Question

The bob of a pendulum at rest is given a sharp hit to impact a horizontal velocity $\sqrt{3.5gl}$, where $l$ is the length of the pendulum. Find the height at which tension in the string is zero.

• A. $l$
• B. $\frac{3l}{2}$
• C. $2l$
• D. $1.866l$

Answer 1

The correct option is B $\frac{3l}{2}$

Let $\theta$ be the angle with vertical at which tension in the string is zero.

From fig.
$T=mg\cos \theta +\frac{m{V}_2^2}{l}$
$T=0\Rightarrow gl\cos \theta +V_2^2=0$
$V_2^2=-gl\cos \theta$
Using energy conservation between $1$ and $2$:
$\Delta KE+\Delta PE=0$
$\frac{1}{2}m{V}_2^2-\frac{1}{2}m{V}_1^2+mgl(1-\cos \theta )=0$
$\frac{1}{2}m{V}_1^2-\frac{1}{2}m{V}_2^2=mgl(1-\cos \theta )$
$V_1^2-V_2^2=2gl(1-\cos \theta )$
$3.5gl+gl\cos \theta =2gl-2gl\cos \theta$
$1.5gl=-3gl\cos \theta$
$-0.5=\cos \theta$
Height at which tension is zero,
$=L(1-\cos \theta )$
$=L(1+0.5)=\frac{3L}{2}$
$h=\frac{3L}{2}$