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2 Super Hard Rotational Motion Problems

  1. Dec 3, 2003 #1
    1.) A car rounds a banked curve where the radius of curvature for the road is R, the banking angle is 0(theta), and the coefficient of static friction is [mu]. (a) Determine the range of speeds the car can have without slipping up or down the road. (b) What is the range of speeds possible if R = 100m, 0(theta)=10, and [mu]=0.10 (slippery conditions)????

    2.) In a popular amusement park ride, a rotating cylinder of radium 3.00m is set in rotation at an angular speed of 5.00rad/s. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rdier's clothing and the wall is needed to keep the rider from slipping?????? (hint: recall that the magnitude of the max force of static friction is equal to [mu]n, where n is the normal force - in this case, the force causing the centripetal acceleration.

  2. jcsd
  3. Dec 3, 2003 #2

    Doc Al

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    Staff: Mentor

    Well, take a crack at them and show your work. Demonstrate that you know something about Fnet = ma, centripetal acceleration, and friction.

    Start, as always, by identifying all the forces acting on the objects in question. (Car in 1; rider in 2) Have fun!
  4. Dec 3, 2003 #3


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    Science Advisor

    Don't look all that "super-hard" to me- just apply the formulas that you already know:

    1. F= ma and, for a friction coefficient of μ, F= mμ. Since the road is banked at angle θ, there is a force down the slope of mg cos(&theta). In order not to slip downward, You must have
    ma> mg cos(&theta)- m&mu; and in order not to slip upward, you must have ma< mg cos(&theta)+ m&mu; You also should know the formula for the acceleration of a car going around a circle at constant speed. Put that in for a and solve for v.

    2. Friction force is &mu; time "normal force". In this case the normal force is ma where a is the "acceleration" due to the rotation around a circle at constant speed (you'll need that formula again).
    The friction force, &mu;(ma) must be at least the force of gravity, mg. Solve &mu;ma= mg for &mu;

    Yeah, It's just awful when people expect you to actually learn how to apply formulas!
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