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Pulley energy conservation

  1. Nov 25, 2006 #1
    This is a multiple part problem but I'd like to have some help on one part at a time. Here is the general question:

    A pulley with a radius of R=0.50m and a mass of M=4.0kg is mounted on a frictionless axle. The pulley is a uniform solid disk with a rotational inertia of I=1/2MR^2. A block with a mass m=2.0kg hangs from a string wrapped around the pulley. When the system is released from rest the block accelerates down. Use g=10m/s^2.
    a) After dropping through a height h, the block's speed is 2.0m/s. Use energy conservation to find h.
    b)Apply Newton's Second Law, and Newton's Second Law for Rotation. Solve your equations to find the acceleration of the block.
    c)Use your value from (b) and one or more constant acceleration equations to find h, the distance the block has dropped when its speed reaches 2.0m/s. Does it agree with your answer from (a)?

    So for part (a)...
    the energy is KE + PE so it is 1/2Iw^2 + mgh which is...1/2(1/2(4.0kg)(0.50m)^2)0^2 + (2.0kg)(10m/s^2)h
    the final energy is 1/2(2.0kg)(2.0m/s)^2 + (2.0kg)(10m/s^2)(0m)
    so setting them equal to each other you get...
    20h=4 so h=4/20, simplified is h=1/5
    Is that right? If not please give me some hints, and I will work on part (b). Thanks!
  2. jcsd
  3. Nov 25, 2006 #2


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    What did you do with the kinetic energy of rotation of the pulley at the final position? It is not zero!! You seem to do it as if you must include the kinetic energy of rotation of the pulley at the initial position only and the kinetic energy of translation of the falling mass at the end only. Actually, both must be included at the initial AND at the final positions. So the full equation is really
    [tex] {1 \over 2} I \omega_i^2 + {1 \over 2} m v_i^2 + m g y_i = {1 \over 2} I \omega_f^2 + {1 \over 2} m v_f^2 + m g y_f [/tex]
    In your case, both the pulley and the mass are initial at rest. But they are both in motion at the end. If you set the origin of the y axis to be at the final position, you get [itex] y_f=0 [/itex] and [itex] y_i = h [/itex]. So you have
    [tex] mg h = {1 \over 2} I \omega_f^2 + {1 \over 2} m v_f^2 [/tex]

    Hope this helps.

    Last edited: Nov 25, 2006
  4. Nov 26, 2006 #3
    how do i figure out the wf of the pulley? I think that's where I'm getting confused. Thanks!
  5. Nov 26, 2006 #4
    Is the wf of the pulley v/r? so is it (2m/s)/(0.50m)=4.0rad/s
    which makes the equation...
    (this is going to look messy sorry)
    (2.0kg)(10m/s^2)h = [1/2(1/2(4.0kg)(0.50m)^2](4.0rad/s)^2 + [1/2(2.0kg)(2.0m/s)^2]
    20h = (4) + (4)
    h= 8/20
    h = 0.40m
    Last edited: Nov 26, 2006
  6. Nov 26, 2006 #5
    Now for part (b)...
    Newton's Second Law is F=ma, and Newton's Second Law for Rotation is T=I@ (torque=moment of inertia*alpha)
    I know that I=mr^2 and alpha=a/r so...torque=mra
    But then if F=ma and torque=mra, I'm not sure where to go with it..?
    If a=F/m and a=torque/mr then ??
  7. Nov 26, 2006 #6


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    The torque on the pulley comes from the tension in the string, and the tension is part of the force acting on the mass. Write the equations for each object in terms of the unknown tension and acceleration. Solve the two equations for the two unknowns.
  8. Nov 26, 2006 #7
    So I need to figure out the torque and the tension? I'm confused. Do I need to figure out what's acting on the pulley and the block? If so how do I put those together?
  9. Nov 26, 2006 #8


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    Draw a FBD for the mass. The forces acting are gravity and tension. Apply Newton's second law. Draw a FBD for the disk. There is a torque related to the tension. Apply the rotational analog of Newton's second law. You now have two equations involving the tension, the linear acceleration of the mass, and the angular acceleration of the disk. You already know how to replace one of the accelerations with the other. Whichever way you choose to do that you are left with two equations and two unknowns. Solve them.
    Last edited: Nov 27, 2006
  10. Nov 26, 2006 #9
    Ok so for the block i got a=F/m and for the pulley i did torque=I*alpha and torque=tension*r so alpha=Tr/I
    But I'm still not understanding where to go with it...?sorry
    Is the F the tension on the block? is that (mv^2)/r?
    Or do I do the block is a=T/m and alpha=Tr/I but isn't alpha just a/r?
  11. Nov 26, 2006 #10
    Or do i do a=T/m and alpha=Tr/I so then i plug in the T=ma to get alpha=(mar)/(1/2mr^2) = (2a)/r

    so the acceleration of the block is a=(alpha*r)/2
  12. Nov 26, 2006 #11


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    F is the sum of two forces acting on the block, gravity and tension. There is no mv²/r in this problem. We do not care about the forces that are holding the disk together.

    You were good up to the point where I ended the quote, except that is not the correct moment of inertia of a disk.
    Last edited: Nov 27, 2006
  13. Nov 26, 2006 #12
    Isn't the problem with a solid disk? Thats why I used the inertia I=1/2mr^2

    For the block, is the F=-mg+T?
    and for the pulley disk is it torque=I(a/r)= (1/2mr^2)(a/r) = 1/2mra?

    And then would the net forces acting on the block equal the torque of the pulley?
    So it would be -mg+T = 1/2mra and solve for a to get a=2(-mg+T)/mr
    Last edited: Nov 26, 2006
  14. Nov 26, 2006 #13


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    ½mr² is good. You wrote mv²/r before. Maybe you did not mean that. Yes, the force on the block is T - mg or mg - T if you take downward as the positive direction, but by Newton's second law the force is ma. Why do you have the ½mr in your block equation? Taking downward positive is convenient becasue you expect downward acceleration. Your torgue equation is fine.
  15. Nov 26, 2006 #14
    Sorry I've been looking at this problem for so long I'm confusing myself. Thanks for being patient.
    Ok so the force on the block is T-mg and that is equal to ma? so then a of the block is a= (T/m)-g

    But then do i set that equal to 1/2mra and solve for ...what?
  16. Nov 27, 2006 #15
    and then what?
  17. Nov 27, 2006 #16


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    You need to be consistent with the sign convention. When you wrote torque = I(a/r) = ½mr²(a/r) = ½mra you were assuming a was positve downward. When you write T - mg = ma, you are assuming a is positve upward. You need a to be positive in only one direction, or you need to have two different a where one is the negative of the other (not recommended). So let's write your last equation as

    mg - T = ma
    g - T/M = a

    Now substitute this expression for a into the other equation you have relating T and a, namely torque = Tr (you wrote that several posts back) and Tr = ½Mra or T = ½Ma. Alternatively, you can solve the equation above for T instead of a, and substitute T into the Tr = ½mra. So either you have

    T = ½M(g - T/M)


    T = mg - ma
    mg - ma = ½Ma

    You can solve these equations for T and a. Usually one just writes one of these equations, solves it, and then goes back to either of the two original equations and plugs in the now-known value to solve for the other variable. That is exactly equivalent to doing the algebra I did to get these two equations. It's just doing things in a little different order.

    Be careful to keep the M and m distinct. Somewhere you merged them into the one symbol, m, but the mass of the disk is not the same as the mass of the block.
    Last edited: Nov 27, 2006
  18. Nov 27, 2006 #17
    so then you should get a = (2mg)/(M+2m) and when you solve for a you get 5m/s^2, right?
  19. Nov 27, 2006 #18
    ok so part (d): Use your value from (c) and one or more constant-acceleration equations to find h,
    the distance the block has dropped when its speed reaches 2.0 m/s. Does it agree with your answer from (a)?

    so i used w = wi + alpha*t
    t = w/alpha
    t = 4/5 = 0.8 sec

    then i plugged that into x = xi + (wi)t = 1/2*alpha*t^2
    xi = 0
    wi = 0
    so, = 1/2(5)(.8)^2

    but my answer comes out to 1.6m

    are (a) and (c) suppsed to have the same answers??? Am i doing something wrong?

    Thanks so much!!!
  20. Nov 27, 2006 #19


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    M = 2m, so a = 2mg/4m = g/2 = 5m/s²

    Looks good!
  21. Nov 27, 2006 #20
    For the last part I used the equation v=vo +at and solved for t=(v-vo)/a = (2m/s-0m/s)/(5m/s^2) = 0.40s
    Then I plugged that into the equation h=ho+vot+1/2at^2 to get h=0+0+1/2(5m/s^2)(0.40s)^2 = 0.40m and that is the same as I got in part (a) YAY! Thank you!
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