Angular + Linear Velocity problem

  • #1
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 havent 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)[itex]^{2}[/itex] + mg(h-initial) = (1/2)(I-final)(omega-final)[itex]^{2}[/itex] + 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)[itex]^{2}[/itex] + mg(h-initial) = (1/2)(I-final)(omega-final)[itex]^{2}[/itex] + mg(h-final) + (1/2)m(v)[itex]^{2}[/itex]

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
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
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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?
 
  • #3
are you referring to v = rω?
 
  • #5
so would it look like:

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

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!
 

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