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Time period of small oscillation- disk and particle system [closed]

1 vote


$y_{COM} = \dfrac{R}{3} = \dfrac {25}3$

$I = \dfrac 12 (2)R^2 + 1R^2 = 2\times 25 = 50 $

Now, T is given by: $T =2\pi \sqrt{\dfrac{I}{mgl}}$ where l is the distance of the centre of mass from the centre of rotation

So $T =2\pi \sqrt{\dfrac{50}{3 \times 10 \times \dfrac{25}{3}}} = 2\pi\sqrt{\dfrac{1}{5}}$

No option :(

closed with the note: Got my mistake
asked Jan 4 in Physics Problems by Reststack (422 points)
closed Jan 6 by Reststack
There is a mistake in your first line.