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Rotation help

  1. Mar 25, 2004 #1
    A bicycle has wheels of 1.2m diameter. the bicyclist accelerates from rest with constant acceleration to 24km/h in 14.0s. What is the angular acceleration of the wheels?

    If the bicycle is going forwards relative to the ground with a speed of 24km/h, then, all points on the tread are moving around the wheel with a speed (s) of 24km/h relative to the axel.

    s = 24km/h = 6.7m/s (I know the significant figures rules but I dont know where to round)
    r = 1.2m/2 = 0.6m

    w=6.7/0.6 = 11.2 rad/s ( is my rounding correct here?)

    a=v^2/r=(6.7)^2/0.6= 74.8m/s^2

    Is the reasoning correct?

    Thank you for your help.
  2. jcsd
  3. Mar 25, 2004 #2
    A number in the problem has three digits (12.0), so that is what your answer should have at most, also.

    The number 6.7/0.6 ought to have just 2 figures, but since the first number really should have 3 figures in the first place, and the second number is accurate to 3 figures, . . .

    Tangential acceleration = angular acceleration * radius
    is another formula you could use to check your work, which I only eyeballed. Looks OK to me.
  4. Mar 25, 2004 #3


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    This is corect

    This is correct

    The question asks for angular acceleration.

    a = (delta w)/t
    a = (11.2)/(14)
    a = 0.8 rad/s^2
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