Question

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.)*

Answer #1

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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