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Inclined disk rotating on the ground

  1. Jul 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A disk of radius R and mass m rotates on a floor with coefficient of friction μ and makes a circle of radius r0 at constant speed v0.
    What is the friction force between the disk and the floor.
    What's the inclination angle α to the vertical.
    It's suggested in the question to use cylindrical coordinate system in which ##\hat{r}## is from the center of the circle to the point of contact of the disk and ##\hat{\theta}## is in the tangential direction and ##\hat{z}## is upwards.

    2. Relevant equations
    Centrifugal acceleration: ##a=\frac{v^2}{r}##

    3. The attempt at a solution
    The friction force:
    $$F=m\frac{v_0^2}{r_0}$$
    The inclination angle. moments round point A=moments round the center o, but i am not sure i am allowed to choose 2 different points:
    $$mgR\sin\alpha=m\frac{v_0^2}{r_0}R\cos\alpha\rightarrow \tan\alpha=\frac{v_0^2}{gr_0}$$
    I don't understand why and how i have to use cylindrical coordinate system since i solved easily. i suspect it's incorrect.
    If i want to take moments of both the friction force f and the weight round the same point i choose point B on the top but both forces rotate the disk counterclockwise.
     

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  2. jcsd
  3. Jul 1, 2015 #2

    Nathanael

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    Are you assuming the net torque is zero? I must admit, angular momentum is one of my weak points, but I think there ought to be a net tangential torque (like the torque which causes a gyroscope to precess). Although the angular momentum is constant in magnitude, it is changing direction.
     
  4. Jul 1, 2015 #3

    haruspex

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    The mass centre of the disc is not at radius r0.
    I agree.
    Karol, consider a gyroscope balanced on a point but leaning. There is a torque from gravity tending to topple it, and no 'centrifugal force' in this case to balance it, yet it does not topple. Instead, it precesses. You need to calculate the precession rate and what net torque is required to produce it.
     
  5. Jul 2, 2015 #4
    I (as usual) forgot to write that we can assume R<<r0
    The torque: ##\frac{dL}{dt}=mRg\sin\alpha##
    The horizontal component of the angular momentum:
    $$L_{(hor)}\cos\alpha=I\omega\cos\alpha=\frac{1}{2}mR^2\cdot \frac{v_0}{R}\cos\alpha$$
    The vector of change in the horizontal component of the torque:
    $$dL=L\cos\alpha\cdot \Omega dt=\frac{1}{2}mR^2\cdot \frac{v_0}{R}\cos\alpha\cdot \frac{v_0}{r_0}dt$$
    $$\rightarrow \frac{dL}{dt}=\frac{mRv_0^2\cos\alpha}{2r_0}$$
    But we know ##\frac{dL}{dt}=mRg\sin\alpha## so:
    $$mRg\sin\alpha=\frac{mRv_0^2\cos\alpha}{2r_0}\;\rightarrow\;\tan\alpha=\frac{v_0^2}{2gr_0}$$
     
    Last edited: Jul 2, 2015
  6. Jul 3, 2015 #5

    haruspex

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    In your first solution you considered the torque needed to provide centripetal force, in the solution just above you've considered the torque needed to provide the precession rate. You need both.
     
  7. Jul 3, 2015 #6
    The precession is only changing the angular momentum's direction, right? and that's what i have calculated, i guess.
    $$mRg\sin\alpha=\frac{mRv_0^2\cos\alpha}{2r_0}+m\frac{v_0^2}{r_0}R\cos\alpha$$
    $$\tan\alpha=\frac{3v_0^2}{2gr_0}$$
    But why am i allowed to take moments round 2 different points in the same equation (for the centripetal force)?
     
  8. Jul 3, 2015 #7

    haruspex

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    I agree with the answer but don't understand the question. Which two points?
     
  9. Jul 3, 2015 #8
    In the original OP i made moments round points A and O in order to find the angle α: i took the torque of the weight round the point of contact A and equaled it to the torque of the friction force round the center point O. am i allowed to consider, in the same equation, moments round two different points?
    It's interesting that if i consider the centrifugal force instead of the friction force i can take moments of both forces round the same point A, but i want to solve in the inertial frame and there there is only the friction.
     
  10. Jul 3, 2015 #9

    haruspex

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    Yes, I thought of it in terms of a centrifugal force - not something I usually do.
    The alternative is to say that the torque produced by the combination of (equal and opposite) weight and normal force, in combination with the 'lever arm' (height of the mass centre) produces the centripetal force.
     
  11. Jul 3, 2015 #10
    Thanks
     
  12. Jul 3, 2015 #11

    SammyS

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    I like pictures displayed when possible. Just my quirk .

    snap1-jpg.85418.jpg
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    snap2-jpg.85419.jpg
     
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