Effective acceleration due to gravity in non-inertial frame

  • #1
Lost1ne
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Take some sort of system accelerating with respect to an inertial reference frame: let's take a spherical mass on the end of a string forming a simple pendulum with the ceiling of a car, and allow that car to accelerate uniformly.

Could someone share with me how they interpret the concept of a geffective where we take the vector sum of the fictitious force and the gravitational force acting on the mass? I don't feel as if I'm understanding it at a level that I would like to.
 

Answers and Replies

  • #3
Lost1ne
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I'm missing something elementary here. If F_apparent = F_true - mA and A comes from F_true = mA, then we may reach the conclusion that a_apparent equals zero always. I've realized that I also used your way of thinking in the article where we add on the fictitious forces and then make the claim that our object of interest is not accelerating in our non-inertial reference frame. But, looking back at this, doesn't it not make sense for a_apparent to undisputedly equal zero? We view numerous objects accelerating through our non-inertial reference frame daily.
 
  • #4
A.T.
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...and A comes from F_true = mA,
No, F_net = ma.
 
  • #5
Lost1ne
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No, F_net = ma.
Aha. "a" is the measured acceleration of our object/system of interest from our inertial reference frame, and "A" is the acceleration of our *non-inertial reference frame with respect to our *inertial reference frame.
 
  • #6
A.T.
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Aha. "a" is the measured acceleration of our object/system of interest from our inertial reference frame, and "A" is the acceleration of our *non-inertial reference frame with respect to our *inertial reference frame.
Lets stick to lower case with subsctpt:

Fnet = Ftrue + Fapparent = mabody

Where for a linearly accelerating reference frame:

Fapparent = -maframe
 
  • #7
kuruman
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But, looking back at this, doesn't it not make sense for a_apparent to undisputedly equal zero? We view numerous objects accelerating through our non-inertial reference frame daily.
That is all correct. One is (by definition) at rest and remains at rest with respect to one's frame of reference. If you are in an elevator you will not be able to tell the difference between the following two cases: (a) The elevator is at rest near the surface of the Earth; (b) the elevator is in free space accelerating with acceleration g (relative to an inertial frame) in a direction from your feet to your head. In either case if you are standing on a bathroom scale, it will display what you know to be your true weight.
 

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