Question

The figure shows a rod arranged at an angle of ${30}^{\circ }$ from horizontal. Two strings are fixed to rod as shown. Attached to the two strings is the mass m as shown. The rod is rotate maintaining its direction in space, so that m travels in a circular path. The strings are of equal length and make angle of ${60}^{\circ }$ with the rod as shown.
Calculate the minimum value of the tangential speed
($m/s$) of the mass such that the string with tension $T_2$ does not brcome slack when the mass is directly above the rod. Take lenght of string as $l=2.4m$.

Answer 1

Radius of the circle $r=r\cos {30}^{\circ }=\frac{2.4\sqrt{3}}{2}$

Equate the force into horizontal direction,

$T_2\cos {30}^{\circ }+T_1\cos {30}^{\circ }+mg\cos {30}^{\circ }$

$=\frac{m{v}^2}{r}\dots \left(i\right)$

Equate the force into vertical direction,

$T_2\sin {30}^{\circ }+mg\sin {30}^{\circ }=T_1\sin {30}^{\circ }$

for just slack condition, $T_2=0$

$T_1=mg$

put the value in equation$\left(i\right)$

$mg\cos {30}^{\circ }+mg\cos {30}^{\circ }=\frac{m{v}^2}{r}$

$\sqrt{3}mg=\frac{m{v}^2}{r}$

$v^2=\sqrt{3}\times 10\times \frac{2.4\sqrt{3}}{2}$

$v=6m/s$

Final answer: 6 m/s