Right. That did it. Thanks a lot
Here is another problem I've worked on where I need some closure:
A skier of mass m starts from rest at the top of a solid sphere of radius r and slides down its frictionless surface. (a) At what angle q will the skier leave the sphere? (b) If friction were present, would the skier fly off at a greater or lesser angle?
(a) There are three forces acting on m, it's weight, the normal force N and the centripetal force F. Picking up and right as the positive directions, the centripetal force vector F
can be written as F
= (-F sin(q), -F cos(q)) and N
= (N sin(q), N cos(q)). As the skier slides down the sphere N = F until the skier flies off the sphere and so N = 0. Using Newton's second law gives:
In the horizontal direction: ma = N sin(q) - F sin(q) = -F sin(q).
In the vertical direction: ma = N cos(q) - F cos(q) - mg => F cos(q) = -mg.
Since F = mv^2/r, then F cos(q) = -mg => cos(q) = -rg/v^2.
Let mgh be the potential energy of the skier initially so that when the skier leaves the sphere, his/her potential energy is zero and his/her kinetic energy is 1/2mv^2. r = h + r cos(q) => h = r(1 - cos(q)). Using conservation of energy, mgh = 1/2mv^2 => v^2 = 2gh = 2gr(1 - cos(q)). Substituing this in the equation above yields:
cos(q) = -rmg/v^2 = -rg/(2gr(1 - cos(q))) => cos(q) - cos^2(q) = -1/2. Solving this gives cos(q) = -(1 + sqrt(3))/2, -(1 - sqrt(3))/2. The former answer makes no sense, so cos(q) = -(1 - sqrt(3))/2 => q is roughly 70 degrees.
(b) Since friction reduces the velocity of the skier, it will take a greater angle for the skier to fly of the sphere.