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Linear Speed

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data
    A solid ball is at the top of a 2 meter inclined plane. Assuming friction is negligible, what is its linear speed (m/s) as it reaches the bottom?


    2. Relevant equations
    PE=mgh
    KE=.5mv^2
    RKE=.5Iw^2
    I of ball=.4mr^2
    w=v/r

    3. The attempt at a solution
    Pretty much what i did was set PE=KE+RKE
    which is mgh=.5mv^2+.5(.4mr^2)+w^2 masses cancel out and so does r when v/r is replaced with w

    equation is now gh=.5v^2+.5(.4)v^2
    if i solve for v i get: 2v^2=2gh/.4= v^2=gh/.4

    thus v=sqrt(9.81*2/.4)

    v=7.0 m/s

    i really have no idea if im doing this right...
     
  2. jcsd
  3. Feb 3, 2010 #2

    kuruman

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    In such problems, I find statements like "assume that friction is negligible" confusing.

    If friction is really negligible, then the ball slides down the incline without rolling. In that case, it might as well be replaced by a sliding block and there is no ω.

    If by "assume that friction is negligible" the intent is to indicate that the ball is actually rolling without slipping but frictional losses are negligible (i.e. mechanical energy is conserved), then the problem should state so more clearly.

    Your method is correct if you assume that the latter interpretation applies.
     
  4. Feb 3, 2010 #3
    Yes, your second statement was correct. As for the problem not being specific it's my teacher not me so talk to him :). As for the answer apparently mine is wrong. So if you have any ideas about something I've missed I'd be glad to hear it
     
  5. Feb 4, 2010 #4

    kuruman

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    The first line is correct. Can you show me how you get from the first line to v2=gh/0.4?
     
  6. Feb 4, 2010 #5
    From gh=.5v^2+.5(.4)v^2 I multiplied both sides by 2 to get rid of the .5: 2gh=v^2+.4v^2. Then I divided by .4 and added the v^2s together: 2gh/.4=2v^2. Divided both sides by 2 again to get: gh/.4=v^2.
     
  7. Feb 4, 2010 #6

    kuruman

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    Not correct. First you add the v2, then you divide by whatever multiplies the v2. In other words, you factor out the v2 term to get v2(1+0.4).
     
  8. Feb 4, 2010 #7
    So that means i have 2gh=v^2(1+.4). Then divide by (1+.4): 2gh/(1+.4)=v^2 and vinally solve for v. which is v=sqrt(2*9.81*2)/(1+.4)

    so v=5.3 m/s! Thanks a lot :)
     
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