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Work - Force - Energy

  1. Oct 20, 2007 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m is tied to a string which goes through a hole in a smooth horizontal table. The particle moves in uniform circular motion with speed v0. The radius of the cricle is R0. By puling the string very slowly, the radius of the circular motion is reduced tao R1.
    Show that the work done by the string force is equal to the change of Kinetic energy.

    2. Relevant equations
    F=Tension(T) = mv^2 /R
    K ( kinetic energy ) =1/2mv^2
    K1-K0 ( final - initial kinetic energy ) = 1/2m ( v1^2 - v0^2 )
    where v1 = R0*v0/R1 ( using angular momentum conservation ).


    3. The attempt at a solution
    W=∆K ( work energy theorem )
    W = ∫Fdr , where F = m*v*v / r ( centripetal force )
    I think I need some substitution to get the integral to give the ∆E...
     
  2. jcsd
  3. Oct 20, 2007 #2

    Astronuc

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    Staff: Mentor

    With the integrand of F = m*v*v/r multiplied by dr, one has units of kg m2/s2 (assuming SI) which is units of kinetic energy.
     
  4. Oct 20, 2007 #3
    I know that the units are consistent. I have found the change in Kinetic energy, which is deltaK = 1/2 *v0^2 * [( R0^2 / R1^2 ) - 1 ].
    But when I integrate mv1^2/R1 dR ( between R0 and R1 ), I get the very same expression but without the 1/2 factor at the start.. What am I doing wrong?
     
  5. Oct 20, 2007 #4

    Doc Al

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    Staff: Mentor

    What's the force as a function of R?
     
  6. Oct 21, 2007 #5
    I solved the integral and I got the same expression as in for the difference in kinetic energy, so it's gotta be right. Thanks for Your help!
     
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