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Rotational Motion FInding the total ENergy

  1. Aug 12, 2004 #1
    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:
     
  2. jcsd
  3. Aug 12, 2004 #2
    I need help ASAP

    Is anyone out there????

    I would appreciate the help :confused:
     
  4. Aug 13, 2004 #3

    Galileo

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    Science Advisor
    Homework Helper

    Looks ok to me. (I got 3.86 m/s, by rounding off)
     
  5. Aug 13, 2004 #4

    Doc Al

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    Staff: Mentor

    rotational KE

    I didn't check your arithmetic, but I have some comments.
    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].
    Yes, if you assume energy is conserved, which seems reasonable for this problem.
     
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