A solid sphere of mass 1.) kg and diameter 0.3 m is suspended from a wire.If the twisting couple per unit twist for the wire is 6 x 10^-3 N-m / radian, then the time period of small oscillations will be?

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Solution

Dear Student ,
The mass of the sphere , M = 1 kg and radius of the sphere , r= 0.3 m , the twisting couple per unit twist of the wire is ,
C = $6 \times 10^{- 3} N - m / r a d i a n$
Now we know that the time period of small oscillation is ,
$T = 2 π \sqrt{\frac{I}{C}} w h e r e I = M o m e n t o f i n e r t i a o f t h e s p h e r e N o w t h e m o m e n t o f i n e r t i a o f t h e s o l i d s p h e r e a b o u t i t s d i a m e t e r i s, I = \frac{2}{5} M r^{2} = \frac{2}{5} \times 1 \times {(0 \cdot 3)}^{2} = 0 \cdot 036 K g - m^{2} N o w t h e t i m e p e r i o d f o r s m a l l o s c i l l a t i o n i s, T = 2 π \sqrt{\frac{I}{C}} = 2 π \sqrt{\frac{0 \cdot 036}{6 \times 10^{- 3}}} = 15 \cdot 39 s e c o n d s$
Regards

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