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Q.

A solid sphere of mass 500 g and radius 5 cm is rotated about one of its diameter with angular speed of $10 rad s^{- 1}$ . If the moment of inertia of the sphere about its tangent is $x \times 10^{- 2}$ times its angular momentum about the diameter. Then the value of x will be ……..

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answer is 35.

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Detailed Solution

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$I_{tangent} = \frac{7}{5} {MR}^{2}$

Angular momentum about diameter:

$L = (\frac{2}{5} {MR}^{2}) ω$

$\frac{I_{tangent}}{L} = \frac{\frac{7}{5} {MR}^{2}}{\frac{2}{5} {MR}^{2} ω} = \frac{7}{2 ω}$

$I_{Tangent} = \frac{7 L}{2 ω} = (35 \times 10^{- 2}) L$

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A solid sphere of mass 500 g and radius 5 cm is rotated about one of its diameter with angular speed of 10 rad s-1. If the moment of inertia of the sphere about its tangent is x×10−2 times its angular momentum about the diameter. Then the value of x will be ……..