Q: A uniform rod of length 2 m and mass 500 g is pivoted at one end and is displaced through 60° from the vertical. What is its kinetic energy while passing through the mean position ?
(a) 2.5 J
(b) 5.0 J
(c) 7.5 J
(d) 10 J
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Ans: (a)
Sol: $\displaystyle \frac{1}{2} m v^2 = m g \frac{L}{2} (1 – cos60^o)$
$\displaystyle \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \times \frac{2}{2} (1 – \frac{1}{2})$
$\displaystyle \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \times (\frac{1}{2})$
= 2.5 J