MHT CET · Physics · Rotational Motion
A thin uniform rod of mass ' m ' and length ' L ' is pivoted at one end so that it can rotate in a vertical plane. The free end is held vertically above pivot and then released. The angular acceleration of the rod when it makes an angle ' \(\theta\) ' with the vertical is [consider negligible friction at the pivot] ( \(\mathrm{g}=\) acceleration due to gravity)
- A \(\frac{3 \mathrm{~g} \sin \theta}{2 \mathrm{~L}}\)
- B \(\frac{3 \mathrm{~g} \cos \theta}{2 \mathrm{~L}}\)
- C \(\frac{2 \mathrm{~g} \sin \theta}{3 \mathrm{~L}}\)
- D \(\frac{2 \mathrm{~g} \cos \theta}{3 \mathrm{~L}}\)
Answer & Solution
Correct Answer
(A) \(\frac{3 \mathrm{~g} \sin \theta}{2 \mathrm{~L}}\)
Step-by-step Solution
Detailed explanation
\(I = \frac{1}{3} mL^2\) \(\tau = mg \left(\frac{L}{2} \sin \theta\right)\)
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