First slide

Vertical circular motion

Question

The string of a pendulum of length l is displaced through 90^o from the vertical and released. Then the minimum strength of the string in order to withstand the tension as the pendulum passes through the mean position is

Moderate

A

3 mg
B

4 mg
C

5 mg
D

6 mg

Solution

The velocity at the lowest point is given by
$v = \sqrt{(2 gr)}$
Further, $T - mg = \frac{{mv}^{2}}{r}$
$\begin{array}{l} ∴ T = mg + \frac{{mv}^{2}}{r} = mg + \frac{m (2 gr)}{r} \\ = mg + 2 mg = 3 mg \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App