} + {r^2}} \right]$

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