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Inertia, frictionless pulley, and a weight

  1. Apr 22, 2008 #1
    1. The problem statement, all variables and given/known data
    A frictionless pulley has the shape of a uniform sold disk of mass 2.5 kg and radius .2m. A 1.5kg stone is attached to a very light wire that is wrapped around the rim of the pulley, and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.5J of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?


    2. Relevant equations
    I for solid disk: I = (1/2)(MR^2)



    3. The attempt at a solution

    first i plugged in this equation:
    K = (1/2)IW^2 + (1/2)MV^2
    i replaced W with V/r
    K = (1/2)I(V/r)^2 + (1/2)MV^2
    first i had to find I:
    using formula above: I = (1/2)(2.5)(.2^2) = .05 kgm^2
    then i plugged in everything to find final velocity :
    4.5J = (1/2)(.05kgm^2)(V/.2m)^2 + (1/2)(1.5kg)V^2
    then i solved for V and got V = 1.8 m/s

    ok now im not entirely sure what to do... i need the distance that the rock drops. but to do that i need to find the acceleration.
    radial acceleration is V^2/r = 16.2 rad/s^2

    then i tried V^2 = Vo^2 + 2(a)(X-Xo)
    Vo=0 and Xo=0 so it becomes:
    V^2 = 2(a)(x)

    i have value for V. i know im supposed to be able to figure out A so i can solve for x. but im not sure if im converting from radial acceleration to regular acceleration correctly because i keep getting the wrong answer.

    i thought this was the formula for that:
    a^2 = Arad^2 + Atan^2
    and Atan = Arad(r) right? because im doin something wrong... idk what it is
    the answer is supposed to be 67.3 cm
     
    Last edited: Apr 22, 2008
  2. jcsd
  3. Apr 22, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    4.5J is the KE of the pulley, not the total KE.

    Figure out the total KE. Then ask where that energy came from. (Hint: Conservation of mechanical energy.)
     
  4. Apr 22, 2008 #3
    thanks for the help. i got part (b) which is Kpulley/Ktotal = 4.5/9.89 x 100 = 45.5% :)

    but im still stuck on part (a). since i got the second part right i know that my velocity is right now. the W = 13.4 rad/s and V = Wr = 13.4(.2) = 2.68 m/s.
    this time i tried solving the distance by finding the change in the angle:
    Arad = (W^2)(r) = 36 rad/s^2
    W^2 = 2(Arad)(feta)
    13.4^2 = 2(36)(feta)
    feta = 2.494 rads
    this doesnt come out to 67.3m :(

    edit: SWEET! i figured it out cus of ur mechanical energy hint!
    i did:
    Ktotal = mgh
    9.89 = 1.5(9.8)h
    h = .673m = 67.3 cm

    thanks a lot!
     
    Last edited: Apr 22, 2008
  5. Apr 22, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    I can't quite follow what you're doing. In any case, instead of messing around with speeds, use conservation of energy. Compare the mechancial energy of the system before and after the stone falls.
     
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