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Angular acceleration, friction, and turntables

  1. Jun 22, 2007 #1

    exi

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    1. The problem statement, all variables and given/known data

    A copper block rests 33.8 cm from the center of a turntable. The coefficient of static friction between the block and turntable is 0.52. The table starts from rest and accelerates with an angular acceleration of 0.226 rad/s².

    Acceleration of gravity = 9.8 m/s².

    After what interval, in seconds, will the block start to slip on the turntable?

    2. Relevant equations

    Tangential acceleration = angular acceleration * radius, so At = 0.226 * 0.338 = 0.0764 m/s².

    3. The attempt at a solution

    I've got the turntable drawn out with a radius of 0.338 m, a block on the edge with unknown mass and a tangential acceleration of 0.0764 m/s², and µ = 0.52.

    I'm a little hazy on how to progress; does the lack of mass information pose a problem?
     
  2. jcsd
  3. Jun 22, 2007 #2

    Doc Al

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    Staff: Mentor

    Hint: Consider the total acceleration, not just the tangential component.

    What's the maximum static friction force possible? Thus, what's the maximum acceleration that the friction can create?
     
  4. Jun 22, 2007 #3

    exi

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    If static friction is <= µs*Fn, I would try to take it from there, but I'm at a loss as to how to go about something with an unknown mass. Unfortunately, this question (and several others on this particular assignment) go to a significantly greater depth than what we've seen in lecture, so it's a bit confusing to a relative beginner.
     
  5. Jun 22, 2007 #4

    Doc Al

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    Call the mass "m". Maybe you won't need it. :wink:

    Hint: What's the maximum acceleration that the friction can create without slipping?
     
  6. Jun 22, 2007 #5

    exi

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    Well, friction force would be equal to or less than 5.096*m if I'm going about this the right way, which means... sorry, I'm a little lost. I'm sure I'm overlooking something right in front of me, but I'm having a hard time seeing this clearly at the moment.
     
  7. Jun 22, 2007 #6

    nrqed

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    Now, what is the maximum acceleration the block can have withouit slipping?
     
  8. Jun 22, 2007 #7

    exi

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    Guys, I think we might not have covered that in any great detail in the hurried lectures leading up to the last exam we just had. I'm afraid I'm not sure what the relationship is between frictional force and acceleration. (The other question about tangential acceleration I had posted was also new territory for us.) :frown:
     
  9. Jun 22, 2007 #8

    Doc Al

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    You're doing fine. But I would suggest that you express things symbolically wherever possible and only plug in numbers when you need to calculate an answer. I'd write that as: Max Friction = µmg.

    It's the same relationship between any force and acceleration: Newton's 2nd law.
     
  10. Jun 22, 2007 #9

    exi

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    Okay, so as far as that goes, µmg = ma, which is a = µg. So if that's the acceleration you were hinting towards as far as the most the block can endure without slipping away, how does that play into the rest of the info given in the question?
     
  11. Jun 22, 2007 #10

    Doc Al

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    See my hint in post #2. You found the tangential acceleration, what about the radial (a.k.a centripetal) component? What affects that component?
     
  12. Jun 22, 2007 #11

    exi

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    This is what I've been doing:

    Taking from the above, frictional force where µ = 0.52, once solved through Newton's second, is coming out to be 5.096 m/s². I was under the impression that this would be acting along the radius of the circle, so using ac = v²/r, I found velocity, and tried to use that to find a period via v = 2*pi*r / T - and end up with 1.6182 s, which is not a correct answer.

    Where is my error in thought? I've got 85% of this thing completed, and the rest is due at midnight tonight, so I'm trying to understand what I'm still missing...
     
  13. Jun 22, 2007 #12

    Doc Al

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    That's the maximum total acceleration, not the radial component. Since you know the tangential component, figure out the radial component.

    You're on the right track, just using the wrong value for the centripetal acceleration.
     
  14. Jun 22, 2007 #13

    exi

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    Doing the same thing that I did for another question where I had to find the total acceleration, I'm getting a number that's within thousandths of a point, and working it out again, the "answer" is also within a thousandth or two - probably well within the 1% margin of error we have, so apparently I'm still doing something wrong.

    Tangential acceleration is equal to angular acceleration (rad/s²) * radius (m), right? I was able to sketch things and use straight Pythagorean theorem to find the answer on a similar question... hrmph.
     
  15. Jun 22, 2007 #14

    Doc Al

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    You're right. Accounting for the tangential acceleration makes a negligible correction. (My bad for not checking!)

    I see what you did wrong:

    This is an incorrect approach. (Again, I must have been sleeping before.) You have the speed and the acceleration (the tangential acceleration)--use simple kinematics to find the time it takes to attain that speed.
     
  16. Jun 22, 2007 #15

    exi

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    I've got several kinematics equations in front of me - one-dimensional, projectile motion, some circular motion, ... - but nothing for circular motion that incorporates strictly time in seconds. I suppose I'm missing something useful there.
     
  17. Jun 22, 2007 #16

    Doc Al

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    The tangential acceleration (which you calculated in post #1) works just like any other one-dimensional linear acceleration. (The fact that it's going in a circle doesn't matter.)
     
  18. Jun 22, 2007 #17

    exi

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    Ohhhhhhhhh, that makes perfect sense now; 17.177s makes much more sense. Epiphany, ahoy.

    Thanks so much for walking me through it.
     
  19. Jun 22, 2007 #18

    Doc Al

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    My pleasure. (Sorry if I threw you off a bit before.)
     
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