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Rotational Motion FInding the total ENergy 
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#1
Aug1204, 10:18 AM

P: 12

Particulars:
ball has a radius of 2.5 cm a mass of .125 and is rolling across a table with a speed of .547 m/s, this table is 1.04 m off the ground. It rolls to the edge and down a ramp How fast will it be rolling across the floor? First I found the Gravitational Potential Energy: Ep=mgh Initial of 1.2753 FInal = 0 THen the Linear Kinetic ENergy : 1/2 mv^2 Initial .0187005625 FInal .0625v^2 Elastic Potential Energy: .5k(delta)x^2 0 0 Rotational Kinetic Energy: 1/5mv^2 initial .007480225 FInal .025v^2 Now I need to bring them all togther and solve the final velocity. Is the Sum of the inital energy's = to the SUM of the final energy's? If that's true then 1.30148075 = .0875v^2 so v = 3.85 m/s Is that at all right?? 


#2
Aug1204, 07:01 PM

P: 12

Is anyone out there????
I would appreciate the help 


#3
Aug1304, 02:29 AM

Sci Advisor
HW Helper
P: 2,002

Looks ok to me. (I got 3.86 m/s, by rounding off)



#4
Aug1304, 08:43 AM

Mentor
P: 41,440

Rotational Motion FInding the total ENergy
I didn't check your arithmetic, but I have some comments.
You will also need the "rolling condition": [itex]V = \omega R[/itex]. 


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