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Rotating Friction Discs For Project Design

  1. Dec 11, 2016 #1
    The situation is part of a mechanism I have been working on. It is used for a bicycle to automatically store and release energy with a spring. It is somewhat similar to a car clutch.

    1. The problem statement, all variables and given/known data
    Two friction discs are rotating with different angular velocities and different torques. The discs come in contact with one another and press together with a certain amount of force. If you know:
    • the initial angular velocity and torque of both discs
    • the radius of both discs
    • the normal force pressing the discs together
    • the coefficient of friction for the friction surfaces, both static and kinetic
    could you find the change in the torque and angular velocity of both discs? If possible, how would one write an equation to solve this? If not, what information would be needed?

    2. Relevant equations
    Ff=FNμ
    τiωi=τoωo

    3. The attempt at a solution
    I have attempted to first take the average of the torques of the two discs. The greater the force of friction, the closer the output torque would be to the average. However, if I write that down mathematically, it would require the force of friction to be simply a ratio.

    It seems that the problem may require calculus, which I have never been taught yet. Using only my current understanding of physics, an attempt to solve the problem would most likely be guesswork.
     
  2. jcsd
  3. Dec 11, 2016 #2

    haruspex

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    The discs will not have an initial torque as such. They will have an initial velocity and a moment of inertia.
    There may be a potential torque in each. On the road 'side' of the transmission, if going uphill, then a torque will be required just to maintain speed, so you could count that as an initial torque. On the cyclist side, the potential torque is however hard the cyclist is prepared to push as the clutch is let in.
    The moment of inertia on the cyclist side would be quite small. It only relates to the drive train between the clutch and the ratchet. (Or do I have the order of components wrong?)

    The frictional force acts tangentially. Assuming the normal force is spread evenly, each small area element can achieve the same frictional force. The further from the centre, the greater the area of the band, and the greater the torque per unit force, so it rises quadratically:
    ##\tau=\int_0^RP\mu 2\pi r^2.dr=\frac 23\pi\mu PR^3##, where P is the normal pressure.
     
  4. Dec 11, 2016 #3
    For the equation, I'm unsure what R means. I can only assume it relates to the radius.
    I know that pressure is force divided by surface area. Therefore, would P=Fn/πr^2 in this scenario?

    For my design, one of the friction discs connects to the main spring. The other is connected to a gear. The torque and angular velocity of this gear is based on the spinning of the bicycle freewheel in comparison to that which is turning it. If the bicycle wheel is spinning fast while the pedaling is slow, the gear will go faster. The rotation of the gear is converted to the normal force with the use of a thread. As the gear rotates, it will also move forward.
    The friction between the friction discs is not manually controlled by the cyclist as it would for a car clutch; this design and a car clutch are only vaguely similar.
    There are more components to my design, but this should explain the problem.
     
    Last edited: Dec 11, 2016
  5. Dec 12, 2016 #4

    haruspex

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    Yes. I was using it to distinguish from the variable of integration, r, but i could have omitted it.
    I should clarify that the equation gives the maximum torque.
    Yes.
    Sure, but as I wrote, there is no actual torque until the clutch is let in. There is angular velocity, moment of inertia, and the product of the two, angular momentum.
    If the angular velocities are different then when the clutch is let in it will generate equal and opposite torques to the two sides of the system.
    Let the two sides be represented by angular velocities ω1, ω2 and moments of inertia I1, I2.
    At first, μ will be kinetic, and we will get angular accelerations ##\frac {2 \pi r^3 \mu_kP}{3I_1}## and ##\frac {2 \pi r^3 \mu_kP}{3I_2}## tending to bring the angular velocities together. This will last for time t where ##t\frac 23 \pi r^3 \mu_kP\left(\frac 1{I_1}+\frac 1{I_2}\right)=|\omega_1-\omega_2|##.
    The work lost as heat in the process is ##\frac{I_1I_2}{2(I_1+I_2)}(\omega_1-\omega_2)^2##.

    Does this help?
     
  6. Dec 12, 2016 #5
    This has been of great help- however, I am slightly confused about the moment of inertia. If the moment of inertia is dependent on the mass, would I therefore need to include the mass of the rotating discs into the equation? If not, how would it be calculated?

    Also, I'm assuming that 2πr3μkP is just 2rFNμk.
     
  7. Dec 12, 2016 #6

    haruspex

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    Yes.
    Yes.
     
  8. Dec 12, 2016 #7
    I'm also having trouble in understanding time in the equation.

    How would you know how much time it takes before the kinetic friction becomes static friction and the velocities become equivalent? Or am I misunderstanding the equation?
     
  9. Dec 12, 2016 #8

    haruspex

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    Initially, the discs will slip against each other. That means we can write down the frictional torques based on kinetic friction, normal force and disc radius.
    Knowing the moments of inertia, we can then find the angular accelerations. Knowing the initial difference in angular speed and the total angular acceleration (the two accelerations will be opposite, because the torques are opposite, so the relative acceleration is the sum of their magnitudes) we can find the time for the relative angular velocity to reach zero.
     
    Last edited: Dec 14, 2016
  10. Dec 13, 2016 #9
    All right. I was thinking of the time as an input value rather than what the equation was solving for. I just want to recap to make sure I understand everything properly.

    In order to know torque generated, you use the equation T=2/3rFNμk. The torque would be equal and opposite for both sides.
    As the force is applied, the discs will slow down by (2FNμk)/(3mr). (I'm using I=mr2 for this).
    Eventually, the velocities will become equivalent (I'm guessing it will be the average of the two initial velocities). The time required is found by the equation t2/3rFNμk(1/mr2 +1/mr2).=|v1-v2|.

    I really appreciate all the help, by the way. I admit that I only have a fairly basic understanding of mechanics (most of it self-taught),so this is taking longer than it might have otherwise.
     
    Last edited: Dec 13, 2016
  11. Dec 14, 2016 #10

    haruspex

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    The r in the torque equation is the radius of the clutch disc. I don't know what components of your system contribute to the moment of inertia, but I would guess there is more than just the clutch.
    Only if the moment of inertia is the same each side.
     
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