Ball rolling down an incline: How to calculate final velocity?

[jti3066]

Homework Statement

A ball of mass 2.80 kg and radius 0.153 m is released from rest on a plane inclined at an angle θ = 41.0° with respect to the horizontal. How fast is the ball moving (in m/s) after it has rolled a distance d=1.60 m? Assume that the ball rolls without slipping, and that its moment of inertia about its center of mass is 1.80E-2 kg·m2.

Homework Equations

I = Icm + M(d^2)...parallel-axis theorem

wf ^2 = wi^2 + 2a(thetaF - thetaI)...constant angular acceleration

theta = S/r

v=wr

The Attempt at a Solution

I think I need to use the parallel-axis theorem to solve for one step in the problem, so...

I = 0.0180 + 2.8(1.6^2) = 7.186 kg * m^2

Next I found how many "radians" the ball travels in 1.6 m...

C = 2"pi"r = 2"pi" * .153m = .936 m (circumference of ball)

1rad = .936 m

1.6m * (1rad/.936m) = 1.7 radians

I am not even sure how to solve this problem...I did the above equations so I could see if I was going on the right track by a process of elimination...usually I don't worry about the actual numbers until the end of the problem...i.e. if I just use the units in the equations and my "answer" is in the correct units it usually means I am on the correct path...Please help me solve this problem...

 

[rl.bhat]

Apply the conservation of energy.
The potential energy = Translational kinetic energy + rotational kinetic energy.
Angle is given, displacement along the inclined plane is given. From that find height through which the ball rolls.

 

[jti3066]

Ok...thanks

mg(d*sin(theta)= .5m *v^2 + .5I(v/r)^2