Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 cm. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is \(\frac{x}{20}\) kg m3 , where the value of x is ____.
Correct Answer: 53
Approach Solution - 1
Given:
Mass of each wheel
m = 2 Kg,
R = 50 cm= 0.5 m,
d = 75 cm = 0.75 m
Step 1: Formula for Moment of Inertia
The moment of inertia \( I \) for the system is given by: \[ I = \left( \frac{2}{5} m R^2 + m d^2 \right) \times 2 \]
Step 2: Substituting the given values
Substituting the values \( m = 2 \, \text{kg}, R = 0.5 \, \text{m}, d = 0.75 \, \text{m} \) into the equation: \[ I = 2 \left( \frac{2}{5} \times 2 \times \left( \frac{1}{2} \right)^2 + 2 \times \left( \frac{3}{4} \right)^2 \right) \]
Step 3: Simplifying the equation
\[ I = 2 \left( \frac{2}{5} \times 2 \times \frac{1}{4} + 2 \times \frac{9}{16} \right) \]
\[ I = 2 \left( \frac{1}{10} + \frac{9}{8} \right) = 2 \times \frac{53}{40} = \frac{53}{20} \, \text{kg} \cdot \text{m}^2 \]
Final Answer:
\[ X = 53 \, \text{kg} \cdot \text{m}^2 \]
Approach Solution -2
The moment of inertia \( I \) of each sphere about the central axis (using the parallel axis theorem) is:
\[ I_{\text{total}} = 2 \left( I_{\text{sphere}} + md^2 \right). \]For a solid sphere:
\[ I_{\text{sphere}} = \frac{2}{5}mR^2 = \frac{2}{5} \times 2 \times (0.5)^2 = 0.2 \, \text{kg m}^2. \]Distance \( d \) from the center of each sphere to the midpoint of the rod is \( 0.75 \, \text{m} \).
So,
\[ I_{\text{total}} = 2 \left( 0.2 + 2 \times (0.75)^2 \right) = 2 \left( 0.2 + 1.125 \right) = \frac{53}{20} \, \text{kg m}^2. \]Thus, \( x = 53 \).
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