First slide

Vertical circular motion

Question

A bob of mass m is attached to one end of a string of length l. This is kept in the horizontal position and released from the rest. At what angle from vertical will the tension in the string be equal to the weight in magnitude ?

Moderate

A

$\cos^{- 1} (\frac{1}{2})$
B

$\cos^{- 1} (\frac{1}{3})$
C

$\cos^{- 1} (1 / \sqrt{2})$
D

$\cos^{- 1} (1 / \sqrt{3})$

Solution

From figure, $T - mgcos θ = \frac{{mv}^{2}}{l}$
$∴ T = mgcos θ + ({mv}^{2} / l)$
$\begin{array}{l} or mg = mgcos θ + ({mv}^{2} / l) \\ or mg = mgcos θ + (\frac{m}{l}) (2 g) (l - h) \\ ∴ 1 = \cos θ + 2 \cos θ \\ 1 = 3 \cos θ \\ ∴ \cos θ = 1 / 3 or θ = \cos^{- 1} (1 / 3) \end{array}$

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