Question

A 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 $\frac{mg}{2}$ when it is inclined at an angle ${30}^0$ to the horizontal. Find the value of $u$.

Answer 1

$T=\frac{mg}{2}butT=\frac{m{u}^2}{r}\therefore \frac{mg}{2}=\frac{m{u}^2}{r}=u^2=\frac{rg}{2}usingenergybalance,\frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2+mg\left(rsin{30}^{\circ }\right)\therefore v^2=u^2+2gr\times \frac{1}{2}=\frac{rg}{2}+rg=\frac{3rg}{2}\therefore v=\sqrt{\frac{3rg}{2}}$