The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is n

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Solution

Q.
The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is n times higher than the moment of inertia of the given ring. Here, n = __________.

Consider all the bodies have equal masses. [2025]

Ans.
(4)

$I_{1} = \frac{M R_{1}^{2}}{4}, I_{2} = \frac{M R_{2}^{2}}{2}, I_{3} = \frac{2 M R_{1}^{2}}{5}$

According to problem

$\frac{I_{1}}{I_{2}} = 2.5 \Rightarrow \frac{\frac{M R_{1}^{2}}{4}}{\frac{M R_{2}^{2}}{2}} = \frac{5}{2} \Rightarrow \frac{R_{1}^{2}}{R_{2}^{2}} = 5$ ... (i)

Now we are provided with information that

$\frac{I_{3}}{I_{2}} = n \Rightarrow \frac{\frac{2 M R_{1}^{2}}{5}}{\frac{M R_{2}^{2}}{2}} = n \Rightarrow \frac{4 R_{1}^{2}}{5 R_{2}^{2}} = n$ ... (ii)

From Eq. (i) and (ii) $\Rightarrow$ n = 4

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