Find the moment of inertia about the given axis.
Two solid spheres are arranged as shown.
Given:
\[
m_1 = 10\,\text{kg},\; R_1 = 20\,\text{cm}; \qquad
m_2 = 5\,\text{kg},\; R_2 = 10\,\text{cm}
\]
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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
Concept: The moment of inertia of a rigid body about a given axis is calculated using the following formulas: **Moment of inertia of a solid sphere about its centre:** \[ I_{\text{cm}} = \frac{2}{5}MR^2 \] **Parallel axis theorem**: \[ I = I_{\text{cm}} + Md^2 \] Where: - \(I_{\text{cm}}\) is the moment of inertia about the centre of mass, - \(M\) is the mass of the body, - \(R\) is the radius of the body, and - \(d\) is the distance between the centre of mass and the new axis of rotation. In this case, we are dealing with two spheres with axes of rotation at a distance equal to their radius from their centres. Step 1: Moment of inertia of sphere \(m_1\). Given: - \(R_1 = 20\,\text{cm} = 0.20\,\text{m}\) - \(m_1 = 10\,\text{kg}\) Using the **parallel axis theorem** for sphere \(m_1\), the moment of inertia is: \[ I_1 = \frac{2}{5}m_1 R_1^2 + m_1 R_1^2 \] Simplifying: \[ I_1 = \left(\frac{2}{5} + 1\right) 10 \times (0.20)^2 \] \[ I_1 = \frac{7}{5} \times 10 \times 0.04 = 0.56\,\text{kg m}^2 \] Step 2: Moment of inertia of sphere \(m_2\). Given: - \(R_2 = 10\,\text{cm} = 0.10\,\text{m}\) - \(m_2 = 5\,\text{kg}\) For sphere \(m_2\), using the parallel axis theorem again: \[ I_2 = \frac{2}{5}m_2 R_2^2 + m_2 R_2^2 \] Simplifying: \[ I_2 = \left(\frac{2}{5} + 1\right) 5 \times (0.10)^2 \] \[ I_2 = \frac{7}{5} \times 5 \times 0.01 = 0.07\,\text{kg m}^2 \] Step 3: Total moment of inertia. The total moment of inertia of the system is the sum of the moments of inertia of the two spheres: \[ I_{\text{total}} = I_1 + I_2 \] \[ I_{\text{total}} = 0.56 + 0.07 = 0.63\,\text{kg m}^2 \] Thus, the total moment of inertia is: \[ \boxed{I = 0.63\,\text{kg m}^2} \]
