A disc of radius R and mass M is pivoted at the rim and set for small oscillations about an axis perpendicular to plane of disc. If simple pendulum have same time period as of disc, the length of pendulum should be 

 

Answers (1)

T = 2\pi\sqrt{\frac{I}{MgL}} = 2\pi\sqrt{\frac{l}{g} } \\*\Rightarrow l = \frac{I}{ML} = \frac{\frac{3}{2}MR^{2}}{MR} = \frac{3}{2}MR

 

Time period of physical pendulum -

T= 2\pi \sqrt{\frac{I}{mgl}}

- wherein

I= moment of inertia about point of hinge

l= Seperation between point of suspension & centre of mass

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