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Q.

A uniform disc of mass m and radius R is pivoted smoothly at P. If a uniform thin ring of mass m and radius R is welded at the lowest point of the disc, find the period of SHM of the system (disc+ ring)

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a

$2 π \sqrt{\frac{23 R}{2 g}}$

b

$π \sqrt{\frac{23 R}{2 g}}$

c

$2 π \sqrt{\frac{23 R}{g}}$

d

$\frac{π}{2} \sqrt{\frac{23 R}{2 g}}$

answer is A.

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Detailed Solution

The time period of a physical pendulum is
$T = 2 π \sqrt{\frac{I}{M g l}}$
Now, $M = m + m = 2 m$
$l = \frac{m (R) + m (3 R)}{m + m} = 2 R l = (\frac{m R^{2}}{2} + m R^{2}) + (m R^{2} + m {(3 R)}^{2}) = m R^{2} [\frac{1}{2} + 1 + 1 + 9] = \frac{23}{2} m R^{2} N o w, T = 2 π \sqrt{\frac{\frac{23}{2} m R^{2}}{(2 m) g (2 R)}} = π \sqrt{\frac{23 R}{2 g}}$

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