Question

A particle P of mass m attached to a vertical axis by two strings AP and BP of length 1m each. The separation $AB=l$. P rotates around the axis with an angular velocity $\omega$. The tension in the two strings are $T_1$ and $T_2$.

• A. $T_1=T_2$
• B. $T_1+T_2=m{\omega }^{2}l$
• C. $T_1-T_2=2mg$
• D. BP will remain taut only if $\omega \ge \sqrt{2g/l}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $T_1+T_2=m{\omega }^{2}l$
C BP will remain taut only if $\omega \ge \sqrt{2g/l}$
D $T_1-T_2=2mg$

As the triangle $ABP$ is equilateral. all the angles are of ${60}^o$.

Balancing the forces in vertical direction,

$T_{1}sin{30}^o-T_{2}sin{30}^o=mg\Rightarrow T_1-T_2=2mg-\left(1\right)$

The radius of rotation of circle, $r=lcos{30}^o$

Balancing forces in horizontal direction,

$T_{1}cos{30}^o+T_{2}cos{30}^o=m{\omega }^{2}r\Rightarrow T_1+T_2=m{\omega }^{2}l-\left(2\right)$

For BP to be taut, $T_2$ should be greater than 0.

From $\left(1\right)-\left(2\right)$,

$2T_2=m{\omega }^{2}l-2mg>0\Rightarrow \omega \ge \sqrt{2g/l}$