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posted by  Fl7675 on 11/7/2007 7:55:17 PM  |  status: Closed  

rotational kinetic energy

Course Textbook Chapter Problem
Algebra Based Physics N/A N/A N/A
Question Details:
A solid bowling ball with a radius of 10.9 cm and a mass of 6.9 kg rolls along a bowling alley at a linear speed of 3.5 m/s.
(a) What is its translational kinetic energy?
correct check mark J
(b) What is its rotational kinetic energy?
wrong check mark J
Can't figure out the rotational kinetic energy for anything... can someone solve it for me and show steps? thanks!

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posted by ktqu on 11/7/2007 8:09:33 PM  |  status: Live
Asker's Rating: Lifesaver   
Fl7675's comment:
"thank you!"
Response Details:
Rotational kinetic energy is equal to 1/2*Iω2.
ω is the angular velocity and is equal to v/r.
I is the moment of inertia. The moment of inertia for a solid sphere is 2/5*MR2
 
ω = (3.5/0.109m) = 32.1 rad/s
I = 2/5 (6.9kg)(0.109m)2 = 0.0328 kg m2
 
KEr = 1/2 (0.0328)(32.1)2 = 16.9 J
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