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e(ho0n3
e(ho0n3 is offline
#3
May23-04, 10:08 PM
P: 1,370
Quote Quote by Doc Al
Wrong! The acceleration is not constant. As the spring is compressed, the force it exerts increases. The maximum force on the elevator (and thus the maximum acceleration) will be when the compression is maximum. (Apply F = ma at that point.) Combine this with conservation of energy and you will get the book's answer.
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.

e(ho0n3