Question

A bod of mass M is suspended by a massless string of length L.The horizontal velocity V at position A is just sufficient to make real the point B.The angle $\theta$ at which the speed of the bob is half of that it A satisfies:

• A. $\theta =\frac{\pi }{4}$
• B. $\frac{\pi }{4}<\theta <\frac{\pi }{2}$
• C. $\frac{\pi }{2}<\theta <\frac{3\pi }{4}$
• D. $\frac{3\pi }{4}<\theta <\pi$

Answer 1

The correct option is D $\frac{3\pi }{4}<\theta <\pi$

Use conservation of energy,

$\frac{1}{2}m{v}_o^2=\frac{1}{2}m{v}^2+mgl(1-\cos \theta )-------\left(i\right)$

where $v_o=$horizontal velocity

$mg\left(2l\right)=\frac{1}{2}m{v}_o^2-\frac{1}{2}m{v}_{top}^2-------\left(ii\right)$

Since $v_o$ is just sufficient

$\frac{m{v}_{top}^2}{l}=T+mgT=0,v_{top}=\sqrt{gl}$

Then equation $\left(ii\right)$,

$⟹v_o=\sqrt{5gl}$

According to the question, $v=\frac{v_o}{2}$

From equation $\left(i\right),$

$\frac{1}{2}m5gl=\frac{1}{2}m\left(\frac{5gl}{4}\right)+mgl(1-\cos \theta )⟹\frac{20mgl-5mgl}{8}=mgl(1-\cos \theta )⟹\cos \theta =\frac{7}{8}$

Hence, $\frac{3\pi }{4}<\theta <\pi$.