Question

A is the bottom most point of a particle describing a vertical circle of radius R. At point B, the acceleration of the particle is $g\sqrt{11}$. Find the velocity of the particle at top point C :

• A. ${\left[gR\right(\sqrt{10}-2\left)\right]}^{\frac{1}{2}}$
• B. ${\left[gR\right]}^{\frac{1}{2}}$
• C. ${\left[gR\right(\sqrt{11}-2\left)\right]}^{\frac{1}{2}}$
• D. None of these

Answer 1

The correct option is A ${\left[gR\right(\sqrt{10}-2\left)\right]}^{\frac{1}{2}}$

Net acceleration at point B is given by
$a=(a_C^2+a_T^2{)}^{\frac{1}{2}}=g\sqrt{11}$
But $a_T=g$
$a_C^2=11g^2-g^2=10g^2$
$a_c=g\sqrt{10}=\frac{V_B^2}{R}$
$V_B^2=gR\sqrt{10}$
Using kinematics equation,
$V_C^2=V_B^2-2gR=gR(\sqrt{10}-2)$
$V_C=\sqrt{gR(\sqrt{10}-2)}$