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Non-Uniform circular motion with friction

  1. Nov 4, 2007 #1
    1. The problem statement, all variables and given/known data
    A small block of mass m slides on a horizontal frictionless surface as it travels around the inside of a hoop of radius R. The coefficient of friction between the block and the wall is mu; therefore, the speed v of the block decreases. In terms of m, R. mu, and v, find expressions for each of the following.
    b. The block's tangential acceleration dv/dt
    c. The time required to reduce the speed of the block from an initial value v0 to vo/3
    (A diagram is given here on page 4: www.swcp.com/~gants/calendars/ap%20physics/sept%20docs/(A)%20Newton's%20LawsC.doc)

    2. Relevant equations
    dv/dt = aT = tangential acceleration
    aA = angular acceleration
    aC = centripetal acceleration
    v/r=w (angular velocity)

    3. The attempt at a solution
    First I tried to use aT = r * aA = r * d^2 (theta) / dt^2 but I don't think that's right because none of those values are given.
    Then I tried to do
    Fnet = m * (aC^2 + aT^2)^(1/2) = ((m*v^2/r)^2+(m*aT - mu*m*v^2/r))^(1/2)
    but I think that's wrong because I'm taking taking the forces and finding the vector sum of them and I don't know if that's allowed like I can do with the (aC^2 + aT^2)^(1/2) to get acceleration.

    Perhaps would it help for me to know that dv/dt = dv/dx * dx/dt = v*dv/dx?? I'm kind of lost, any help would be greatly appreciated.
    Last edited: Nov 4, 2007
  2. jcsd
  3. Nov 4, 2007 #2

    Doc Al

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    Staff: Mentor

    The only tangential force is friction. What does that equal? (Hint: What's the normal force?)
  4. Nov 4, 2007 #3
    Is this correct?

    F = m * dv/dt = mu*m*v^2/r

    dv/dt = mu*v^2/r
  5. Nov 4, 2007 #4

    Doc Al

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    Looks good to me.
  6. Nov 4, 2007 #5
    Yay. :smile: Thank you so much.
  7. Nov 4, 2007 #6
    Wait, tangential acceleration is supposed to be negative, right? So is it supposed to be -mu*v^2/r or is something done wrong?

    And just to make sure, for part c where it asks about the time required to reduce speed from v0 to 1/3*v0, is it:
    dv/dt = a = -mu*v^2/r

    2/3*v / delta(t) = -mu*v^2/r

    (-2*r)/(3*mu*v) = delta(t)

    But that doesn't make sense because the change in time cannot be negative.

    -EDIT: never mind about the time being negative part. haha i know the change in velocity has to be -2/3 because its 1/3 v - 1 v. and the negative of the negative 2/3 makes it positive.
    but can you still check everything else?
    Last edited: Nov 4, 2007
  8. Nov 4, 2007 #7

    Doc Al

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    Right. It should be negative.

    That should be:
    dv/dt = -mu*v^2/r

    This is not correct. Don't assume the acceleration is constant (the velocity is not). Integrate! ([itex]dv/dt \neq \Delta v / \Delta t[/itex])
  9. Nov 4, 2007 #8
    Is this right?

    dv/dt = a = -mu*v^2/r

    integral(dv/v^2) = integral (-mu/r dt)

    -1/v = -mu/r * delta(t)

    delta (t) = r/(v*mu)

    And isn't acceleration constant? I know velocity is not.
  10. Nov 4, 2007 #9

    Doc Al

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    Evaluate the integral properly:
    [tex]\int_{v_0}^{v_0/3}(1/v^2) dv = -\int\mu/r dt[/tex]

    How can it be constant if it depends on velocity?
    Last edited: Nov 4, 2007
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