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Classical Mechanics: Inertial Reference Frames

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data

    Classical Mechanics: John Taylor

    (1.27) The hallmark of an inertial reference frame is that any object which is subject to a zero net force will travel in a straight line at a constant speed. To illustrate this, consider the following experiment: I am standing on the ground (which we shall take to be an inertial frame) beside a perfectly flat horizontal turntable, rotating with constant angular velocity ω. I lean over and shove a frictionless puck so that it slides across the turntable, straight through the center. The puck is subject to zero net force and, as seen from my inertial frame, travels in a straight line. Describe the puck's path as observed by someone sitting at rest on the turntable. This requires careful thought, but you should be able to get a qualitative picture. For a quantitative picture, it helps to use polar coordinates; see Problem 1.46.

    2. Relevant equations

    [tex]a_r=\ddot{r}[/tex]
    [tex]a_φ=\ddot{φ}[/tex]
    [tex]v_r=\dot{r}[/tex]
    [tex]v_φ=\dot{φ}[/tex]

    3. The attempt at a solution

    To make this problem as simple as possible (I think), I started the puck at the center of the circle, which I took as the origin, and chose the puck to travel along the [itex]φ=0[/itex] axis with a constant velocity [itex]v_o[/itex] in the [itex]S[/itex] frame. In this frame, the puck will travel along a straight line from [itex](r,φ)=(0,0)[/itex] to [itex](r,φ)=(R,0)[/itex], where [itex]R[/itex] is the radius of the turntable. Therefore, in the [itex]S[/itex] frame,
    [tex]a_r=a_φ=0[/tex]
    [tex]v_r=v_o[/tex]
    [tex]v_φ=0[/tex]
    [tex]r=v_ot[/tex]
    [tex]φ=0[/tex]

    In the [itex]S'[/itex] frame, the puck will start at the origin and travel southeast with a curved trajectory from [itex](r',φ')=(0,0)[/itex] to [itex](r',φ')=(R,φ_{final})[/itex]. In this frame, the puck is seen to have some centripetal acceleration since the direction of its linear velocity changes, but no tangential acceleration since its speed remains constant. Therefore, this frame is non-inertial because the puck has a zero net force on it but is seen to accelerate. In this frame,
    [tex]a'_r=-ω^2r'[/tex]
    [tex]a'_φ=0[/tex]
    [tex]v'_φ=-ω[/tex]
    [tex]r'=\cos(-ωt)[/tex]
    [tex]φ'=-ωt[/tex]

    I am not quite sure if this is correct, and I would like to gain some insight. Please show me any misconceptions I have or incorrect math processes that I have taken.
     
  2. jcsd
  3. Apr 7, 2016 #2

    PeroK

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    Those equations are not right. The way I would look at it (starting at the origin was a good idea) is to take the velocity of the puck in the ##S## frame as ##v##. Then, in the ##S## frame at time ##t##, the puck is at:

    ##r = vt, \phi = 0##

    Now, where is the puck in the ##S'## frame at time ##t##?
     
  4. Apr 7, 2016 #3
    I seem to have forgotten about the velocity of the puck in the [itex]S[/itex] frame for my calculations in the [itex]S'[/itex] frame.

    Here is another attempt at the solution:

    In the radial direction, the puck travels at a constant velocity [itex]v[/itex], so the radial position of the puck in the [itex]S'[/itex] frame at time [itex]t[/itex] will be,
    [tex]r'=vt[/tex]

    In the angular direction, the puck travels with a constant angular velocity [itex]-ω[/itex], so the angular position of the puck will be,
    [tex]φ'=-ωt[/tex]

    Is this correct? If so, wouldn't this mean that the puck doesn't have an acceleration in the [itex]S'[/itex] frame, and hence the frame would also be an inertial frame?
     
  5. Apr 7, 2016 #4

    PeroK

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    That's the next question. ##\ddot{r'} = 0 ## and ##\ddot{\phi'} = 0## so how can there be acceleration? Hint: think about circular motion at constant speed.
     
  6. Apr 7, 2016 #5
    In the [itex]S'[/itex] frame, the puck has a constant speed but is changing direction, so it has a centripetal acceleration. How can the puck have [itex]\ddot{r'}=0[/itex] if it has a centripetal acceleration?

    Attempt to understand this:

    If I define the position vector of the puck in the [itex]S'[/itex] frame as [itex]\vec{\ddot{p'}}=(\ddot{r'},\ddot{φ'})[/itex], then [itex]|\vec{\ddot{p'}}|=0[/itex], but [itex]\vec{\ddot{p'}}\neq{0}[/itex].

    I assume this isn't correct, because it doesn't seem possible for the magnitude of a vector to equal zero while the vector itself isn't zero. I must be misunderstanding.
     
  7. Apr 7, 2016 #6

    PeroK

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    To cut to the chase, the acceleration in polar coordinates is given by:

    ##\ddot{\vec{r}} = (\ddot{r} - r\dot{\phi}^2)\vec{\hat{r}} + (r\ddot{\phi} + 2\dot{r} \dot{\phi})\vec{\hat{\phi}}##

    So, even with ##\ddot{r} = \ddot{\phi} = 0## you still have terms contributing to the acceleration. The first ##-r\dot{\phi}^2## is called the "centrifugal" term and the second ##2\dot{r} \dot{phi}## is the "coriolis" term.

    You need to go back to your equations of motion for the puck and calculate the acceleration as above.

    PS Taylor must cover this somewhere, so it's worth finding that.

    PPS When it comes to acceleration, polar coordinates can make things more complicated than good old Cartesian.
     
  8. Apr 7, 2016 #7
    It seems that he covers this concept in more depth in Chapter 9. Perhaps I should revisit this question after understanding this chapter.

    Before I suspend this problem until further notice, let me see if I'm on the correct path.

    Initially, I had my acceleration in the angular direction wrong. I should have had,
    [tex]a'_r=-r'ω^2[/tex]
    [tex]a'_φ=-2\dot{r'}ω[/tex]

    Would these give me the following differential equations?
    [tex]\ddot{r'}=-r'ω^2[/tex]
    [tex]\ddot{φ'}=-2\dot{r'}ω[/tex]

    But this doesn't seem right, because we already said that [itex]\ddot{r'}=\ddot{φ'}=0[/itex]

    Also, would you suggest trying to solve this problem in Cartesian rather than polar coordinates?
     
  9. Apr 7, 2016 #8

    PeroK

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    You're missing something here. At any point, there are "radial" and "angular" unit vectors, denoted by ##\vec{\hat{r}}## and ##\vec{\hat{\phi}}##. Note that these change with position and that's essentially what makes the formula for acceleration I quoted in post #6 complicated. Now, the acceleration of a particle at a point is a vector, so it can be resolved into components in these two directions. This is similar to what you do in Cartesian coordinates, except there the unit vectors ##\vec{i}## and ##\vec{j}## are fixed and don't change with position.

    So, at every point in a particle's path you have an instantaneous acceleration (vector) and an instantaneous pair of unit vectors (##\vec{\hat{r}}## and ##\vec{\hat{\phi}}##). The equation in post #6 relates these. Note that this is a vector component.

    Now, the important point. The component of acceleration in the radial direction is not simply ##\ddot{r}##. Again, think of circular motion where ##r## is constant, yet there is acceleration in the (constantly changing) radial direction. Similarly, the component of acceleration in the angular direction is not ##\ddot{\phi}##.

    If you have not covered the equation in post #6 yet, then it may be a good idea to convert your equations in post #3 to Cartesian coordinates and calculate ##\ddot{x}## and ##\ddot{y}##, which you should find are non-zero.
     
  10. Apr 7, 2016 #9
    OK, I see that I need to revisit polar coordinates, because I'm struggling to find the correct differential equations in this coordinate system.

    Let me take your suggestion and convert the variables in post #3 to Cartesian coordinates. The x- and y-positions of the puck in the [itex]S'[/itex] frame are,
    [tex]x'=r'\cos(φ')=vt\cos(-ωt)[/tex]
    [tex]y'=r'\sin(φ')=vt\sin(-ωt)[/tex]

    Taking the double derivative of each of these positions will then give me the acceleration in the x- and y-directions, which are (very much to my liking) non-zero.

    [tex]\ddot{x'}=2ωv\sin(-ωt)-ω^2vt\cos(-ωt)[/tex]
    [tex]\ddot{y'}=-[2ωv\cos(-ωt)+ω^2vt\sin(-ωt)][/tex]

    Does this analysis seem correct you?
     
  11. Apr 7, 2016 #10

    PeroK

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    It looks right.
     
  12. Apr 7, 2016 #11
    Sweet, sweet understanding.

    So, to recap. In the [itex]S[/itex] frame, the puck has a zero net force on it and is seen to have a zero acceleration (post #1). Hence, this frame is inertial. In the [itex]S'[/itex] frame, the x'- and y'-positions are seen to vary sinusoidally, and hence, the acceleration in the x'- and y'-directions will also vary sinusoidally (post #9). Since the puck has a zero net force, but is seen to accelerate, this frame must be non-inertial.

    Does this mean that centrifugal and coriolis forces are fictitious forces that are used to account for the acceleration in non-inertial reference frames?
     
  13. Apr 8, 2016 #12

    PeroK

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    This point came up in another thread recently:

    https://www.physicsforums.com/threads/is-polar-coordinate-system-non-inertial.864543/#post-5431151

    There are two different situations:

    In this case, you have an object with no forces on it. But, when viewed in an non-inertial reference frame, it is seen to accelerate, hence has fictitious forces acting on it in that frame. These are closely related to centripetal and coriolis forces.

    However, if you consider an object in the rotating reference frame moving in a straight line at constant velocity in that frame, then that object is really accelerating and must have real forces acting on it. These are the real centripetal and coriolis forces.
     
  14. Apr 8, 2016 #13
    OK, let me try to put these concepts in my own words to see if I understand them. I'll take the example of a mass on a string traveling around in a circle at constant velocity.

    In an inertial reference frame that is at rest with respect to the mass, there is a real net force on the mass from the string. The tension force from the string, in this specific case, is what we call the real centripetal force.

    In a non-inertial reference frame that is co-rotating with the mass, the mass is at rest, so it must have a zero net-force. In this case the mass would have a fictitious centrifugal force that is equal in magnitude and opposite in direction to the tension force.

    Let's say the string is cut and the mass moves with a constant velocity in the rotating reference frame (which is no longer a co-rotating frame since the mass is no longer rotating, but is still non-inertial). In this frame the mass will have the fictitious centrifugal force pushing it outward AND the fictitious coriolis force on it pushing it along a curved path. Apparently there is third fictitious force that I haven't heard much about called the Euler force. This force will appear to act on the mass if it is moving at a changing speed in the rotating reference frame. I'm not quite sure yet how this force would affect the trajectory of the mass in the non-inertial frame.

    Back to the inertial frame. What if the mass has a changing speed and hence, an angular acceleration. The tension force can't be what caused the angular acceleration because the tension force doesn't exert a torque on the mass. Could we say that the mass has an angular acceleration in the inertial reference frame due to some real force exerted on it such as drag from the medium it's traveling in?

     
  15. Apr 8, 2016 #14

    PeroK

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    This is correct.

    This isn't correct. The mass will move at constant velocity in the inertial frame and have no real forces on it.

    In the non-inertial frame, it will have ficticious centrifugal and coriolis forces on it: to explain its motion in that frame.

    I had never heard of the Euler force, but I looked it up and that applies when the angular velocity of the reference frame is changing. Given that I'm not particularly brilliant, if I can look that up for myself, perhaps you should too?

    I don't understand this. It can't change speed without another force. If it's slowing down, then yes friction or air resistance would be real forces that would do that.
     
  16. Apr 8, 2016 #15
    Yes, I should've said the mass will move with a constant velocity in the inertial frame, and a changing velocity in the non-inertial frame.

    OK, I seem to have a much better understanding of some of these concepts now. Thanks for you help.
     
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