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Rotational Motion FInding the total ENergy

  • Thread starter Schu
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  • #1
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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??
:confused:
 

Answers and Replies

  • #2
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I need help ASAP

Is anyone out there????

I would appreciate the help :confused:
 
  • #3
Galileo
Science Advisor
Homework Helper
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Looks ok to me. (I got 3.86 m/s, by rounding off)
 
  • #4
Doc Al
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rotational KE

I didn't check your arithmetic, but I have some comments.
Schu said:
Rotational Kinetic Energy: 1/5mv^2
The rotational KE is [itex]{KE}_{rot} = 1/2 I \omega^2[/itex].

You will also need the "rolling condition": [itex]V = \omega R[/itex].
Is the Sum of the inital energy's = to the SUM of the final energy's?
Yes, if you assume energy is conserved, which seems reasonable for this problem.
 

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