Two identical solid spheres each of mass 2 kg and radii 10 cm are fixed at the ends of a light rod. The separation between the centres of the spheres is 40 cm. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is __________ \(× 10^{-3} \)kg-\(m^2\)
Correct Answer: 176
Approach Solution - 1
\(\begin{aligned} & I_{s y s}=\left(\frac{2}{5} m r^2+m d^2\right) \times 2 \\ & \Rightarrow I_{s y s}=\left(\frac{2}{5} \times 2 \times 0.01+2 \times 0.04\right) \times 2=(0.008+0.08) \times 2=0.088 \times 2=176 \times 10^{-3} \end{aligned}\)
Approach Solution -2
Moment of Inertia of a System with Two Spheres
Step 1: Moment of Inertia of a Solid Sphere
The moment of inertia of a solid sphere about its diameter is given by the formula:
\( I_{\text{sphere}} = \frac{2}{5}mr^2 \)
where: - \( m \) is the mass of the sphere, - \( r \) is the radius of the sphere.
Step 2: Parallel Axis Theorem
The parallel axis theorem states that the moment of inertia of a body about an axis parallel to and a distance \( d \) away from an axis through its center of mass is given by:
\( I = I_{\text{cm}} + md^2 \)
where: - \( I_{\text{cm}} \) is the moment of inertia about the center of mass axis, - \( d \) is the distance from the center of mass to the new axis.
Step 3: Moment of Inertia of One Sphere about the Midpoint of the Rod
The distance between the center of a sphere and the midpoint of the rod is:
\( d = \frac{40}{2} \, \text{cm} = 20 \, \text{cm} = 0.2 \, \text{m} \)
The radius of each sphere is \( r = 10 \, \text{cm} = 0.1 \, \text{m} \). Using the parallel axis theorem, the moment of inertia of one sphere about the midpoint of the rod is:
\( I_{\text{one}} = \frac{2}{5}mr^2 + md^2 \)
Substitute the values into the equation:
\( I_{\text{one}} = \frac{2}{5}(2)(0.1)^2 + (2)(0.2)^2 = 0.008 + 0.08 = 0.088 \, \text{kg-m}^2 \)
Step 4: Total Moment of Inertia
Since there are two identical spheres, the total moment of inertia of the system is:
\( I_{\text{sys}} = 2 \times I_{\text{one}} = 2 \times 0.088 = 0.176 \, \text{kg-m}^2 = 176 \times 10^{-3} \, \text{kg-m}^2 \)
Conclusion:
The moment of inertia of the system is \( \mathbf{176 \times 10^{-3} \, \text{kg-m}^2} \).
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