Question

A particle of mass m attached to a string of length l is describing circular motion on a smooth plane inclined at an angle $\alpha$ with the horizontal. For the particle to reach the highest point its velocity at the lowest point should exceed :

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Here, h=2l sin ∝

A is the lowest point and B the highest point. At B, in critical case tension is zero. Let velocity of particle at B at this instant be $v_B$. Then

$mgsin\alpha =\frac{m{v}_B^2}{l}\text{o}r{v}_B^2=glsin\alpha Now{v}_A^2=v_B^2+2gh=\left(glsin\alpha \right)+2g\left(2lsin\alpha \right)\therefore v_A=\sqrt{5glsin\alpha }$