Translational Speed of a Bowling Ball

AI Thread Summary
A bowling ball with a mass of 6.95 kg and a radius of 0.194 m rolls down a 49.2-degree slope from a height of 2.21 m, and the discussion revolves around calculating its translational speed upon leaving the incline. The initial calculations incorrectly used the moment of inertia for a solid sphere (2/3), which led to an incorrect speed of 5.1 m/s. After realizing the correct moment of inertia for a solid sphere is actually 2/5, the calculations were adjusted. This correction significantly impacted the final result, leading to a successful resolution of the problem. The discussion highlights the importance of using the correct formulas and values in physics calculations.
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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)


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!
 
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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
 
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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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