Question

The bob of a pendulum at rest is given a hit to impart a horizontal velocity $\sqrt{gl}\text{m/s}$ at the bottommost position of the bob, where $l$ is the length of the string in $\text{metres}$. Find the tangential acceleration at the point where the velocity of the bob is zero. Take $g=10{\text{m/s}}^2$

• A. $5\sqrt{3}{\text{m/s}}^2$
• B. $5{\text{m/s}}^2$
• C. $\frac{5}{\sqrt{3}}{\text{m/s}}^2$
• D. $10{\text{m/s}}^2$

Answer 1

The correct option is A $5\sqrt{3}{\text{m/s}}^2$

Let $\theta$ be the angle with the vertical as shown in figure, at which velocity of the bob is zero i.e $\text{v}=0$.


At the bottommost position, velocity of the particle was $u=\sqrt{gl}$

Since work done by tension $W_T=0$, applying mechanical energy conservation from bottommost position to given position:
$\Delta KE+\Delta PE=0$
$(\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2)+\left(mgl\right[1-\cos \theta ]-0)=0$
Taking datum for $PE=0$ at bottom and $v=0$ at angle $\theta$,
$u^2-v^2=2gl[1-\cos \theta ]$
$gl-0=2gl-2gl\cos \theta$
$gl=2gl\cos \theta$
$\Rightarrow \cos \theta =\frac{1}{2}$
$\therefore \theta ={60}^{\circ }$

From the FBD of pendulum bob, net tangential force
$F_t=mg\sin \theta$
Tangential acceleration of the bob will be:
$a_t=\frac{F_t}{m}=g\sin \theta =10\times \sin {60}^{\circ }$
$\therefore a_t=5\sqrt{3}{\text{m/s}}^2$