Question:

Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 cm. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is \(\frac{x}{20}\) kg m3 , where the value of x is ____.

Updated On: Nov 19, 2024
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Correct Answer: 53

Solution and Explanation

The moment of inertia \( I \) of each sphere about the central axis (using the parallel axis theorem) is:

\[ I_{\text{total}} = 2 \left( I_{\text{sphere}} + md^2 \right). \]

For a solid sphere:

\[ I_{\text{sphere}} = \frac{2}{5}mR^2 = \frac{2}{5} \times 2 \times (0.5)^2 = 0.2 \, \text{kg m}^2. \]

Distance \( d \) from the center of each sphere to the midpoint of the rod is \( 0.75 \, \text{m} \).

So,

\[ I_{\text{total}} = 2 \left( 0.2 + 2 \times (0.75)^2 \right) = 2 \left( 0.2 + 1.125 \right) = \frac{53}{20} \, \text{kg m}^2. \]

Thus, \( x = 53 \).

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