Translational Speed


by Kelschul
Tags: speed, translational
Kelschul
Kelschul is offline
#1
Mar12-07, 09:44 PM
P: 5
A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 7.88 m/s at the bottom of the rise. Find the translational speed at the top.
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Kurdt
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#2
Mar13-07, 07:11 AM
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You really need to show some working before you can receive any help. As a hint you can think about the conservation of energy.
despanie
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#3
May5-07, 03:12 PM
P: 4
Quote Quote by Kelschul View Post
A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 7.88 m/s at the bottom of the rise. Find the translational speed at the top.
HOW IS THE EQUATION SETUP FOR THIS PROBLEM

despanie
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#4
May5-07, 03:23 PM
P: 4
Angry

Translational Speed


Quote Quote by Kelschul View Post
A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 7.88 m/s at the bottom of the rise. Find the translational speed at the top.
kE1+PE1=KE2+PE2
1/2(MKG)(7.880)^2+(MKG)(9.8)(.76M)=1/2MKG(V)^2+MKG(9.8)(X)
despanie
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#5
May5-07, 03:25 PM
P: 4
Quote Quote by despanie View Post
kE1+PE1=KE2+PE2
1/2(MKG)(7.880)^2+(MKG)(9.8)(.76M)=1/2MKG(V)^2+MKG(9.8)(X)
WILL THIS EUATION WORK FOR THE SPEED
hage567
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#6
May5-07, 03:33 PM
HW Helper
P: 1,542
You're on the right track, but you mixed up your gravitational potential energy terms a bit. If you take the bottom of the rise as your reference point, what is the gravitational potential energy of the ball before it goes up the rise?


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