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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Easy2 minPYQ_2020

PhysicsEasynumerical

Four solid spheres each of diameter $\sqrt{5} cm$ and mass $0.5 kg$ are placed with their centres at the corners of a square of side $4 cm$ . The moment of inertia of the system about the diagonal of the square is $N \times 10^{- 4} kg m^{2}$ , then $N$ is

Answer:

9.00

Solution:

$r = \frac{d}{2} = \frac{\sqrt{5}}{2} cm = \frac{\sqrt{5}}{2} \times 10^{- 2} m$

$m = 0.5 kg$

$a = 4 cm = 4 \times 10^{- 2} m$

$I_{XX} = I_{1} + I_{2} + I_{3} + I_{4}$

$= \frac{2}{5} {mr}^{2} + (m {(\frac{a}{\sqrt{2}})}^{2} + \frac{2}{5} {mr}^{2}) + \frac{2}{5} {mr}^{2} + (m {(\frac{a}{\sqrt{2}})}^{2} + \frac{2}{5} {mr}^{2})$

Substituting the values, we get,

$I_{XX} = 9 \times 10^{- 4} kg m^{2}$

Therefore, $N = 9$

Subject:PhysicsTopic:RotationSubtopic:Calculation of M.I.

⏱ 2mℹ️ Source: PYQ_2020

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