First slide

Vertical circular motion

Question

A pendulum bob has a mass $m = \frac{1}{2} k g$ and length of string $l = 1 m .$ The bob is released from position A. Then maximum tension in the string is ____

Moderate

A

10 N
B

$10 \sqrt{3} N$
C

20 N
D

$20 \sqrt{3} N$

Solution

$\begin{array}{l} V_{\max} = \sqrt{2 g l (1 - \cos θ)} = \sqrt{2 \times 10 \times 1 \times (1 - \cos 60^{0})} m / s \\ = \sqrt{10} m / s \\ ∴ T_{\max} = m g + \frac{m V_{\max}^{2}}{l} = (\frac{1}{2} \times 10 + \frac{\frac{1}{2} \times 10}{1}) N = 10 N \end{array}$

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