Angular + Linear Velocity problem

[GeorgeCostanz]

I kno this is pretty simple and the answer is probably staring me in the face, but I'm lost for some reason. Our teacher gave back our tests so i have the answer, but I'm not sure how to get it. I haven't had a chance to speak to my teacher yet, but i intend to tomorrow if no one helps me first.

Homework Statement

A bowling ball [(mass = 7 kg)(radius = .50m)] rolls without slipping down a 3m high ramp. Starting from rest, find the velocity at the bottom of the ramp.

Homework Equations

Bowling ball inertia = I = (2/5)(M)(R)^2 = (0.7)

The Attempt at a Solution

I used conservation of angular momentum

(1/2)(I-initial)(omega-initial)$^{2}$ + mg(h-initial) = (1/2)(I-final)(omega-final)$^{2}$ + mg(h-final)

after calculating i got omega-final = 24 (actually 24.25, but he instructed us to round to the nearest whole number)
i converted that to v = 12

which is wrong

i was supposed to set it up as:

(1/2)(I-initial)(omega-initial)$^{2}$ + mg(h-initial) = (1/2)(I-final)(omega-final)$^{2}$ + mg(h-final) + (1/2)m(v)$^{2}$

and v = 6.5 is the answer

i was under the impression we were supposed to find the angular velocity at the bottom of the ramp and convert to linear velocity

i'm not sure how he got v = 6.5

 

[Doc Al]

In your solution, you ignored the translational motion of the ball. The ball ends up with both rotational speed and translational speed. Hint: How are those two speeds related for rolling without slipping?

 

[GeorgeCostanz]

are you referring to v = rω?

 

[Doc Al]

are you referring to v = rω?

Yes.

 

[GeorgeCostanz]

so would it look like:

(1/2)(I-initial)(omega-initial)$^{2}$ + mg(h-initial) = (1/2)(I-final)(omega-final)$^{2}$ + mg(h-final) + (1/2)m(r)$^{2}$(omega-final)$^{2}$

just worked it out and got the right answer
i actually set it up like that last night and worked it out but just now realized i made a simple error
hate when that happens

thanks doc!