# Homework Help: Simple Pinball Spring Problem

1. Dec 2, 2005

### Nm

Let's say I have a pinball (solid sphere) with a given mass and radius, and it is held back at a certain distance from equilibrium on a spring. The spring constant k is given, and the ball is then fired onto the pinball machine surface, which is tilted. How would I find the maximum height the pinball rolls to (above its initial location.)? I know I should use energy conservation, but is this the correct formula: 0.5m(vi^2) + 0.5k(yi^2) = 0.5m(vf^2) + 0.5k(yf^2)? Would yf be the same as the height at which the pinball machine is tilted or is it the distance from the spring equilibrium the pinball travels to? Thanks.

2. Dec 2, 2005

### mrjeffy321

The potential energy stored in the spring as the ball is pushed and held back is then converted into kinetic energy as the ball is moving up the machine, then into gravitational potential energy as it moves up the surface.

The potential energy os a spring is,
PE_s = 1/2 k*x^2
The gravitational potential energy is,
PE_g = mgh

so if you set these equal to each other,
1/2*k*x^2 = mgh
now solve for the height h (or y, whatever you want to call it)

3. Dec 2, 2005

### Nm

Thanks for your help. I have another question: When the pinball is at a certain distance above its initial location (on the way up), how would I find its speed? Do I set KE = PE of the spring?

4. Dec 2, 2005

### Leong

in this case,
elastic potential energy of the spring = kinetic energy of the ball + its gravitational potential energy.

assumptions we have to made in order to use the law of conservation of energy:
1. the surface of the machine has no friction.
2.only elastic potential energy, kinetic energy and potential energy are involved. no sound energy is produced.

what i know is that the h in the formula mgh refers to the vertical distance or height. so i think it is important to know the angle the machine is tilted.

5. Dec 2, 2005

### mrjeffy321

Another thing you assume by saying that the,
"elastic potential energy of the spring = kinetic energy of the ball + its gravitational potential energy." is that the ball is not spinning and/or it is a point mass.
I only bring this up because in the original statement of the problem, Nm said that the ball had a given radius (amongst other things). This almost seems to imply that we are suppose to take the rotation of the ball into account. If this is indeed the case, then not only does the ball have gravitational potential energy, and [normal] translational kinetic energy, but also rotational kinetic energy (the "spin" energy if you will).