Question

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

• A. $7\text{rad/s}$
• B. $8.5\text{rad/s}$
• C. $11\text{rad/s}$
• D. $14\text{rad/s}$

Answer 1

The correct option is D $14\text{rad/s}$

Let the angular velocity of the system be $\omega$.
FBD of block:


For horizontal equilibrium of the block,
$T_1\sin {60}^{\circ }+T_2\sin {60}^{\circ }=m{\omega }^{2}r=m{\omega }^{2}l\sin {60}^{\circ }$
$\Rightarrow T_1+T_2=m{\omega }^{2}l....\left(i\right)$

For vertical equilibrium of the block,
$T_1\cos {60}^{\circ }=T_2\cos {60}^{\circ }+mg$
$\Rightarrow T_1-T_2=\frac{mg}{\cos {60}^{\circ }}=2mg....\left(ii\right)$
Dividing $\left(i\right)$ by $\left(ii\right)$, $\frac{T_1+T_2}{T_1-T_2}=\frac{{\omega }^{2}l}{2g}$
$\Rightarrow \frac{\frac{T_1}{T_2}+1}{\frac{T_1}{T_2}-1}=\frac{{\omega }^{2}l}{2g}\Rightarrow \frac{4+1}{4-1}=\frac{{\omega }^{2}l}{2g}$
or ${\omega }^2=\frac{10g}{3l}=\frac{10\times 9.8\times 3}{3\times 50\times {10}^{-2}}=196$
or $\omega =14{\text{rad s}}^{-1}$