A heavy ball of mass m slides without friction down an inclined chute which forms a loop of radius R (fig.).
At what height will the ball leave the chute and to what maximum height will it rise afterwards if it begins to run down the chute without initial velocity from a height h = 2R? Consider the size of the ball as negligible.
Text Solution
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Sol. The condition for the body to be detached from the sphere is F = 0 or mg cos α = 
considering that cos α = 
we find that the body will detached at a distance from the top equal to h 1 = 
R
H 1 = 
R The ball will move from the point C (Fig.) in a parabola with initial velocity
v 0 = 
= 
directed at an angle α which can be determined from the equation
cos α = 
= 
At the maximum elevation of the parabola the velocity of the ball will be equal to the horizontal component of velocity v 0 , i.e.,
v x = v 0 cos α = 
It follows from the law of conservation of energy that at this moment the ball should be at such a distance h 2 along the vertical from the point A the v x 2 = 2gh 2 . Hence h 2 = 
= 
R
And H 2 = 2R – h 2 = 
R
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