Rotational Motion Question 10
Question 10 - 2024 (31 Jan Shift 2)
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 $\frac{x}{20} kg m^{2}$, where the value of $x$ is
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Answer: (53)
Solution:
$I=\left(\frac{2}{5} mR^{2}+md^{2}\right) \times 2$
$I=2\left(\frac{2}{5} \times 2 \times\left(\frac{1}{2}\right)^{2}+2 \times\left(\frac{3}{4}\right)^{2}\right)=\frac{53}{20} kg-m^{2}$
$X=53$
