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Moment of inertia of a thin rod of mass ...

Moment of inertia of a thin rod of mass m and length l about an axis passing through a point l/4 from one end and perpendicular to the rod is

A

`(ml^(2))/(12)`

B

`(ml^(2))/(3)`

C

`(7 ml^(2))/(48)`

D

`(ml^(2))/(9)`

Text Solution

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The correct Answer is:
To find the moment of inertia of a thin rod of mass \( m \) and length \( l \) about an axis passing through a point \( \frac{l}{4} \) from one end and perpendicular to the rod, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Moment of Inertia about the Center of Mass (CM)**: The moment of inertia of a thin rod about its center of mass (which is at a distance \( \frac{l}{2} \) from either end) is given by the formula: \[ I_{CM} = \frac{1}{12} m l^2 \] 2. **Determine the Distance from the Center of Mass to the New Axis**: The new axis is located \( \frac{l}{4} \) from one end. The distance from the center of mass to this new axis can be calculated as follows: - The distance from one end to the center of mass is \( \frac{l}{2} \). - The distance from the center of mass to the new axis is: \[ d = \frac{l}{2} - \frac{l}{4} = \frac{l}{4} \] 3. **Apply the Parallel Axis Theorem**: The parallel axis theorem states that the moment of inertia about a new axis parallel to the one through the center of mass is given by: \[ I = I_{CM} + m d^2 \] Substituting the values we have: \[ I = \frac{1}{12} m l^2 + m \left(\frac{l}{4}\right)^2 \] 4. **Calculate \( m d^2 \)**: Now calculate \( m d^2 \): \[ m d^2 = m \left(\frac{l}{4}\right)^2 = m \frac{l^2}{16} \] 5. **Combine the Terms**: Substitute \( I_{CM} \) and \( m d^2 \) into the equation: \[ I = \frac{1}{12} m l^2 + m \frac{l^2}{16} \] 6. **Find a Common Denominator**: The common denominator for \( 12 \) and \( 16 \) is \( 48 \): - Convert \( \frac{1}{12} \) to \( \frac{4}{48} \) - Convert \( \frac{1}{16} \) to \( \frac{3}{48} \) Therefore: \[ I = \frac{4}{48} m l^2 + \frac{3}{48} m l^2 = \frac{7}{48} m l^2 \] 7. **Final Result**: The moment of inertia of the thin rod about the specified axis is: \[ I = \frac{7}{48} m l^2 \]
To find the moment of inertia of a thin rod of mass \( m \) and length \( l \) about an axis passing through a point \( \frac{l}{4} \) from one end and perpendicular to the rod, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Moment of Inertia about the Center of Mass (CM)**: The moment of inertia of a thin rod about its center of mass (which is at a distance \( \frac{l}{2} \) from either end) is given by the formula: \[ I_{CM} = \frac{1}{12} m l^2 ...
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