# Ball rolling down an incline

1. Oct 25, 2007

### delecticious

1. The problem statement, all variables and given/known data
A ball of mass 2.80 kg and radius 0.148 m is released from rest on a plane inclined at an angle θ = 38.0° with respect to the horizontal. How fast is the ball moving (in m/s) after it has rolled a distance d=1.55 m? Assume that the ball rolls without slipping, and that its moment of inertia about its center of mass is 1.50E-2 kg·m2.

2. Relevant equations

KE - kinetic energy
KEr - rotational kinetic energy
PE - potential kinetic energy

KE0 + KEr0 + PE0 = KEf + KErf + PEf

3. The attempt at a solution

I thought I had it right but it seems I went wrong somewhere and I'm assuming that it's probably where the potential energy is concerned. Since, the entire hypotenuse isn't given and all that is given is the distance d, I readjusted the axis for potential energy and made the point where the ball goes down to in the picture the ground level so that PE at that point equals 0. The initial KE is 0 as well as the initial rotational KE, and since I made the final PE equal 0 I got an equation PE0 = KEf + KErf. To find the height for the initial PE I did sin(theta) times the d. I guessing that that's the wrong thing to do. Anyone know the right way to go about it?

2. Oct 25, 2007

### delecticious

help?

3. Oct 25, 2007

### Shooting Star

You have done everything correctly. Repeat the calculations to see if you've made any arithmetical mistake.