Question

A particle suspended from a fixed point, by a light inextensible thread of length L is projected horizontally from its lowest position with velocity $\sqrt{\frac{7gL}{2}}$. The thread will slack after swinging through an angle $\theta$, such that $\theta$ equal.

• A. ${30}^o$
• B. ${135}^o$
• C. ${120}^o$
• D. ${150}^o$

Answer 1

Answer: (C)

${120}^o$

As shown in figure, $h=L+\sin \varphi$

By energy conservation,

$\frac{1}{2}m{v}^2=mgh+\frac{1}{2}m{v}^{\prime 2}$

$\frac{1}{2}\times (\sqrt{\frac{7gl}{2}}{)}^2=g(L+L\sin \varphi )+\frac{1}{2}{v}^{\prime 2}$

$v^{\prime 2}=-2gL-2gL\sin \varphi +\frac{7}{2}gL$

$=\frac{3}{2}gL-2gL\sin \varphi$

If string slack then,

$mg\sin \varphi =\frac{m{v}^{\prime 2}}{L}$

$g\sin \varphi =\frac{1}{L}(v^{\prime }{)}^2=\frac{1}{L}(\frac{3}{2}gL-2gL\sin \varphi )$

$g\sin \varphi =\frac{3}{2}g-2g\sin \varphi$

$3g\sin \varphi =\frac{3}{2}g$

$\sin \varphi =\frac{1}{2}$

$\varphi =\frac{\pi }{6}$

$\theta =\frac{\pi }{2}+\frac{\pi }{6}=\frac{4\pi }{6}={120}^{\circ }$