rotational-motion

Question: The string of a simple pendulum is replaced by a uniform rod of length L and mass M. If the mass of the bob of the pendulum is m, then for small oscillations its time period would be (assume radius of bob r << L)

1) $2\pi \sqrt{\frac{2(M+3m)\,L}{3(M+2m)\,g}}$

2) $2\pi \sqrt{\frac{(M+2m)\,L}{3(M+3m)\,g}}$

3) $2\pi \sqrt{\left( \frac{2M}{3m} \right)\,\frac{L}{g}}$

4) $2\pi \sqrt{\left( \frac{M+m}{M+3m} \right)\,\frac{L}{g}}$

Solution:

Explanation: No Explanation

compound pendulum

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