# Ball rolling down a slope

1. Apr 16, 2009

### nisse pisse

1. The problem statement, all variables and given/known data

If you put a ball somewhere on a slope and it rolls down, how far will it go when the ramp is at some hight over the ground and the output angle is something.Do not take in account friction or air resistance.

2. Relevant equations

3. The attempt at a solution

When u put the ball in the slope it should have the possible energy of mgh relative to the output, and at the putput it should be (mv^2)/2.
So if the output angle should be 0 the output speed should be (2gh)^0.5.
Then the distance should be the ((2gh)^0.5)*(2s/g)^0.5 where s is the height from the output from the slope to the ground.
I wonder how the output angle will effect this, the ball will get some speed vertically aswell.
Some tips or help would be nice :)

Last edited: Apr 16, 2009
2. Apr 16, 2009

### Jack21222

I don't quite understand this problem, because if you're ignoring air resistance and friction, wouldn't it just go on forever? Is the question really asking for the distance the ball will roll?

3. Apr 16, 2009

### nisse pisse

Im sry, I mean when the ball first touches the ground.The slope is above the ground at some height (s) .

4. Sep 29, 2009

### woodie37

Suppose the ball starts at height, h, on the incline, inclined at angle x. We can make a list of facts first

-energy is conserved and transfered
-gravitational potential energy is converted into linear and rotational kinetic energy
-assuming that friction is strong enough that the ball does not slip, then it will not lose energy to friction
-the total change in energy is equal to the change in gravitational potential energy, rotational kinetic, and linear kinetic, and they must sum to zero as total energy is conserved

so we have 0 = $$\Delta$$U + $$\Delta$$K + $$\Delta$$R, where U is the potential energy on the incline, K is the linear kinetic energy and R is rotational kinetic.

so we have 0 = mg(0 - h) + 0.5m(v$$^{2}$$ - o) + 0.5I$$\omega$$$$^{2}$$
2gh = v$$^{2}$$ + Iv$$^{2}$$/r$$^{2}$$

Solving for v we get

2gh = (1 + I/r$$^{2}$$)v$$^{2}$$
v$$^{2}$$ = 2gh/(1 +I/r$$^{2}$$)

Note that the angle x does not matter. This is because the length of the slope on which the ball rolls down does not affect the final velocity. Also note that v = r$$\omega$$[tex], which tells us that the ratio of linear velocity and angular velocity is constant, independent of length of movement along the slope.