Question 409: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

$$\begin{aligned}
& \mathrm{I}=\left(\frac{2}{5} \mathrm{mR}^2+\mathrm{md}^2ight) 	imes 2 \
& \mathrm{I}=2\left(\frac{2}{5} 	imes 2 	imes\left(\frac{1}{2}ight)^2+2 	imes\left(\frac{3}{4}ight)^2ight)=\frac{53}{20} \mathrm{~kg}-\mathrm{m}^2 \
& \mathrm{X}=53
\end{aligned}$$

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type