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Uniform Circular Motion of dice

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Fuzzy dice hang from the rear-view mirror of a car rounding a curve. If the curve has a radius of 275 meters and the dice are hanging at an angle of 12° from the vertical, how fast is the car going?

    2. Relevant equations
    a = v2/R
    for an acceleration with a constant magnitude, as in uniform circular motion. I took this to mean that a in this equation is the direction of the acceleration. i.e., the Cartesian coordinates.

    3. The attempt at a solution
    In uniform circular motion, the magnitude of the acceleration is always the same and pointing inwards. Does this mean the magnitude of the acceleration is equal to R? So the magnitude of acceleration would be 275 m/s2 in this case? I'm not sure how to begin this problem but I could solve it if that was true. But I think R and the magnitude of the acceleration are just related, not necessarily equal, but I don't really understand how.
  2. jcsd
  3. Sep 18, 2010 #2


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    No, R is not 275 m/s2, it is 275 m. Plus as you already wrote a=v2/R, and that is not equal to R.
    Try drawing a free-body diagram for the dice. What forces act on the dice?
  4. Sep 18, 2010 #3
    Consider the forces acting on the dice when it is inclined at an angle with the vertical.
    There is tension in the string, its weight vertically downwards and lastly the centripetal force.
    Equate horizontal and vertical forces and solve the 2 equations.
    You will get the the answer.
  5. Sep 19, 2010 #4
    So I tried something but it didn't really work out:
    F = centripetal force
    T = tension force
    Fg = force gravity = mg
    Theta = 12 degrees, measured from the vertical

    Fnet,x = F + T sin theta
    Fnet,y = T cos theta - mg

    so if I arrange the equations like so:
    T sin theta = -F
    T cos theta = mg

    T sin theta = Tx and T cos theta = Ty and by trig, Tx/Ty = tan theta
    so I said

    tan theta = -F/mg
    and since F = centripetal force, which is also F = m * (v2/R)
    I then can say
    tan theta = -m(v2/R) / mg
    The negative sign becomes a problem, since I must take the square root of the term to solve for v. So should my F be pointing in the opposite direction of T, which would then make everything work out nicely? Or is my method just flawed? I rather feel like T and F should be pointing in the same direction (both towards the center of the circle), but I'm not sure what I did wrong.
  6. Sep 19, 2010 #5


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    You're pretty close.
    The y equation is okay, but we have to think about the x equation. It's best to think about what is physically causing the forces: there is only the tension force T and gravitational force mg.

    The centripetal force F is the net force and is the vector sum of the two forces I just mentioned. Another way to say that is: Fnet is equal to mv2/R, acting in the x-direction:

    mv2/R = Fnet,x = T sin(θ)
    0 = Fnet,y = T cos(θ) - mg

    (On the left side of each equation, I have written what we know must be the net force, due to the circular motion of the dice. On the right side are the forces that come from some physical cause--string tension or gravity).

    That one should be slightly different; no minus sign.
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