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Homework Help: Circular Motion

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data
    Ignoring the motion of the Earth around the Sun and the motion of the Sun through space, find:

    a) the angular velocity
    b) the linear speed
    c) the centripetal acceleration

    of a body resting on the ground at the equator.

    What would the length of a day be if the angular speed of the Earth's rotation on its axis were to increase until the body became effectively weightless? Radius of Earth = 6.4 x 106m

    2. Relevant equations

    1: ω=θ/t
    2: v=2πr/T and v=ωr
    3: a=v2/r and a=rω²

    3. The attempt at a solution

    a) ω=θ/t
    We have 24 hours for one complete rotation of the Earth = 24x60x60=86400s
    One complete rotation = 360° = 2π(rads)
    Therefore, 2π/86400s=73x10-6rad s-1 (2s.f.)

    b) v=ωr

    c) a=rω2
    a=(6.4x106)x(73x10-6)2=0.0341056m s-1
    =34.1x10-3m s-1

    In order to answer the last part of the question, I guess we must consider what angular speed would cause the body to become effectively weightless. We don't have a mass of the body (kg) and therefore, can't calculate its weight.
    I'm lost. Please advise and check first part of question.
    Much appreciated as always :bugeye:
  2. jcsd
  3. Feb 7, 2010 #2


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    You don't need the mass of the body. If the Earth rotates so fast that a 70 kg father becomes weightless, so does his child who has a mass of only 20 kg. Bodies become weightless when the acceleration of gravity is equal to the centripetal acceleration.

    The other parts of the question look OK although I did not perform the numerical calculations.
  4. Feb 7, 2010 #3


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    Could you solve the problem if the body had 1 kg mass?

  5. Feb 7, 2010 #4
    When the acceleration of gravity is equal to the centripetal acceleration:
    Centripetal acceleration = 10m s-2

    a=rω², so ω=±√a/r = √10m s-2/6.4 x 106m
    = 1.25x10-3rad s-1 (is that unit correct?)

    Therefore, 2π x 1.25x10-3 = 7.9x10-3s
    A really quick day!

    Is that right?
  6. Feb 7, 2010 #5


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    The unit is correct.

    What equation is this? Please write it symbolically so that I can see where it is coming from.
  7. Feb 8, 2010 #6


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    [tex]\omega=\frac{2\pi}{T}=1.25 \cdot 10^{-3} s^{-1}\rightarrow T=?[/tex]

  8. Feb 8, 2010 #7
    This is the new centripetal acceleration:
    ω=±√a/r = √10m s-2/(6.4 x 106m)
    = 1.25x10-3rad s-1

    I think I may have messed up there and multiplied 2π by ω (2π x 1.25x10-3 = 7.9x10-3s). Instead I guess I should have divided by ω.
    Using ω=2π/T would give: 2π/(1.25x10-3rad s-1=1600π or 5026.5s (1d.p.)
  9. Feb 9, 2010 #8


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    looks better, doesn't it?

  10. Feb 17, 2010 #9
    Looks really good to me. Still a very quick day.
    Glad I can get more sunshine than that.
    Thanks everyone :smile:
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