Question

A small block is connected to one end of two identical massless strings of length $16\frac{2}{3}$ each with their other ends fixed to a vertical rod. If the ratio of tensions $T_1/T_2$ is 4: 1, then what will be the angular velocity of the block? Take$g=9.8m{s}^{-2}$

Answer 1

For horizontal,
$T_{1}sin{60}^0+T_{2}sin{60}^0=m{\omega }^{2}r=m{\omega }^{2}lsin{60}^0$
$\Rightarrow T_1+T_2=m{\omega }^{2}l$ (i)
For vertical equilibrium of the block,
$T_{1}cos{60}^0=T_{2}cos{60}^0+mg$
$\Rightarrow T_1-T_2=\frac{mg}{cos{60}^0}=2mg$ (ii)
Dividing (i) by (ii), $\frac{T_1+T_2}{T_1-T_2}=\frac{{\omega }^{2}l}{2g}$
$\Rightarrow \frac{\frac{T_1}{T_1}+1}{\frac{T_1}{T_2}-1}=\frac{{\omega }^{2}l}{2g}\Rightarrow \frac{4+1}{4-1}=\frac{{\omega }^{2}l}{2g}$
or ${\omega }^2=\frac{10g}{3l}=\frac{10\times 9.8\times 3}{3\times 50\times {10}^{-2}}=196$
or $\omega =14rad{s}^{-1}$