rotational dynamics hard

Question

A uniform rod of length L and mass M pivoted at one end released from horizontal. Angular speed at vertical position is (g acceleration):

RankGuru · Accepted Answer

Use Conservation of Energy:
1. **Initial Position (Horizontal)**: $K_i = 0$. $U_i = M g (L/2)$ (taking the pivot as vertical reference, center of mass is $L/2$ above ground).
2. **Final Position (Vertical)**: $U_f = 0$. $K_f = \frac{1}{2} I \omega^2$.
- Moment of inertia for a rod pivoted at one end: $I = \frac{1}{3} M L^2$.
Equation:
$M g \frac{L}{2} = \frac{1}{2} \left( \frac{1}{3} M L^2 \right) \omega^2$
$M g \frac{L}{2} = \frac{1}{6} M L^2 \omega^2$
$\omega^2 = \frac{6 M g L}{2 M L^2} = \frac{3 g}{L}$
$\omega = \sqrt{\frac{3 g}{L}}$.

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rotational dynamics hard

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