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Mass on Turntable

  1. Oct 8, 2006 #1
    Information given:

    A small metal cylinder rests on a circular turntable that is rotating at a constant speed as illustrated in the diagram View Figure .

    The small metal cylinder has a mass of 0.20 \rm kg, the coefficient of static friction between the cylinder and the turntable is 0.080, and the cylinder is located 0.15 \rm m from the center of the turntable.

    Take the magnitude of the acceleration due to gravity to be 9.81 \rm m/s^2


    What is the maximum speed v_max that the cylinder can move along its circular path without slipping off the turntable?
    Express your answer numerically in meters per second to two significant figures.

    Attached Files:

  2. jcsd
  3. Oct 8, 2006 #2

    Chi Meson

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  4. May 8, 2010 #3
    The centripetal force Fc, for a body of mass m traveling with speed v in a circle of radius r is,

    Fc = (mv^2)/r.................eqn 1

    The object to slip off the turntable if the centripetal force overcomes the maximum force due to static friction F_max:

    F_s = (u_s)N = (u_s)mg................eqn 2

    u_s is the coefficient of static friction.
    g = 9.8 m/s^2
    m = mass

    Using Eq. 1 and Eq. 2 (Fc=Fs) we solve for the maximum speed vmax.

    Thus v_max = sqrt((u_s)mg)

    you can do the math.
  5. Feb 22, 2012 #4
    The above process is correct but when u solve eqn1 n 2 u get
    Vmax = sqrt(μs*r*g) = =0.34 m/s
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