Σ F = N − mg ∴ N = mg when they are at the top, where N is normal force.
When they start to slide:
you get mgcosθ - N = mv2/r and they will "fall off" when the N value equals zero, right?
and solve for your velocity. Then you can use conservation of energy
(your "height" will be in terms of r and θ to solve for the angle.
he person
leaves the surface of the slide at the instant the normal force is
equal to the centripetal force. After leaving the slide, the normal
force goes to zero. In this case it is not necessary to define PE =
zero at h =0.
At the top of the slide, before the person begins their journey,
the force of gravity acting downward is balanced by the normal force
acting upward.
ΣF = N -mg
Thus N = mg
As the person moves down the slide, a component of gravity mgcos(theta) supplies the centripetal force. The normal force is:
mgcos(theta) -N = mv^2 / r
N = mgcos(theta) - mv^2/r
Finally at the instant the person leaves the surfae of the slide N =0
mv^2 / r = mgcos(theta)
v^2 = grcos(theta)
Conservation of Mechanical Energy:
mghi = 1/2mv^2 +mghf
mgr = 1/2mv^2 + mg4cos(theta) [in terms of r, and theta]
mgr = 1/2mgrcos(theta) +mgrcos(theta) [v^2 = grcos(theta)]
1 = 1/2cos(theta) +cos(theta) = 3/2 cos(theta)
theta = cos-1 (2/3) = 48 degrees