Rotational Motion

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Question 33 of 280

A solid sphere of mass 500g500g radius 5cm5cm is rotated about one of its diameter with angular speed of 10rads−110rad{s}^{-1}. If the moment of inertia of the sphere about its tangent is x×10−2x\times {10}^{-2} times its angular momentum about the diameter. Then the value of xx will be

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Answer: 35

The angular momentum about the diameter is
Ldiameter=25MR2ω;{L}_{\text{diameter}}=\dfrac{2}{5}M{R}^{2}\omega ;
The moment of inertia of a sphere is ICM=25MR2{I}_{CM}=\dfrac{2}{5}M{R}^{2}. The moment of inertia about the tangent is
Itangent=ICM+MR2=25MR2+MR2=75MR2{I}_{\text{tangent}}={I}_{CM}+M{R}^{2}=\dfrac{2}{5}M{R}^{2}+M{R}^{2}=\dfrac{7}{5}M{R}^{2}.
It is given that ω=10rads−1\omega =10rad{s}^{-1}
Using the given relation between the moment of inertia and the angular momentum,
75MR2=x×10−2×25MR2ω⇒x=3.5×102ω=3.5×10=35\dfrac{7}{5}M{R}^{2}=x\times {10}^{-2}\times \dfrac{2}{5}M{R}^{2}\omega \Rightarrow x=\dfrac{3.5\times {10}^{2}}{\omega }=3.5\times 10=35

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