Question 426: Centre of Mass and Rotational Motion

Question

Two identical solid spheres each of mass $$2 \mathrm{~kg}$$ and radii $$10 \mathrm{~cm}$$ are fixed at the ends of a light rod. The separation between the centres of the spheres is $$40 \mathrm{~cm}$$. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is __________ $$	imes 10^{-3} \mathrm{~kg}~\mathrm{m}^{2}$$

Accepted Answer

The problem requires calculating the moment of inertia of the system consisting of two identical solid spheres fixed at the ends of a light rod. We need to find the moment of inertia about an axis perpendicular to the rod and passing through its midpoint.
First, letÃ¢ÂÂs identify the moment of inertia of each solid sphere about its own center, which is given by the formula:
$$I_{	ext{sphere}} = \frac{2}{5} m r^2$$
Where:
$$m$$ is the mass of the sphere = $$2 \mathrm{~kg}$$
$$r$$ is the radius of the sphere = $$0.1 \mathrm{~m}$$
Substituting the values:
$$I_{	ext{sphere}} = \frac{2}{5} 	imes 2 \mathrm{~kg} 	imes (0.1 \mathrm{~m})^2 = \frac{4}{5} 	imes 0.01 \mathrm{~kg}~\mathrm{m}^2 = 0.008 \mathrm{~kg}~\mathrm{m}^2$$
Now, we need the moment of inertia of the two spheres about the axis passing through the midpoint of the rod. This requires using the parallel axis theorem, which states:
$$I_{	ext{total}} = I_{	ext{sphere}} + m d^2$$
Where:
$$d$$ is the distance from the center of the sphere to the axis through the rodÃ¢ÂÂs midpoint = $$0.2 \mathrm{~m}$$
Calculating the additional inertia due to the parallel axis theorem for one sphere:
$$I_{	ext{parallel}} = m d^2 = 2 \mathrm{~kg} 	imes (0.2 \mathrm{~m})^2 = 2 \mathrm{~kg} 	imes 0.04 \mathrm{~m}^2 = 0.08 \mathrm{~kg}~\mathrm{m}^2$$
The total moment of inertia for one sphere about the midpoint of the rod is:
$$I_{	ext{one sphere, total}} = I_{	ext{sphere}} + I_{	ext{parallel}} = 0.008 \mathrm{~kg}~\mathrm{m}^2 + 0.08 \mathrm{~kg}~\mathrm{m}^2 = 0.088 \mathrm{~kg}~\mathrm{m}^2$$
Since there are two identical spheres, the total moment of inertia of the system is:
$$I_{	ext{system}} = 2 	imes 0.088 \mathrm{~kg}~\mathrm{m}^2 = 0.176 \mathrm{~kg}~\mathrm{m}^2$$
Converting the result to the given form:
$$0.176 \mathrm{~kg}~\mathrm{m}^2 = 176 	imes 10^{-3} \mathrm{~kg}~\mathrm{m}^2$$
So, the moment of inertia of the system about the given axis is:
$$176 	imes 10^{-3} \mathrm{~kg}~\mathrm{m}^2$$

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type