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Physics - Circular Motion

  1. Sep 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Could someone please help me with this quesiton

    4. The earth revolves about its axis every 24 hours. Find the magnitude of the average acceleration of a point on

    the equator over a 5-hour time interval. (The radius of the earth is 6.38 x 106 m.)


    2. Relevant equations

    A = V^2/R

    A = 4pi^2*R/T^2

    A = 4pi^2 * R * f*2

    3. The attempt at a solution

    ok im not quite sure what to do but this is what i have done so far

    T = 24h or 86400s
    R = 6.38 x 10^6

    A= 4pi^2 * (6.38 x 10^6)/(24h)^2
    A = 437278.31m/h^2

    so i am guessing that's the acceleration per hour

    for 7 hours it would be

    437278.31m/h^2 * 7h = 3.06 x10^6m/h
    this is where i get confused im pretty sure im supposed to have m/h^2 in the answer but i have m/h

    please help or explain what i am doing wrong
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 16, 2014 #2

    tjmiller88

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    Gold Member

    So first off, when I think of problems like these, I always always want to get my units in terms of radians, meters, and seconds. Start off by converting your units.

    Second - what type of acceleration is this point experiencing? Is it speeding up or slowing down? Or is it staying at the same speed? Make sure you understand what centripetal acceleration is, and remember that the Earth is rotating at constant speed.

    Here's a hint - this is the equation I would use: ac2*R

    See if that helps.
     
  4. Sep 16, 2014 #3
    @tjmiller88
    sorry i dont really understand the part about it speeding up or slowing down. also ac=ω2*R .. i dont understand how to use that, since the velocity has'nt been given. the one thing that is putting me off is the 5hr interval
     
  5. Sep 16, 2014 #4

    collinsmark

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    Homework Helper
    Gold Member

    Hello tayyabah1415,

    Welcome to Physics Forums! :smile:

    This is a pretty odd problem. But let's try to do our best, given the information we have.

    You can calculate the instantaneous, centripetal acceleration of the point by using your a = v2/R formula. You've already started with that.

    The problem statement asks you to find the average acceleration over a 5 hour time interval. That's an unusual thing to ask for in my opinion, but let's go with anyway.

    You can determine the average acceleration by adding the initial and final acceleration together, and dividing by 2.

    [tex] \vec a_{ave} = \frac{\vec a_i + \vec a_f}{2} [/tex]

    In this problem, the initial and final acceleration vectors have the same magnitude, but different directions. To find the average acceleration over a time interval, you must add them together like vectors, not scalars (you must place the tail of one to the head of the other, then trace out a new vector from tail to head). Then as a final step, divide the magnitude of the resultant vector by 2.

    By the way, according to your problem statement, you are supposed to find the average acceleration over a 5 hour time interval. But in your attempted solution you were using 7 hours. I'm not sure what to make of that. :uhh:
     
  6. Sep 17, 2014 #5

    collinsmark

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    Homework Helper
    Gold Member

    On second thought, the problem might be asking you to calculate the the "average acceleration" in a different way.

    Once could say that,

    [tex] a_{ave} = \frac{\Delta v}{\Delta t} [/tex]

    Once again, the magnitude of the point's velocity does not change over time, but its direction does. Rewriting that in vector form gives

    [tex] \vec a_{ave} = \frac{\vec v_f - \vec v_i}{\Delta t} [/tex]

    Don't forget to treat the velocities like vectors.

    By the way, that's not [quite] the same thing as my advice in the previous post. I guess it depends on how your instructor/textbook defines "average acceleration."
     
    Last edited: Sep 17, 2014
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