Question:
A uniform rod of mass 20 kg and length 1.6 m is pivoted at its one end and can swing freely in the vertical plane. The angular acceleration of the rod just after the rod is released from rest in the horizontal position is:
A uniform rod of mass 20 kg and length 1.6 m is pivoted at its one end and can swing freely in the vertical plane. The angular acceleration of the rod just after the rod is released from rest in the horizontal position is:
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Angular acceleration for a pivoted rod depends on the length; longer rods have lower angular acceleration.
Updated On: Jun 10, 2026
- $\frac{15}{16}g$
- $\frac{17}{16}g$
- $\frac{16}{15}g$
- $\frac{9}{16}g$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Torque ($\tau$) = $I \cdot \alpha$. For a rod pivoted at one end, $I = \frac{ML^2}{3}$.
Step 2: Analysis
Torque due to gravity (acting at $L/2$): $\tau = Mg \cdot \frac{L}{2}$. $\alpha = \frac{\tau}{I} = \frac{Mg(L/2)}{ML^2/3} = \frac{3g}{2L}$. Given $L = 1.6 m$, $\alpha = \frac{3g}{2(1.6)} = \frac{3g}{3.2} = \frac{30g}{32} = \frac{15}{16}g$.
Step 3: Conclusion
The angular acceleration is $\frac{15}{16}g$.
Final Answer: (A)
Torque ($\tau$) = $I \cdot \alpha$. For a rod pivoted at one end, $I = \frac{ML^2}{3}$.
Step 2: Analysis
Torque due to gravity (acting at $L/2$): $\tau = Mg \cdot \frac{L}{2}$. $\alpha = \frac{\tau}{I} = \frac{Mg(L/2)}{ML^2/3} = \frac{3g}{2L}$. Given $L = 1.6 m$, $\alpha = \frac{3g}{2(1.6)} = \frac{3g}{3.2} = \frac{30g}{32} = \frac{15}{16}g$.
Step 3: Conclusion
The angular acceleration is $\frac{15}{16}g$.
Final Answer: (A)
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