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Physics: Torque on a Disk problem - Multiple forces

  1. Nov 14, 2013 #1
    1. The problem statement, all variables and given/known data

    A uniform disk with mass m = 9.27 kg and radius R = 1.42 m lies in the x-y plane and centered at the origin. Three forces act in the +y-direction on the disk: 1) a force 344 N at the edge of the disk on the +x-axis, 2) a force 344 N at the edge of the disk on the –y-axis, and 3) a force 344 N acts at the edge of the disk at an angle θ = 39° above the –x-axis.

    Diagram: https://www.smartphysics.com/Content/smartPhysics/Media/Images/Mechanics/15/torqueondisk.png
    1) What is the magnitude of the torque on the disk about the z axis due to F1?
    2) What is the magnitude of the torque on the disk about the z axis due to F2? = 0, this was correct
    3) What is the magnitude of the torque on the disk about the z axis due to F3? = 379.6, this was correct
    4) What is the x-component of the net torque about the z axis on the disk?
    5) What is the y-component of the net torque about the z axis on the disk?
    6) What is the z-component of the net torque about the z axis on the disk?
    7) What is the magnitude of the angular acceleration about the z axis of the disk?
    8) If the disk starts from rest, what is the rotational energy of the disk after the forces have been applied for t = 1.8 s?


    2. Relevant equations
    torque = F * r * sin(theta)
    torque = I * alpha


    3. The attempt at a solution

    1) torque = F * r * sin(theta)
    so torque = (344N) * (1.42) * sin(90)
    the answer I get is 488N*m but it says it's incorrect. I'm also not sure how to convert the sin(90) to the z axis instead of the x-y axis.

    4) I can't actually find this until I get 1 correct, however, it would be the sum of all torque, so
    Ʃtorque = torquey + torquex + torquez where torquez and torquey are to be transferred into the y direction, but multiplying by sin(90)?

    5/6, same method as above

    7) The equations I used are torque = I * alpha, torque = F * r, I = 1/2 * mdisk * radius2

    rearranging, I * alpha = F * r so
    alpha = F * r / I where
    I = 1/2 * mdisk * radius2
    to end up with

    alpha = F*R/(1/2 * mdisk * radius2)
    alpha = 344 * 1.42 / (1/2 * 9.27 * 1.422)
    alpha = 52.267 rad/s2 <- seems too high

    8) Krotational = (1/2) * I * ω2 where ω = alpha * t and I = 1/2 * mdisk * radius2

    so K = (1/2) * (1/2 * mdisk * radius2) * (alpha * t)2
    K = 551.278J


    Thanks for the help
     
  2. jcsd
  3. Nov 14, 2013 #2

    Simon Bridge

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    Did you sketch out the situation?
    I'm guessing the on-axis forces are pointing along the axis, and the one at the -x axis has components in the +y and -x direction.

    It would be helpful if the question were clearer about the directions.

    You appear to be applying the equation without understanding it.
    Divide all your forces into radial and tangential components.

    The radial component ##F_r## points along the line joining the center or rotation to the point of application of the force.
    The tangential component ##F_{\!\perp}## is perpendicular to that.
    The magnitude of the torque is given by: $$\tau = rF_{\!\perp}$$ ... where ##r## is the distance from the center to the point of application.

    You can also get the direction by $$\vec{\tau}=\vec{r}\times\vec{F}$$
     
  4. Nov 14, 2013 #3
    I did sketch it out, however it still makes no sense due to the confusing directions. I'm not quite sure what you mean by Fperpendicular. How would I find that? And why would there be a difference? The force, to my understanding is already tangent to the disk at force 1 judging by the wording and the way my sketch and the picture that I linked look.
     
    Last edited: Nov 14, 2013
  5. Nov 14, 2013 #4

    Simon Bridge

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    (my bf) Excuse - must have had my eyes closed.
    Sorry I thought you were doing something else.

    Now I get it - all the forces are in the +y direction ... the last one is not on an axis.

    I'm guessing you are submitting this to a computer?
    In that case it's probably just being fussy - try using more decimal places and try using a minus sign - though, usually, anti-clockwork torques (viewed down the axis) are positive.

    Your other questions:
    To work out which direction the torque points - remember that the torque is a cross product.
    Are all the forces in the x-y plane?
    Use the right-hand-screw rule on the vectors.
     
  6. Nov 14, 2013 #5
    Apparantly the computer can't tell the difference between 488.48 and 488.5. I was trying to figure out what I was doing wrong, just a fussy website. As it stands, after reading the question four times, the reason they say about the z-axis, to my understanding is that the disk is rotating around the z axis, so all of the forces act in the x-y plane. My problem is, is that I don't quite know how to find the sum of all torque on a rotating axis, when the two torque are against each other. F3 being clockwise, and F1 being counter-clockwise.
     
  7. Nov 14, 2013 #6

    Simon Bridge

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    You use the right-hand screw rule to determine if the torque is positive (pointing up the z axis) or negative (pointing down the z axis).



    On that video - r is blue, F is green, so that ##\vec \tau = \vec r \times \vec F##
     
    Last edited by a moderator: Sep 25, 2014
  8. Nov 14, 2013 #7
    I just realized I answered my own question in my question. Thank you for the help Simon
     
  9. Nov 14, 2013 #8

    Simon Bridge

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    Well done - that is often the case :)
     
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