Question

Two wires $AC$ and $BC$ are each tied to a sphere at $C$. $AB=0.2\text{m}$ the sphere is made to revolve in a horizontal circle at a constant speed $v$. Then, when both wires are taut.

• A. $v_{min}=1\text{m/s}$
• B. $v_{min}=2\text{m/s}$
• C. $v_{max}=\sqrt{3}\text{m/s}$
• D. $v_{max}=4\text{m/s}$

Answer 1

Answer: (A), (C)

The correct options are
A $v_{min}=1\text{m/s}$
C $v_{max}=\sqrt{3}\text{m/s}$

For vertical equilibrium,
$T_1\cos {30}^{\circ }+T_2\cos {60}^{\circ }=mg$
$\sqrt{3}{T}_1+T_2=2mg...\left(1\right)$
The net force along the horizontal direction provides the centripetal force for circular motion.
$T_1\cos {30}^{\circ }+T_2\cos {60}^{\circ }=\frac{m{v}^2}{r}$
$T_1+\sqrt{3}{T}_2=\frac{2m{v}^2}{r}...\left(2\right)$
From (1) and (2), $T_2=\frac{\sqrt{3}m{v}^2}{r}-mg$,$T_1=\sqrt{3}mg-\frac{m{v}^2}{r}$

Here $AB=BC=0.2,r=(\frac{\sqrt{3}}{2}\times 0.2)\text{m}$
The string is taut, if
$T_2>0,v_{min}=\sqrt{\frac{gr}{\sqrt{3}}}=1\text{m/s}$
$T_1>0,v_{max}=\sqrt{\sqrt{3}gr}=\sqrt{\sqrt{3}\times 10\times \sqrt{3}\times 0.1}=\sqrt{3}\text{m/s}$