First slide

Moment of intertia

Question

The moment of inertia of sphere is 20 kg- $m^{2}$ about the diameter. The moment of inertia about any tangent will be-

Moderate

A

$70 kg - m^{2}$
B

$35 kg - m^{2} 35 kg - m^{2}$
C

$50 kg - m^{2}$
D

$20 kg - m^{2}$

Solution

According to the theorem of parallel axes, the have
$I = I_{G} + {Ma}^{2} = \frac{2}{5} {MR}^{2} + {MR}^{2} (∵ a = R) = \frac{7}{5} {MR}^{2} Given that \frac{2}{5} {MR}^{2} = 20 {or MR}^{2} = \frac{20 \times 5}{2} = 50, ∴ I = \frac{7}{5} \times 50 = 70 kg - m^{2}$

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