Question

A heavy particle hanging from a fixed point by a light in extensible string of length l is projected horizontally with speed $\sqrt{\left(gl\right)}$. Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string is equal to the weight of the particle.

• A. $v=\sqrt{\frac{gl}{3}},cos\theta =\frac{2}{3}$
• B. $v=\sqrt{\frac{gl}{3}},cos\theta =\frac{1}{3}$
• C. $v=\sqrt{\frac{2gl}{3}},cos\theta =\frac{1}{3}$
• D. $v=\sqrt{\frac{gl}{3}},cos\theta =\frac{2}{5}$

Answer 1

The correct option is A $v=\sqrt{\frac{gl}{3}},cos\theta =\frac{2}{3}$


Assume tension in the string becomes equal to the weight of the particle when particle reaches the point B and deflection of the string from vertical is $\theta$. Resolving mg along the string and perpendicular to the string, we get net radial force on the particle at B i.e.
$F_R=T-mgcos\theta$ (i)
If v is the speed of the particle at B, then $F_R=\frac{m{v}^2}{l}$ (ii)
From (i) and (ii), we get $T-mgcos\theta =\frac{m{v}^2}{l}$ (iii)
Since at B, T = mg $mg(1-cos\theta )=\frac{m{v}^2}{l}{v}^2=gl(1-cos\theta )$ (iv)
Conserving the energy of the particle at point A and B, we have
$\frac{1}{2}m{v}_0^2=mgl(1-cos\theta )+\frac{1}{2}m{v}^2$
Where $v_0=\sqrt{gl}$ and $v=\sqrt{gl(1-cos\theta )}$
$gl=2gl(1-cos\theta )+gl(1-cos\theta )$, $cos\theta =\frac{2}{3}$
Putting the value of $cos\theta$ in equation (iv) we get
$v=\sqrt{\frac{gl}{3}}$