  Question

A particle is attached with a string of length $$l$$ which is fixed at point O on an inclined plane. What minimum velocity should be given to the particle along the incline so that it may complete a circle on inclined plane? (The plane is smooth and initially particle was resting on the inclined plane.) A
5gl  B
5gl2  C
53gl2  D
4gl  Solution

The correct option is A $$\sqrt{\dfrac{5gl}{2}}$$For a particle to complete vertical circle,$$\dfrac{1}{2}mv_{bottom}^2 = mg\times 2l + \dfrac{1}{2}mv_{top}^2$$$$\dfrac{mv_{top}^2}{l} = mg$$Solving, $$v_{bottom} = \sqrt{5gl}$$In the given problem, the circle is inclined at $$30^o$$ to the horizontal which changes the effective acceleration due to gravity. Only the component of $$g$$ along the plane has an effect on centripetal acceleration. The component of $$g$$ perpendicular to the plane is balanced by the normal reaction. $$\therefore { g }_{ eff }=gsin\theta =gsin{ 30 }^{ 0 }=g/2\\ { v }_{ min }=\sqrt { 5{ g }_{ eff }l } =\sqrt { \dfrac { 5gl }{ 2 } }$$Physics

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