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Physics Problem: Centripetal Force

  1. Oct 3, 2004 #1
    A stone has a mass of 3.0*10^-3 kg and is wedged into the tread of an automobile tire, as the drawing shows. The coefficient of static friction between the stone and each side of the tread channel is 0.72. When the tire surface is rotating at 20 m/s, the stone flies out of the tread. The magnitude FN of the normal force that each side of the tread channel exerts on the stone is 1.8 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire.

    I tried using the equation: r= m*v^2 divided by static friction coefficient times mg, but I got the wrong answer. Please Help!!
  2. jcsd
  3. Oct 3, 2004 #2
    ahem ...

    -- AI
  4. Oct 3, 2004 #3

    Doc Al

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    Staff: Mentor

    What force is providing the centripetal acceleration? Once you find the centripetal force, then use the equation [itex]F = m v^2/r[/itex] to find the radius.
  5. Oct 3, 2004 #4
    A Pitfall Prevention:
    The force causing centripetal acceleration is called centripetal force in some textbooks, but this is a pitfall for students. Giving the force causing circular motion a name -centripetal force- leads many students to consider this a new kind of force, rather than a new role for force. A common mistake in force diagrams is to draw in all of the usual forces and then add another vector for the centripetal force. But it is not a separate force it is simply one of our familiar forces acting in the role of causing a circular motion. For the motion of the earth around the sun for example, the “centripetal force” is gravity. For a rock whirled on the end of a string, the “centripetal force” is the tension in the string. For an amusement park patron pressed against the inner wall of a rapidly rotating circular room, the “centripetal force” is the normal force from the wall. So I don’t think you should use the phrase centripetal force.
  6. Oct 4, 2004 #5

    Doc Al

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    Staff: Mentor

    Faiza raises a good point, that some folks act as if "centripetal force" were a new kind of force, like gravity. (I did not see any evidence of that misconception here, though.)

    Calling a force "centripetal" just means that it acts towards the center of some circular motion. Conceptually, it is no different than talking about horizontal or vertical forces.
  7. Oct 4, 2004 #6
    Yes that is a problem (so much so that some books nowdays even mention it in a box) that can be solved by emphasizing the equation of motion along the radial direction for curvilinear motion is

    [tex]\sum F_{r} = m(\ddot{r} - r\dot{\theta}^2)[/tex]

    which reduces to

    [tex]\sum F_{r} = -mr\dot{\theta}^2[/tex]

    in case of uniform circular motion (radius = constant). This essentially means that the sum of all forces acting on the body along the radial direction must equal the product of the mass and the radial acceleration (as it must be by Newton's Laws). Additionally

    [tex]a_{r} = -r\dot{\theta}^2[/tex]

    is precisely the radial acceleration term. So a useful interpretation is that the vector sum of the radial forces equals the product of the mass and the radial acceleration (the centripetal acceleration in case of uniform circular motion).

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