A small mass particle is given an initial velocity v 0 tangent to the horizontal rim of a smooth hemispherical bowl at a radius r 0 from the vertical centerline, as shown at point A . As the particle slides past point B, a distance h below A and a distance r from the vertical centerline, its velocity v makes an angle θ with the horizontal tangent to the bowl through B. Determine θ .
Text Solution
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Sol. The forces on the particle are its weight and the normal reaction exerted by the smooth surface of the bowl. Neither force exerts a moment about the axis O–O, so that angular momentum is conserved about that axis. Thus,
[(H O ) 1 = (H O ) 2 ] mv 0 r 0 = mvr cos θ
Also, energy is conserved so that E 1 = E 2 . Thus
[T 1 + 
= T 2 + 
] 
mv 0 2 +mgh = 
mv 2 + 0
v = 
Eliminating v and substituting r 2 = r 0 2 – h 2 give
v 0 r 0 = 
cos θ
θ = cos –1 
Ans.
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