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Clarifications regarding frame of reference.

  1. Mar 28, 2012 #1
    I have just completed my class 10th. I was presently reading the class 11th H.C Verma physics textbook and here is what I confronted which seems to me as if I am unable to fathom :

    Now I can't understand that how come acceleration of book with respect to earth isn't 0 ?! Please see bold part of quote above. Also please provide an intuitive understanding of frame of reference. Also , please explain the quote a bit more comprehensively.

    Thanks in advace.

  2. jcsd
  3. Mar 28, 2012 #2


    Staff: Mentor

    as you look at the book on the table (dr Suess book?) does it move? if no then the sum total of forces on the book are zero by virtue of f=ma when a=0 then f=0
  4. Mar 28, 2012 #3

    D H

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    Staff Emeritus
    Science Advisor

    The bolded parts of that quote do not say that the acceleration of the book with respect to the Earth is zero. Here's the pertinent part of that quote from the text:
    (Aside: I truly doubt the book says "The sum of two forces ain't 0")

    There are other forces acting on the book. For one, there's the buoyant force exerted by the atmosphere on the book. There's also the gravitational force exerted by the Sun (and by the Moon, and Jupiter, and every other object in the universe) on the book.

    To keep it simple, I'll ignore these other forces. Assume the only objects of interest are the Earth, the table which is resting solidly on the Earth, and the book which is resting solidly on the table.

    Key question: What is the book doing from the perspective of an inertial observer?

    The answer is that the book is undergoing uniform circular motion. The Earth is rotating on it's axis, one revolution per day. Suppose the book has a mass of 1 kg and suppose the table is at the equator at sea level, 6378 kilometers from the center of the Earth. The book is undergoing uniform circular motion with an angular velocity of one revolution per day (one revolution per sidereal day to be pedantic). Calling this angular velocity ω, [itex]\omega = \frac{2\pi}{\text{sidereal day}} \approx 72.92\,\text{microradians}/\text{second}[/itex], the acceleration of the book as observed by our inertial observer is
    [tex]a = \frac{v^2} r = r\omega^2 \approx 0.0339\,\text{meters}/\text{second}^2[/tex]
    This means the net force on this 1 kg book is this about 0.0339 newtons -- not zero. It can't be zero. If it was zero the book wouldn't be undergoing uniform circular motion. This 0.0339 newton net force is small compared to the force due to gravity, which for a 1 kg book is about 9.81 newtons. That it is small why we can ignore the rotation of the Earth -- at least in some problems.
  5. Mar 28, 2012 #4
    Very well explained ! Thanks.

    That's really mind-bashing of the book's author. Well he mustn't say that acceleration is not zero. I am not saying that he is wrong , of course he is cent percent correct. But still he must have given some intuitive idea of the word "why ?" regarding this discussion.
  6. Mar 30, 2012 #5
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