Question

Find the moment of inertia about the given axis. Two solid spheres are arranged as shown. Given: \[ m_1 = 10\,\text{kg},\; R_1 = 20\,\text{cm}; \qquad m_2 = 5\,\text{kg},\; R_2 = 10\,\text{cm} \]

• A. \(0.63\,\text{kg m}^2\)
• B. \(0.61\,\text{kg m}^2\)
• C. \(0.62\,\text{kg m}^2\)
• D. \(0.60\,\text{kg m}^2\)

Hint:

For composite rigid bodies:
Always calculate moment of inertia of each part separately
Use parallel axis theorem when axis does not pass through centre
Convert all quantities to SI units before substitution

Answer 1

The Correct Option is A

Concept:
The moment of inertia of a rigid body about a given axis can be found using:
Moment of inertia of a solid sphere about its centre: \[ I_{\text{cm}} = \frac{2}{5}MR^2 \]
Parallel axis theorem: \[ I = I_{\text{cm}} + Md^2 \] Here, the axis of rotation is vertical and passes tangentially between the two spheres. Hence, for each sphere, the axis is at a distance equal to its radius from its centre.
Step 1: Moment of inertia of sphere \(m_1\). Convert radius to metres: \[ R_1 = 20\,\text{cm} = 0.20\,\text{m} \] Using parallel axis theorem: \[ I_1 = \frac{2}{5}m_1 R_1^2 + m_1 R_1^2 \] \[ I_1 = \left(\frac{2}{5}+1\right)10(0.20)^2 \] \[ I_1 = \frac{7}{5}\times 10 \times 0.04 = 0.56\,\text{kg m}^2 \]
Step 2: Moment of inertia of sphere \(m_2\). Convert radius to metres: \[ R_2 = 10\,\text{cm} = 0.10\,\text{m} \] \[ I_2 = \frac{2}{5}m_2 R_2^2 + m_2 R_2^2 \] \[ I_2 = \left(\frac{2}{5}+1\right)5(0.10)^2 \] \[ I_2 = \frac{7}{5}\times 5 \times 0.01 = 0.07\,\text{kg m}^2 \]
Step 3: Total moment of inertia. \[ I_{\text{total}} = I_1 + I_2 = 0.56 + 0.07 = 0.63\,\text{kg m}^2 \] \[ \boxed{I = 0.63\,\text{kg m}^2} \]