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Rigid Object Under net torque question

  1. Dec 7, 2016 #1
    1. The problem statement, all variables and given/known data
    An electric motor turns a flywheel through a drive belt that joins a pulley n the motor and a pulley that is rigidly attached to the flywheel. The flywheel is a solid disk with a mass of 80.0kg and a radius R = 0.625m. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of r = 0.230m. The tension Tu in the upper (taut) segment of the belt is 135N, and the flywheel as a clockwise angular acceleration of 1.67rad/s^2. Find the tension in the lower (slack) segment of the belt.

    10-p-037.gif

    2. Relevant equations

    I don't even know how to start solving this. Our textbook goes over the basic definitions of torque, but I have no idea how to apply this to a circumstance such as this to find Tension.

    3. The attempt at a solution
    Well, since I don't even know how to start, I guess I would find the torque as a result of the tension forces? I would appreciate if you could explain how I would start this question, as our textbook is next to useless here.
     

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  2. jcsd
  3. Dec 7, 2016 #2

    TSny

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    Yes, you're on the right track with using the concept of torque. How would you find the torque due to the upper tension?

    Note that the problem also gives you information about the angular acceleration of the flywheel. So, you need to relate torque and angular acceleration. Your textbook should state the relation.
     
  4. Dec 7, 2016 #3
    So, I use the τ = Iα, where apparently according to an answer I saw somewhere else, I = 1/2mr^2, and the angular acceleration is the one given (1.67rad/s^2)?

    Could you explain why I = 1/2mr^2?. I understand I = mr^2 which is from the relation of torque to Force (τ = Ft*r), but why the 1/2?
     
  5. Dec 7, 2016 #4

    TSny

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    Yes. The torque on the left side of the equation is the net torque due to all forces.

    The formula ##I = mr^2## is for a point mass rotating in a circle of radius ##r##. The mass of the flywheel is distributed over the shape of the object rather than having the mass concentrated in a point. The flywheel is said to have a "continuous mass distribution". Using calculus, it is possible to derive formulas for ##I## for some common shapes of objects. For a uniform solid disk rotating about an axis that is perpendicular to the disk and passing through the center, it turns out that ##I = \frac{1}{2} M R^2## where ##M## is the total mass and ##R## is the radius of the disk.

    Your textbook probably has a table that lists the formulas for ##I## for different shapes.
     
  6. Dec 8, 2016 #5
    Okay so I read up on calculations of moment of inertia, and this is what my textbook says:

    1. I = ∫r2 dm

    2. dm = ρdV = pL(2πr)dr

    3. I = ∫r2*(ρL2πr)dr

    4. 2ρLπ∫r3dr (where the integral is defined between 0 and R)

    5. ρ = M/V = M/πR2L

    6. I = (1/2)MR2

    My question: I sort of get why one would take the integral over the entire radius, but I'm not clear as to what exactly this does. Could you explain exactly why the integral is taken from 0 to R?
     
  7. Dec 8, 2016 #6

    TSny

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    The integration represents summing over infinitesimal rings that make up the disk. The rings vary in radius from 0 to R.
    See if this helps
     
  8. Dec 8, 2016 #7
    Oh okay, I see. So it's adding up all the changes in volume as a result of a combination of small rings formed by the change in radius of dr, and this would need to be done from the origin of the axis (0), to the radius of the ring with total radius R. Is that right?
     
  9. Dec 8, 2016 #8

    TSny

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    Yes, that's right.
     
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