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Translational Speed of a Bowling Ball

  • Thread starter deltas
  • Start date
7
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1. A bowling ball with a mass of 6.95 kg and a radius of 0.194 m starts from rest at a height of 2.21 m and rolls down a 49.2 degree slope. What is the translational speed of the ball when it leaves the incline?


2. Unsure! Here's what I've got, though:
ω = v/r
mgh=1/2mv2+1/2Iω2
And so: ω= √(2mgh/(mr2+I))
I of a solid sphere = 2/3(m)(r2)


3. The Attempt at a Solution
I = 2/3(6.95)(.1942) = 0.174
ω= √(2(6.95)(9.8)(2.21)/((6.95)(.1942+0.174)) = √(301.05)/(0.436) = 26.29
v = ωr
v = 26.20 X 0.194 = 5.1
Alas, 5.1 is not the correct answer. I'm sure I need to factor in the angle somewhere, but I don't know how! Help, please!
 
Are you sure the moment of inertia for a solid sphere is 2/3?
I thought that was for a hollow sphere.
Try 2/5?

http://img80.imageshack.us/img80/5226/picture2hr.png [Broken]
 
Last edited by a moderator:
7
0
Oh my goodness, has this been my problem the whole time?

I tried it, it was! You are such a lifesaver, thank you so much! I couldn't for the life of me figure out what I was doing wrong! Thanks again!
 

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