Question

Consider a particle is moving with a minimum speed ' $v$' and is at highest point of vertical circle of radius ' $R$ '. If the radius of the circle doubled the corresponding minimum speed will be

• A. $v$
• B. $\frac{v}{\sqrt{2}}$
• C. $\sqrt{3} v$
• D. $\sqrt{2} v$

Answer 1

Answer: (D)

When a particle moves along a vertical circle, then it has its minimum speed at the highest point of the vertical circle and in this case it just completes the loop. The minimum speed is
$v_{\min }=\sqrt{g R} $
$\Rightarrow v_{\min } \propto \sqrt{R}$
So, $\frac{\left(v_{\min }\right)_{1}}{\left(v_{\min }\right)_{2}}=\sqrt{\frac{R_{1}}{R_{2}}} $
$\Rightarrow \frac{v}{\left(v_{\min }\right)_{2}}=\sqrt{\frac{R}{2 R}}$
$\Rightarrow \frac{v}{\left(v_{\min }\right)_{2}}=\frac{1}{\sqrt{2}} $
$\Rightarrow \left(v_{\min }\right)_{2}=\sqrt{2} v$