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Friction - Coin on Rotating LP (solving for mu)

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    You can use the arrow keys to watch frame by frame (later used to measure time)

    2. Relevant equations
    F_friction = [tex]\mu[/tex](f_normal)
    F_c = mv^2/r

    3. The attempt at a solution
    So in the first attempt radius is 7 inches. We can find time by counting the number of frames it takes the coin to make 1 revolution (by using the arrow keys), and then dividing by 15 (since it's at 15 frames per second). Dividing the circumference by time would give us velocity. From here I wasn't sure where to go, so using the second half of the video I drew an incline. We were given the right side of the incline , 10 in., and the hypotenuse, 15 in., so I calculated theta, 41.81 degrees. Then I knew the force parallel to the incline was mgsin(theta), and Fnormal was mgcos(theta). Again, I wasn't sure from here after but here is how I solved for mu:
    From the point where it stood and then fell was 9 inches (estimation), and the time it took was .4 seconds (6/15). Since velocity is displacement/time, I plugged in and got .05715 m/s. Then i did v/t to solve for acceleration and got a value of .142875 m/s^2.
    I set up another equation, where [tex]\sum[/tex]F = (F_parallel - F_friction) + (F_grav - F_normal)
    I reduced this to [tex]\mu[/tex] = ((a/g) + cos(theta) - 1 - sin(theta))/(-cos(theta))
    Plugging in gave me [tex]\mu[/tex] = 1.255
    We did this in class, but I forgot exactly how we solved for mu. A friend of mine got .3 as an answer.
    Can someone please explain the appropriate method, or guide me toward it.
  2. jcsd
  3. Nov 9, 2008 #2
    We know that:

    [tex]\textbf{F}_f=\textbf{F}_c=m\textbf{a}_c=m\omega^2\textbf{r}=\mu\textbf{N}\rightarrow \mu=\omega^2\textbf{r}/\textbf{g}[/tex],

    where [tex]\omega[/tex] is given by the angular displacement, [tex]\theta[/tex], over any given time, [tex]t[/tex]. From this, we find:

    [tex]\mu=\Delta\theta^2\textbf{r}/(\textbf{g}\Delta t^2)[/tex].
  4. Nov 9, 2008 #3
    one slight problem...we aren't allowed to use mw^2, and also haven't touched upon angular momentum (ap physics: b)
    There has to be another way to solve this problem...
  5. Nov 9, 2008 #4
  6. Nov 9, 2008 #5
    The problem is that I can't (a) use w in the problem (since we haven't been taught it yet) and (b) we haven't learned angular displacement so the whole change and in theta will not work.

    Using what you had I simplified it to mu = (v^2) / (rgcos(theta))
    Checking to see if it works
  7. Nov 9, 2008 #6
    The equation you have should be right. I couldn't look at the video so IDK about your trig. usage but if it's at a slant, then yeah, you're right.
  8. Nov 9, 2008 #7
    o you couldnt look at the video ... maybe that's why you were using those complex formula's (atleast to me lol).
    Anyway I solved for it and got a value of .543, which seems more appropriate.
    Ok so I'm still unsure on if my velocity is correct. I'll quickly explain the video. The LP moves in a circular motion and a coin is placed on it. With radius of 7 inches and 9 inches the coin stays on the lp, but at a radius of 11 inches the coin falls of. The total distance from the center to the edge is 15 inches. Then they place the lp against a ruler and incline it, so that the right angle is to the right and the point of the triangle is facing the left. At a height of 10 inches the coin slowly falls off. Wouldn't the velocity be different if it were on an incline rather than moving in circular motion...

    And if it was how would I solve for mu lol ...
  9. Nov 9, 2008 #8
    What's an LP?
  10. Nov 9, 2008 #9
    Those old disks, like rly rly old cd's. (that have the rod coming and rotating the "cd's".)
  11. Nov 9, 2008 #10
    I can't see the system. :-\
  12. Nov 9, 2008 #11
  13. Nov 9, 2008 #12
  14. Nov 9, 2008 #13
    Imagine the books being the ruler and the long brown thing to be the cd.
  15. Nov 9, 2008 #14
    When it's rotating, [tex]F_f=F_c=ma_c=mv^2/r=\mu N \rightarrow \mu=v^2/(rg)[/tex].

    When it's being inclined, [tex]F_f=F_{g,x}=mg\cos{\theta}=\mu N=\mu mg\sin{\theta} \rightarrow \mu=\tan{\theta}[/tex], I think.
  16. Nov 9, 2008 #15
    Ok so when it's on an incline it's .3535, which is correct.
    Using the rotating method (which is what we have to do) they don't match.
    EDIT: I calculated velocity, by doing the (2pier/time)
  17. Nov 9, 2008 #16
    How would you calculate the velocity (and I believe in this case it's the angular velocity)?
  18. Nov 9, 2008 #17
    Can we get more people on this??
    I think this problem is pretty challenging, since we're not given mass or acceleration and are finding for mu.
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