Question

A bob of mass m is suspended by a massless string of length l. The hon'zontal velocity v at position a is just sufcient to make it reach the point
b. The angle 0 at which the speed of the bob is half of that at a

Answer 1

Question:

A bob of mass M is suspended by a massless  string of length L. The horizontal velocity v at  position A is just sufficient to make it reach point  B. The angle θ at which the speed of the bob is  half of that at A satisfies.

$\sf (1) \ \frac{3\pi}{4} < \theta < \pi$

$\sf (2) \ \theta = \frac{\pi}{4}$

$\sf (3) \ \frac{\pi}{4} < \theta < \frac{\pi}{2}$

$\sf (4) \ \frac{\pi}{2} < \theta < \frac{3\pi}{4}$

Answer:

$\boxed{\sf (1) \ \frac{3\pi}{4} < \theta < \pi}$

Given:

Mass of bob = M

Length of massless string = l

Horizontal velocity at position A which is sufficient to make the bob reach the point B = V

To Find:

Angle θ at which the speed of the bob is half that at A

Explanation:

Horizontal velocity (V) at the lower most point i.e. A to reach point B on the vertical circle = $\sf \sqrt{5gl}$

By using conservation of energy we can write:

$\sf \implies KE_1 + PE_1 = KE_2 + PE_2$

$\sf \implies \frac{1}{2} M(\sqrt{5gl} )^{2} +0=\frac{1}{2} M(\frac{\sqrt{5gl}}{2} )^{2} + Mgl(1-cos \theta)$

$\sf \implies \frac{1}{2} \cancel{M}(5\cancel{gl}) =\frac{1}{2}\cancel{ M}(\frac{5\cancel{gl}}{4} ) + \cancel{Mgl}(1-cos \theta)$

$\sf \implies \frac{5}{2} =\frac{1}{2} \times \frac{5}{4} + (1-cos \theta)$

$\sf \implies \frac{5}{2} = \frac{5}{8} + (1 - cos \theta)$

$\sf \implies \frac{5}{2}= \frac{5}{8} +(1- cos \theta)$$\sf \implies 1- cos \theta = \frac{5}{2} -\frac{5}{8}$

$\sf \implies 1- cos \theta = \frac{20-5}{8}$

$\sf \implies 1- cos \theta = \frac{15}{8}$

$\sf \implies cos \theta =1- \frac{15}{8}$

$\sf \implies cos \theta = \frac{8-15 }{8}$

$\sf \implies cos \theta = -\frac{7 }{8}$

$\sf \implies \theta = cos^{-1}(- \frac{7 }{8} )$

$\sf \implies \theta \approx 151^\circ$

$\therefore$

$\sf \frac{3\pi}{4} < \theta < \pi$