# Rotational Energy and Linear Momentum Problem

1. Nov 16, 2013

### Arixal

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

In the figure here, a small, solid, uniform ball is to be shot from point P so that it rolls smoothly along a horizontal path, up along a ramp, and onto a plateau. Then it leaves the plateau horizontally to land on a game board, at a horizontal distance d from the right edge of the plateau. The vertical heights are h1 = 4.5 cm and h2 = 1.00 cm. With what speed must the ball be shot at point P for it to land at d = 3.5 cm?

2. Relevant equations

E'=E
Translational KE = mv^2/2
Rotational KE = Iω^2/2
y = y0 + v0T + aT^2/2

3. The attempt at a solution

(0.5)mv0^2 = (0.5)mv'^2 + mgh1
Which simplifies to v0 = sqrt(2gh1 + v'^2) (1)

To find v': d = v'T => v' = d/T
To find T: h2 = (0.5)gT^2 => T = sqrt(2h2/2)

So, v' = d/sqrt(2h2/2) which is approximately 0.77 m/s.

Plugging this into (1) v0 = 1.217 m/s.

However, this is not the right answer.

So, I then attempted it again this time taking into account that the moment of inertia of a solid sphere is (2/5)mR^2.

(0.5)(2/5)(mR^2)ω0^2 = (0.5)(2/5)(mR^2)ω'^2 + mgh1

v = rω
(1/5)(mR^2)v0^2/R^2 = (1/5)(mR^2)v'^2/R^2 + mgh1
v0^2 = v'^2 + 5gh1
v0 = sqrt(v'^2 + 5gh1)

Then v' is the same in this attempt as the previous, so v0 = 1.67 m/s.
However, this is also not the right answer.

It appears there is not enough information to take into account energy lost to friction so I'm assuming that's negligible. So, I'm at a loss.

2. Nov 16, 2013

### lucasem_

You also have to take into account the linear energy at P.
$$K_P=\frac{1}{2}mv_i^2 + \frac{1}{2}I\omega_i^2$$
Also at the top, consider the energies for gravity, movement, and rotation.
$$mgh_1+\frac{1}{2}mv_2^2+\frac{1}{2}I\omega_2^2$$
When the ball then falls, It must have a final energy greater than $mgh_2$ (so it could be rolling along and make it a distance d away, not just survive the fall.) You have already shown that you know I for a sphere to be $\frac{2}{5}mr^2$, and $\omega$ can be calculated as $v/r$.
When you put the equations together you will be able to calculate the correct answer.

3. Nov 16, 2013

### Arixal

Oh shoot, that's obvious. I don't know why when I remembered to take into account rotational kinetic energy I dropped off translational kinetic energy.

Thank you.