A person starts from rest at the top of a large frictionless spherical surface, and slides into the water below at what angle θ does the person leave the surface? (Hint: When the person leaves the surface, the normal force is zero.)
let the person leaves at angle theta
mg cos(theta) - N = mv^2 / r
since at this theta , N = 0
=> mg cos(theta) = mv^2 / r
=> g cos(theta) = v^2 / r
=> v^2 = gr cos(theta) -------------eq1
also
dKE + dPE = 0;
=> 1/2mv^2 - mg r ( 1 -cos(theta) ) = 0
=> 1/2 v^2 = g r ( 1 -cos(theta) )
using eq1 ,
=> 1/2 gr cos(theta) = gr ( 1 - cos(theta) )
=> 1/2 cos(theta) = ( 1 - cos(theta) )
=> 3/2 cos(theta) = 1
=> cos(theta) = 2 /3
=> theta = 48.18 degree
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