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Newton's Law on Circular Motion

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Block mass "m" is sliding on frictionless horizontal surface while traveling inside of hoop radius "R". Coefficient of f between block and wall is "u", therefore, speed of block is decreasing. In terms of "R", "v, block's veloc.", "m", and "u", find expressions for:

    1. frictional force on the block:

    2. block's tangential acceleration (dv/dt).

    3. From #2 find time required to reduce speed of block from its original velocity to one-third of its original velocity.


    2. Relevant equations
    fk = un
    F = ma
    ar = v^2/r <-- ar is radial acceleration


    3. The attempt at a solution

    My process for 1:
    fk = un
    ar = v^2/R
    F = n = ma
    n = m(v^2/R)

    fk = (umv^2)/R

    2:
    F = ma
    I have no clue on this one.
    Would it be from #1, use
    v^2 = fR/um
    v = sqrt[(fR)/(um)]
    And take the derivative of that? And it will be my tangential acceleration?
    but the thing is, you have to give the answer in terms of m, R, u, and v only. Can't put in terms of f - friction

    3: I assume to be easy with #2? It's just simple kinematics right?

    I found a previous inquiry on this problem in the forums but it wasn't very clear, so I wanted to ask it myself.
     
    Last edited: Oct 30, 2009
  2. jcsd
  3. Oct 30, 2009 #2

    ehild

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    Your answer to question 1 is correct, but fk is negative with respect to the velocity of the block. It is also parallel with the velocity, a tangential force, acting along the tangent of the circle, and causing tangential acceleration. Now, how do you get acceleration from force?

    ehild
     
  4. Oct 31, 2009 #3
    You get acceleration from force by using F = ma

    So would I:
    fk = ma
    a = fk / m?
    Could you please give me some kind of hint, something that triggers my thinking? Because you're posing the question that I'm essentially asking.

    If a = fk / m
    i would simply have:
    fk = -(umv^2)/R

    a = -(uv^2)/R

    Is that correct? I feel that is too simple. Especially because the question poses: "Find the block's tangential acceleration dv/dt", kind of hints towards taking some kind of derivative of a velocity equation? Please tell me what you think.
     
  5. Nov 1, 2009 #4
    ahhh please help me someone.. I need help on this by Wednesday =(
     
  6. Nov 1, 2009 #5

    Doc Al

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    Yes.
    That hint will come in handy in part 3.
     
  7. Nov 1, 2009 #6

    rl.bhat

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    You have written
    fk = μ*mv^2/R.
    Since v is slowing down, fk is not a constant.
    So d(fk)/dt = (μm/R)*2v*dv/dt.
    Here dv/dt is the tangential acceleration.
     
  8. Nov 1, 2009 #7
    Yes! Thank you for answering =DDD

    Now, I did what you said, which is also what I originally asked if I should do and I have
    dv/dt = -(R/(2umv))(d(fk)/dt)

    The issue I have, which I presented when I originally asked if I should do that, is with the fact that the answer must be in terms of m, R, u, and v. I have in terms of fk...

    That is where my problems are, because I have nowhere to go that I know of. I can't take the derivative of fk to plug in, because I will simply end up canceling everything out by using the same equation.
    Help mee!! I'm so hopeless on this problem right now..
     
  9. Nov 1, 2009 #8

    Doc Al

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    If you're still talking about part 2, you already solved it in post #3.
     
  10. Nov 1, 2009 #9
    Oh.. okay awesome!

    But I don't get what you mean by the hint about dv/dt will come into play for part c) because how I am solving for c) is quite simple:

    with a = -(uv^2)/R

    Vf = Vi + at
    (1/3)Vo = Vo + (-uv^2/R)t
    I assume that Vo = v and therefore
    t = (2R)/(3uv)

    Is that right? Thanks sooo much! So helpful
     
  11. Nov 1, 2009 #10

    Doc Al

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    No, not right. You can't use Vf = Vi + at, since the acceleration is not constant. (That's why it seemed so easy!)

    That's where the hint comes in:
    a = dv/dt = -(uv^2)/R

    Use a bit of calculus.
     
  12. Nov 1, 2009 #11
    Doc I've given it thought but I really have no clue what to do, what direction to go in.

    Should I integrate the equation? I don't know what would that even do though.
     
    Last edited: Nov 1, 2009
  13. Nov 1, 2009 #12
    should I bring up dt and bring over v^2 to the left and all its coefficients, and integrate that from Vo to Vo/3?

    like such:

    dv/dt = -uv^2/R
    (-R/u)(1/v^2)dv = dt
    integrate that equation?? It doesn't seem to make much sense though. I'm so clueless..
     
    Last edited: Nov 1, 2009
  14. Nov 2, 2009 #13

    Doc Al

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    Yep. (Where you leave the constants is up to you.)
    Yep.
    Why not?
     
  15. Nov 2, 2009 #14
    If I integrate it should I be integrating it from Vo to Vo/3? Or do I do something else with the integrated equation for t. I think I will integrate from Vo to Vo/3 that seems logical.

    Also, could you give me a brief explanation as to why doing:

    fk = -(umv^2)/R
    v^2 = -fR/um
    2v(dv/dt) = -[df/dt]R/um
    isolate dv/dt

    is incorrect for solving for acceleration? Both seem right to me mathematically, so I don't understand how I would know which one to do, say if I were doing an exam & had no physics forums help.
     
    Last edited: Nov 2, 2009
  16. Nov 2, 2009 #15

    Doc Al

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    Makes sense to me.

    But what does this accomplish? All it does is express the acceleration in terms of the derivative of the unknown friction force. Just going around in circles.

    Note that the instructions tell you to express your answers in terms of R, v, m, and u. This doesn't get you any closer to that goal.
     
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