JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Rotational Mechanics

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Two identical solid spheres each of mass 2 kg and radii 10 cm are fixed at the ends of a light rod. The separation between the centres of the spheres is 40 cm. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is

____

\times 1 0^{- 3} kg - m^{2}

Answer: 176

Solution

Using the parallel axis theorem, the moment of inertia of each sphere about the axis is

I = I_{c m} + m d^{2} = \frac{2}{5} m r^{2} + m d^{2}

, where

d = 0.20

m is the distance from the center to each sphere. For two spheres:

I_{t o t a l} = 2 (\frac{2}{5} \times 2 \times (0.1)^{2} + 2 \times (0.20)^{2}) = 2 (0.008 + 0.08) = 2 \times 0.088 = 0.176

kg·m²

= 176 \times 1 0^{- 3}

kg·m².

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Integer answer

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