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Angular Momentum of a counterweight

  1. Apr 5, 2008 #1
    [SOLVED] Angular Momentum

    1. The problem statement, all variables and given/known data
    A 4.60 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The spool is a uniform solid cylinder of radius 8.00 cm and mass 1.30 kg.
    (a) What is the net torque on the system about the point O?
    Magnitude
    (b) When the counterweight has a speed v, the pulley has an angular speed = v/R. Determine the total angular momentum of the system about O.
    (c) Using the fact that = dL/dt and your result from (b), calculate the acceleration of the counterweight.
    2. Relevant equations
    [tex] L = r X P[/tex]
    [tex] \tau = r X F[/tex]
    [tex] L = I\omega[/tex]

    3. The attempt at a solution
    I was able to solve the first part:
    [tex] \tau = rF [/tex]
    [tex] \tau = (0.08)(4.6)(9.8) = 3.61 Nm[/tex]

    I am not sure how to solve part B though, here is what I tried but it was wrong:

    [tex] L_{net} = rp + I\omega [/tex]
    [tex] L_{net} = r X mv + I*\frac{v}{r} [/tex]
    [tex] L_{net} = v (r X m + \frac{I}{r})[/tex]

    I factored out the v because it doesn't give you the velocity, and the answer space looks like this:

    ________ X v kg *m^2/s
     
  2. jcsd
  3. Apr 5, 2008 #2

    Doc Al

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    Staff: Mentor

    Looks OK to me--keep going.

    Hint: What's the rotational inertia of a solid cylinder?
     
  4. Apr 5, 2008 #3
    Rotational inertia of a solid cylinder is 0.5MR^2

    With my equation there, at the rXm, do I have to do the cross product?
     
  5. Apr 5, 2008 #4

    Doc Al

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    Staff: Mentor

    Good.

    You can only take cross products of vectors, so r X m doesn't make sense. (I was wondering if you meant that X as a cross product or just as multiplication.) The time to do the cross product is before you factor out the speed.

    In the definition of angular momentum of a particle, [itex]\vec{L} = \vec{r}\times m\vec{v}[/itex], [itex]\vec{r}[/itex] is the position vector of the particle. The cross product turns out to be easy to evaluate in terms of r (the radius). So what is the angular momentum of the particle?
     
  6. Apr 5, 2008 #5
    I don't know. I am a little unclear about how to do cross products.
     
  7. Apr 6, 2008 #6
    Can someone tell me how I would go about cross producting [itex] \vec{L} = \vec{r}\times m\vec{v}
    [/itex]
     
  8. Apr 7, 2008 #7
    1. The problem statement, all variables and given/known data
    A 4.60 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The spool is a uniform solid cylinder of radius 8.00 cm and mass 1.30 kg.
    (a) What is the net torque on the system about the point O?
    Magnitude
    (b) When the counterweight has a speed v, the pulley has an angular speed = v/R. Determine the total angular momentum of the system about O.
    (c) Using the fact that = dL/dt and your result from (b), calculate the acceleration of the counterweight.


    2. Relevant equations
    [tex] L = r X P [/tex]
    [tex] \tau = r X F [/tex]
    [tex] L = I\omega [/tex]
    3. The attempt at a solution
    I was able to solve the first part:
    [tex] \tau = rF [/tex]
    [tex] \tau = (0.08)(4.6)(9.8) = 3.61 Nm [/tex]

    I am not sure how to solve part B though,

    [tex] L_{net} = rp + I\omega [/tex]
    [tex] L_{net} = (r)mv + I*\frac{v}{r} [/tex]

    [tex] L_{net} = v (rm + \frac{I}{r}) [/tex]
    [tex]L_{net} = v [ (.08)(4.6) + \frac{1}{2}(1.30)(.08)]} [/tex]
    [tex]L_{net} = v(0.42)[/tex]

    I factored out the v because it doesn't give you the velocity, and the answer space looks like this:

    ________ X v kg *m^2/s

    Is that how I would go about part B? Thanks
     
    Last edited by a moderator: Apr 7, 2008
  9. Apr 7, 2008 #8

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
  10. Apr 7, 2008 #9
    Can you help me with part c?
     
  11. Apr 7, 2008 #10

    Doc Al

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    Staff: Mentor

    What's the full statement of part c? (Your initial post has something missing.) You'll need to take the derivative of L (that's what dL/dt means).
     
  12. Apr 7, 2008 #11
    sorry, yes I know i have to take the derivative...
    Here is the post:
    Using the fact that t = dL/dt and your result from (b), calculate the acceleration of the counterweight.

    so would it be: 3.61 = dL/dt and dL/dt= 0.42dv/dt and dv/dt is acceleration so would it be a= 3.61*0.42
     
  13. Apr 7, 2008 #12

    Doc Al

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    Staff: Mentor

    Good.
    Divide, not multiply, by 0.42.
     
  14. Apr 7, 2008 #13
    haha yeah i just realized that.

    Thank ye
     
  15. Mar 26, 2011 #14
    Re: [SOLVED] Angular Momentum

    just wondering what units the final answer is in - if the counterweight is just accelerating downwards, do we still write the answer in rads/sec^2 ?
     
  16. Mar 27, 2011 #15
    Re: [SOLVED] Angular Momentum

    or would it be in N*s/kg?

    (N*m)/(kg*m^2/s)
     
  17. Mar 27, 2011 #16

    Doc Al

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    Staff: Mentor

    Re: [SOLVED] Angular Momentum

    The final answer for what question? rads/sec^2 are units for angular acceleration; the counterweight has linear acceleration, which has units of m/s^2.
     
  18. Mar 27, 2011 #17
    Re: [SOLVED] Angular Momentum

    thankyou! :) i was thinking angular acceleration *facepalm* :)
     
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