Question

A particle enters into a smooth frictionless circular loop of radius R, at point P as shown below. If g is the acceleration due gravity, then the minimum speed with which the particle should enter the circular loop such it can complete one full circular revolution is

• A. $\sqrt{5Rg}$
• B. $\sqrt{3Rg}$
• C. $\sqrt{2Rg}$
• D. $\infty$

Answer 1

The correct option is A $\sqrt{5Rg}$



$\frac{m{v}^2}{R}=mg\text{(at point A)}$

$\therefore v=\sqrt{Rg}$

By applying the energy principle,

$\frac{1}{2}m{v}^2=mg\times 2R+\frac{1}{2}m\times Rg$

$=\frac{5}{2}mgR$

$\therefore v=\sqrt{5gR}$