Question

The length of a simple pendulum is '$L$'. Its bob from rest position is projected horizontally with a velocity $\sqrt{\frac{7gL}{2}}$. The maximum angular displacement of bob, such that the string does not slack, is:

• A. ${30}^o$
• B. ${60}^o$
• C. ${120}^o$
• D. ${150}^o$

Answer 1

Answer: (C)

${120}^o$

At point where tension become just zero, the centripetal force must be balanced by gravitational force
$mgcos\theta =\frac{m{v}_1^2}{L}$ ------ (1)
BY conservation of energy
$\frac{1}{2}m\frac{7gL}{2}=\frac{m{v}_1^2}{2}+mgL(1+cos\theta )$
$\sqrt{2gL}\sqrt{\frac{3}{4}-cos\theta }=v_1$
Put $v_1$ in (1)
$mg$$cos\theta =$$\frac{m\times 2gL}{L}(\frac{3}{4}-cos\theta ).$
$cos\theta =\frac{3}{2}-2cos\theta$
$cos\theta =\frac{1}{2}$
$\theta ={60}^0$
$\therefore$Total maximum angular displacement
$={180}^0-{60}^0$
$={120}^0$