Question

A particle of mass $m$ is suspended by a string of length $l$ from a fixed rigid support. A sufficient horizontal velocity $v_0=\sqrt{3gl}$ is imparted to it suddenly. Calculate the angle made by the string with the vertical when the acceleration of the particle is inclined to the string by ${45}^{\circ }.$

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

Answer 1

The correct option is A $\theta =\frac{\pi }{2}$

Given : $u=\sqrt{3gl}$

Let the velocity of the bob be $v$ when it reaches the point C.

From figure, $AB=l-lcos\theta =l(1-cos\theta )$

Using work-energy theorem : $W=\Delta K.E$

$\therefore$ $-mgl(1-cos\theta )=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

OR $-2gl(1-cos\theta )=v^2-\left(3gl\right)$ $⟹v^2=gl+2glcos\theta$

Using $a_r=\frac{v^2}{l}=g+2gcos\theta$

Also tangential acceleration $a_t=gsin\theta$

As the acceleration makes an angle ${45}^o$ with the string, thus $tan{45}^o=\frac{a_t}{a_r}$

$⟹$ $a_r=a_t$

$\therefore$ $g+2gcos\theta =gsin\theta$

$⟹$ $sin\theta -2cos\theta -1=0$ where $\theta =\frac{\pi }{2}$ satisfies the equation.