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
$v_{\min }=\sqrt{gR}$
$\Rightarrow v_{\min }\propto \sqrt{R}$
So, $\frac{{\left(v_{\min }\right)}_1}{{\left(v_{\min }\right)}_2}=\sqrt{\frac{R_1}{R_2}}$
$\Rightarrow \frac{v}{{\left(v_{\min }\right)}_2}=\sqrt{\frac{R}{2R}}$
$\Rightarrow \frac{v}{{\left(v_{\min }\right)}_2}=\frac{1}{\sqrt{2}}$
$\Rightarrow {\left(v_{\min }\right)}_2=\sqrt{2}v$