# Need some closure

1. May 23, 2004

### e(ho0n3

Hi everyone,

I need some closure on this problem:

An engineer is designing a spring to be palced at the bottom of an elevator shaft. If the elevator cable should happen to break when the elevator is at a height h above the top of the spring, calculate the value of the spring constant k so that the passengers undergo an acceleration of no more than 5.0 g when brought to a rest. Let M be the total mass of the elevator and passengers.

Let y be the distance of the elevator from the spring. There are three locations of interest here:
Loc. 1 - When the elevator is at a height y = h.
Loc. 2 - When the elevator is at a height y = 0.
Loc. 3 - When the elevator is at a height of y = -x, i.e. when the elevator is at rest a distance x below the uncompressed height of the spring.

The total mechanical energy at
Loc. 1 is Mgh
Loc. 2 is 1/2Mv^2
Loc. 3 is 1/2kx^2 - Mgx

Now assuming the acceleration of the elevator is constant when on the spring, then
0 = v^2 - 2ax => v^2 = 2ax

Using conservation of energy, then
Mgh = 1/2Mv^2 = Max => x = gh/a
1/2Mv^2 = Max = 1/2kx^2 - Mgx => 1/2kx^2 = M(ax + gx) so
kx = 2M(a + g) => kgh/a = 2M(a + g) => k = 2Ma(a + g)/ (gh)

Since a <= 5g, then k <= 10Mg (6g) / (gh) = 60Mg/h. The book where I got this problem from claims the anwer to be 12Mg/h. I don't think I did anything wrong here though. What gives?

e(ho0n3

2. May 23, 2004

### Staff: Mentor

acceleration is not constant

Right!
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.

3. May 23, 2004

### e(ho0n3

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

Last edited: May 23, 2004
4. May 24, 2004

### Staff: Mentor

Skier on a sphere

You are very welcome. You will get plenty of help here, from me and many other members. But I ask you to do two things to make it easier for everyone:
(1) Start each new problem in new thread (in College Level Help, of course)
(2) Give the thread a descriptive title. For example: "Elevator and spring" or "Skier on a sphere". (If every question is titled "Problem" or "Need help", they all tend to blur together. )
Several things. First, centripetal force is not a separate force! It is just the name given to those forces that pull something towards the center. There are only two forces acting on the skier: The weight, acting down, and the normal force. The skier begins to lose contact with the sphere when the normal force becomes zero. At that point, the only force providing the centripetal force is the component of the weight acting radially.

So... use what you know about centripetal acceleration and combine that with conservation of energy. You'll be able to solve for θ.

Right.

5. May 24, 2004

### e(ho0n3

Re: Skier on Sphere

Gotcha.

Right, I completely forgot about that. OK, so the weight acting in the radial direction has magnitude mgcos(θ) and so
mgcos(θ) = mv^2/r​
Now v^2 = 2gr(1 - cos(θ)) (see my previous posting) and so
cos(θ) = 2(1 - cos(θ)) = 2 - 2cos(θ)​
3cos(θ) = 2 => cos(θ) = 2/3 => θ = 48.2 degrees​

Thanks again,
e(ho0n3