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Angular Acceleration and torque problem

  1. Jul 12, 2014 #1
    1. The problem statement, all variables and given/known data
    A 12 kg mass is attached to a cord that is wrapped around a wheel of radius r = 10 cm (Fig. 8.20). The acceleration of the mass down the friction-less incline is 2.0 m/s^2/. Assuming the axle of the wheel to be friction-less, determine (a) the tension in the cord, (b) the moment of inertia of the wheel , and (c) the angular speed of the wheel 2 s after it begins rotating, starting from rest.


    2. Relevant equations
    F = ma




    3. The attempt at a solution
    I found the tangential force by using F = ma and found it to be 24. I am not sure how to continue to find the angular acceleration and moment of inertia because the question doesn't give the mass of the wheel.
     
  2. jcsd
  3. Jul 12, 2014 #2
  4. Jul 12, 2014 #3

    Nathanael

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    You're not given the mass of the wheel, but you are given the acceleration of the object.

    If you were given the mass of the wheel instead of the acceleration of the object, would you be able to figure out the acceleration?

    If so, then you should be able to figure out the mass given the acceleration.
     
  5. Jul 29, 2014 #4
    OK so I need to find the acceleration of the block down the hill and that will be the acceleration on the wheel? If so how do I do that?
     
  6. Jul 29, 2014 #5

    Nathanael

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    You don't need to find the acceleration of the block down the hill, because the problem has already told you that it is 2 m/s^2

    The linear acceleration of the block is equal to the linear acceleration of the edge of the wheel.
    (This assumes the rope doesn't slip on the pulley, which you need to assume for the problem to be solvable.)

    This gives you the equation:
    a [itex]=\alpha r[/itex]
    Where "a" is the linear acceleration (=2 m/s^2) and "[itex]\alpha[/itex]" is the angular acceleration of the wheel.
     
  7. Jul 30, 2014 #6
    OK I did that and got the last part of the problem right. However, I am still having trouble with the second part of the problem. So

    Ft = ma
    Ft = 12(2)
    Ft = 24
    T = Ft*r
    T = 24 * .1
    T = 2.4
    T = I*a
    T = I * 20
    2.4 = I *20
    I = .12

    The correct answer is 0.234 kg * m2
     
  8. Jul 30, 2014 #7

    haruspex

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    Ft is the net downslope force on the block, right? But you have equated that to the tension, which is an applied upslope force, and neglected gravity.
     
  9. Jul 30, 2014 #8
    Why do I have to use gravity it already tells me that the block is accelerating down the slop at 2 m/s^2?
     
  10. Jul 30, 2014 #9

    Nathanael

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    Because the two forces on the block are gravity and tension (in opposite directions)

    You don't know the tension, but you do know the gravity and you know the net acceleration (and thererfore net force) so you can use that to find the tension
     
  11. Jul 30, 2014 #10
    So do I do
    Fg + A = T?
     
  12. Jul 30, 2014 #11

    Nathanael

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    Fg - T = M*a

    Fg - Ma = T
     
  13. Jul 30, 2014 #12
    How did you get these equations?
     
  14. Jul 30, 2014 #13

    Nathanael

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    Well the second one is just a rearrangement of the first one.

    I forgot to mention: "Fg" is not mg, it's the just the component of mg down the slope.


    I got the equation because "Fg" (the force of gravity down the slope) is pulling the object down the slope, while the tension "T" is pulling the object up the slope.

    The net acceleration is "a" down the slope so the net force is "m*a" (directed down the slope)

    Since there is no other forces acting (the normal force is cancelled by a component of gravity) those two forces ("Fg" and "T") must add to give "ma" (They are in opposite directions, so that's where the negative sign comes from)


    This wasn't worded very clearly, but I hope you understand what I was trying to say
     
  15. Jul 30, 2014 #14

    rcgldr

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    The acceleration of the block is the net force on the block divided by the mass of the block.

    The component of force due to gravity on a frictionless incline = m g sin(θ)
    The opposing force is the tension in the rope = T.

    The net force on the block in the direction of slope of the incline = m g sin(θ) - T

    The acceleration of the block = (m g sin(θ) - T) / m = g sin(θ) - T / m = g sin(37°) - T / 12kg = 2 m / s^2, use this to solve for T. The unknown here is what value to use for g, 9.80665, 9.8, 9.81, ... ?
     
    Last edited: Jul 30, 2014
  16. Jul 30, 2014 #15
    So gravity is not causing the 2 m/s^2 acceleration. I understand how you got the variables but I am still confused on the relationships between them. I know that the force of gravity and the force of the acceleration are both in the same direction and are resisted by the tension. However, the formula is not Fg + ma = T. Why is this?
     
  17. Jul 30, 2014 #16

    Nathanael

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    Because the "force of acceleration" is not really a force in the system.

    You know (from Fnet=ma) that the net force on the block has to be ma, but ma is not really a force acting on the block, it's just the result of combining the other forces (tension and gravity)
     
  18. Jul 30, 2014 #17

    rcgldr

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    Take a look at post #14, which explains the math in more detail.
     
  19. Jul 30, 2014 #18
    OK so I did that and am still getting the question wrong so
    (g sin 37 -T)/12 = 2
    g sin 37 -T = 24
    5.898 - T = 24
    T = -18.102
    The book says the correct answer is 46.8 N
     
  20. Jul 30, 2014 #19

    Nathanael

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    It acidentally double posted this... Is there a way to delete posts?

    @rcgldr, I don't see a delete button. (I've seen it, and used it, before, but it's now nowhere to be found)
     
    Last edited: Jul 31, 2014
  21. Jul 30, 2014 #20

    Nathanael

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    it should be:

    g sin(37°) - T/12 = 24

    (The "g sin(37)" is already divided by 12, because it was originally "mg sin(37)" so you don't need to divide it by 12 again)
     
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