Question

A bob of mass m is projected with a horizontal velocity $v=\sqrt{\frac{gl}{2}}$ as shown in Fig. 8.224. In consequence, it moves in a circular path in a vertical plane by the inextensible string which passes over the smooth fixed peg. Find the maximum angle that the bob swings in the left hand side.

• A. $co{s}^{-1}$ ($\frac{1}{8}$)
• B. $co{s}^{-1}$ ($\frac{1}{2}$)
• C. $co{s}^{-1}$ ($\frac{1}{6}$)
• D. $co{s}^{-1}$ ($\frac{1}{3}$)

Answer 1

Answer: (C)

$co{s}^{-1}$ ($\frac{1}{6}$)

given, $V=\sqrt{\frac{gl}{2}}$

let, the max angle at the lett side $=\theta$

let, the $l/2$ distance below the point $p$ is point of reference.

By conserving Energy we get

$\frac{1}{2}m{V}^2+mgl=mg\frac{3l}{2}(1-\cos \theta )$

or $\frac{V^2}{2}+gl=\frac{3}{2}gl(1-cos\theta )$

or $V^2+2gl=3gl(1-\cos \theta )$

or $\frac{gl}{2}+2gl=3gl(1-\cos \theta )$

or $5gl=6gl(1-\cos \theta )$

or $\frac{5}{6}=1-\cos \theta$

or $\cos \theta =\frac{1}{6}$

or $\theta ={\cos }^{-1}\left(\frac{1}{6}\right)$

Option $\left(C\right)$ is correct