Question

Particle of mass m is tied to one end of a string of length l. The particle is held horizontal with the string taut. It is then projected upward with a velocity u. The tension in the string is when the particle velocity is at an angle 60 to the horizontal. Then find the value of u.

Answer 1

The value of u is

$\boxed{\sqrt{\frac{(2\sqrt{3}+1)lg}{2}}}$

Explanation:

Assuming that when the particle is at an angle $60^\circ$ from the horizontal, its velocity is v

Then

By the conservation of energy

$\frac{1}{2}mv^2+mgl\sin 60^\circ=\frac{1}{2}mu^2$

$\implies v^2+\sqrt{3}gl=u^2$

At that point the tension will provide the centrifugal acceleration

Therefore,

$\frac{mv^2}{l}=mg\sin30^\circ$

$\implies v^2=\frac{lg}{2}$

Thus,

$\frac{gl}{2}+\sqrt{3}lg=u^2$

$\frac{(2\sqrt{3}+1)lg}{2}=u^2$

$\implies u=\sqrt{\frac{(2\sqrt{3}+1)lg}{2}}$

Hope this helps.