Question

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

(1) $T_1-T_2=2mg$
(2) $T_1+T_2=m{\omega }^{2}l$
(3) $T_1=T_2$
(4) $T_1+T_2=mg$

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (B)

Let $T_1$ and $T_2$ be the tensions in $PA$ and $PB$ respectively.

From the figure,
$T_1\sin 30^{\circ }=T_2\sin 30^{\circ }+mg$
or $T_1-T_2=2mg\dots$...(i)
and $T_1\cos 30^{\circ }+T_2\cos 30^{\circ }=m{\omega }^{2}l$
or $T_1+T_2=m{\omega }^{2}l\dots$...(ii)
From Eqs. (i) and (ii)
$T_1=\frac{m}{2}\left[{\omega }^{2}l+2g\right]$
and $T_2=\frac{m}{2}\left[{\omega }^{2}l-2g\right]$