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Centripetal acceleration and KE problem

  1. Jan 23, 2006 #1
    The question and information given is -
    "The formula for the centripetal acceleration experienced by a body moving at speed v in a circle of radius r is ac = v squared / r

    A small mass m slides without friction along a loop-the-loop track. The circular loop has radius R. The mass starts from rest at point P a distance h abouve the bottom of the loop.

    a. What is the KE of m when it reaches the top of the loop?
    b. What is its acceleration at the top of the loop assuming that it stays on the track?
    c. What is the least value of h if m is to reach the top of the loop without leaving the track?"

    PE = mgh at point P, so that would be 9.8mh, and KE = 0
    a. At the top of the loop, how would you figure out the KE? If KE = 1/2mv2, it would be 1/2m and v2 = acr, how would you express this?
    the PE = mg2R, so how would you use this to find the KE?
    b. How would the ac equation be used to figure this out?
    c. Would this answer be 2R, since objects return to their original height in the absence of friction? Or does the loop complicate this?

    I'm sorry about this question, but finals are this week and my brain's slightly fried.

  2. jcsd
  3. Jan 23, 2006 #2


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    Homework Helper

    Obviously, the KE remaining when at the top of the track depends on "h".

    did you draw a free-body diagram of the small mass at the top of the loop?
    What direction are the Forces? These cause the acceleration.
    (you now know a formula for the acceleration component that is perp. to v).

    To JUST stay in contact with the track, F_Normal = 0 .
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