Question

Two wires $A C$ and $B C$ are tied at $C$ of small sphere of mass $5 kg$ which revolves at a constant speed $v$ in the horizontal circle of radius $1.6 m$ . Find the minimum value of $v$ .

Accepted Answer

According to the question, after drawing F.B.D.,

In equilibrium condition along vertical direction,

$T_{1} \cos 30 ° + T_{2} \cos 45 ° = m g \Rightarrow \frac{\sqrt{3} T_{1}}{2} + \frac{T_{2}}{\sqrt{2}} = m g . . . (i)$

nd in horizontal direction,

$T_{1} \sin 30 ° + T_{2} \sin 45 ° = \frac{m v^{2}}{r} \Rightarrow \frac{T_{1}}{2} + \frac{T_{2}}{\sqrt{2}} = \frac{m v^{2}}{r} . . . (i i)$

After solving equations $(i)$ and $(ii)$ ,

$\frac{\sqrt{3} T_{1}}{2} - \frac{T_{1}}{2} = m g - \frac{m v^{2}}{r} \Rightarrow T_{1} = \frac{m g - \frac{m v^{2}}{r}}{\frac{(\sqrt{3} - 1)}{2}}$

But

$T_{1} > 0 \frac{m g - \frac{m v^{2}}{r}}{\frac{\sqrt{3} - 1}{2}} > 0 \Rightarrow m g \geq \frac{m v^{2}}{r} \Rightarrow v \leq \sqrt{r g}$

$\Rightarrow v_{\min} = \sqrt{r g} = \sqrt{1.6 \times 10} = 40 m s^{- 1}$ .

B M Sharma Solutions for Chapter: Circular Motion, Exercise 38: ILLUSTRATION 9.36

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Circular Motion, Exercise 38: ILLUSTRATION 9.36

Questions from B M Sharma Solutions for Chapter: Circular Motion, Exercise 38: ILLUSTRATION 9.36 with Hints & Solutions

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