Question

A small stone of mass $200\text{g}$ is tied to one end of a string of length $80\text{cm}$. Holding the other end in the hand, the stone is whirled in a vertical circle. What is the minimum speed that needs to be imparted at the lowest point such that the stone is just able to complete the vertical circle?(Take $g=10\text{m}/s^2$)

• A. $4\text{m/s}$
• B. $5.13\text{m/s}$
• C. $6.32\text{m/s}$
• D. $7\text{m/s}$

Answer 1

The correct option is C $6.32\text{m/s}$


At the top most point,
$T+mg=\frac{m{v}_A^2}{R}$
If stone is just able to complete vertical ciruclar motion, then at the top most point, $T=0$
$\Rightarrow mg=\frac{m{v}_A^2}{R}$
$\Rightarrow v_A=\sqrt{Rg}$

Applying law of conservation of energy at A & B (Taking reference point at $\text{A}$)
$U_A+K{E}_A=U_B+K{E}_B$
$\Rightarrow mg\left(2R\right)+\frac{1}{2}m{V}_A^2=0+\frac{1}{2}m{V}_B^2$
$\Rightarrow V_B=\sqrt{5gR}$
$=\sqrt{5\times 10\times 0.8}$
$=6.32\text{m/s}$