Question

A small block is connected to one end of two identical massless strings of length $\frac{50}{3}$ cm each with their other ends fixed to a vertical rod. If the ratio of tensions be 4:1, then what will be the angular velocity $\omega$ of the block.

• A. 2 rad/sec
• B. 1.414 rad/sec
• C. 4 rad/sec
• D. none of these

Answer 1

The correct option is D

none of these

For horizontal equilibrium of the block, $T_{1}sin{60}^{\circ }+T_{2}sin{60}^{\circ }=m{\omega }^{2}r$

$T_1+T_2=m{\omega }^{2}l$ ( $\therefore r=lsin{60}^{\circ }$ ) ............ (i)

For vertical equilibrium of the block, $T_{1}cos{60}^{\circ }=T_{2}cos{60}^{\circ }+mg$

$T_1-T_2=2mg$ ............ (ii)

Dividing (i) by (ii) $\frac{T_1+T_2}{T_1-T_2}=\frac{{\omega }^{2}l}{2g}\Rightarrow \frac{\frac{T_1}{T_2}+1}{\frac{T_1}{T_2}-1}=\frac{{\omega }^{2}l}{2g}\Rightarrow \frac{4+1}{4-1}=\frac{{\omega }^{2}l}{2g}$

${\omega }^2=\frac{10g}{3l}={\omega }^2=\frac{10\times 9.8\times 3}{3\times 50\times {10}^{-3}}\Rightarrow \omega =\sqrt{196}$ = 14 rad/s