Question

The bob of a pendulum at rest is given an impulse to impart a horizontal velocity $\sqrt{gl}\text{m/s}$ where $l$ is the length of the pendulum. Find the tangential acceleration at the point where velocity of the bob is zero.

• A. $8.66{\text{m/s}}^2$
• B. $5{\text{m/s}}^2$
• C. $7.07{\text{m/s}}^2$
• D. $10{\text{m/s}}^2$

Answer 1

The correct option is A $8.66{\text{m/s}}^2$

Let $\theta$ be the angle with the vertical at which velocity become zero. Let $V_1$ and $V_2$ be the velocities at points $1$ and $2$ respectively.


Using energy conservation,
$\Delta KE+\Delta PE=0$
$\frac{1}{2}m{V}_2^2-\frac{1}{2}m{V}_1^2+mgl[1-\cos \theta ]=0$
If $V_2=0$,
$0-\frac{1}{2}m{V}_1^2+mgl(1-\cos \theta )=0$
$mgl(1-\cos \theta )=\frac{1}{2}m{V}_1^2$
Putting value of $V_1=\sqrt{gl}$,
$mgl[1-\cos \theta )=\frac{1}{2}mgl$
$[1-\cos \theta ]=0.5$
$\Rightarrow \cos \theta =0.5$ or $\theta ={60}^{\circ }$
From the figure,
Tangental acceleration = $g\sin \theta$
$a_t=10\times \sin {60}^{\circ }$
$=8.66{\text{m/s}}^2$