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Tangential and Centripetal Acceleration in Circular Motion

  1. Jun 28, 2015 #1
    Why is there no component of tangential acceleration in uniform circular motion? I think its because when we apply the WE Theorem (Work-Energy Theorem) as follows:
    Work Done = ΔKE
    => F.S = 1/2*m*(vf2 - vi2),
    =>(ma).s = 1/2*m*(vf2 - vi2)
    => (ma).s = 1/2*m*(0) (since the speed remains constant in uniform circular motion)
    => mascosθ = 0
    => θ = 90°
    That means the angle between a and s is 90°. So in this case the acceleration of the particle is purely centripetal. This can also be seen by the fact that when we compute Δv over a very small time interval Δt, the resulting vector (that we would get from calculating the difference of vectors Δv) would always point towards the centre or in other words, it would be centripetal acceleration. So that's why there is no tangential component in uniform circular motion.

    But how does this component change the direction of the particle so as to keep the velocity tangential at all points? Can someone show it quantitatively using equation in vector form?

    Coming to non-uniform circular motion, the a we get from computing Δv over a very small time interval Δt points as shown in the attachment. We break this a into 2 components, one of which comes out to be centripetal and the other one comes out to be tangential. As I discussed above, the centripetal component cannot increase the velocity, so it has to be the tangential one that does that.

    My question is, firstly how does the combination of centripetal and tangential acceleration work in this case so as to keep the particle moving along the tangent at all points as well as increase its velocity, simultaneously? Can anyone show it quantitatively by using equations of motion in vector form?

    Secondly, when we decompose the acceleration into two components, its upon us as to how we decompose them , meaning that the choice of axis is ours. So usually we take the normal x and y axis, so that y points towards the centre and x would hence automatically point tangentially. But what if we don't do it that way? Then the components we would get would be different and we won't be able to generalize that one of the components cannot increase the speed. It would be very confusing that way.
     

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  2. jcsd
  3. Jun 28, 2015 #2

    DrGreg

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    Method 1. Use[tex]
    \textbf{r} = r \cos \theta \, \textbf{i} + r \sin \theta \, \textbf{j} \, ,
    [/tex]or, if you prefer, the notation[tex]
    \textbf{r} = \begin{bmatrix} r \cos \theta \\ r \sin \theta \end{bmatrix} \, .
    [/tex]Differentiate with respect to time to get velocity [itex]\dot{\textbf{r}}[/itex] and acceleration [itex]\ddot{\textbf{r}}[/itex]. Radius [itex]r[/itex] is constant and, for uniform motion, angular velocity [itex]\dot \theta[/itex] is constant.

    Method 2. Without using components, differentiate the equations[tex]
    \textbf{r} \cdot \textbf{r} = \mbox{constant (for any circular motion)} \\
    \dot{\textbf{r}} \cdot \dot{\textbf{r}} = \mbox{constant (for }\textbf{uniform}\mbox{ circular motion)}
    [/tex]
    Whichever method you use, this should give you information about the relative directions of [itex]\textbf{r}[/itex], [itex]\dot{\textbf{r}}[/itex] and [itex]\ddot{\textbf{r}}[/itex].
     
  4. Jun 28, 2015 #3
    The method you gave doesn't explain how centripetal acceleration changes direction of the particle and how tangential acceleration affects the speed.
     
  5. Jun 29, 2015 #4

    A.T.

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  6. Jun 29, 2015 #5

    DrGreg

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    Method 2 won't, but method 1 will if you take it far enough.

    Note that once you have expressions for [itex]\textbf{r}[/itex], [itex]\dot{\textbf{r}}[/itex], and [itex]\ddot{\textbf{r}}[/itex], you can simplify by putting [itex]\theta = 0[/itex], on grounds of symmetry; what happens at that angle ought to be no different from what happens at any other angle.

    The answers you get should be in terms of [itex]r[/itex], [itex]\dot{\theta}[/itex] and [itex]\ddot{\theta}[/itex] (assuming constant [itex]r[/itex]).
     
    Last edited: Jun 29, 2015
  7. Jun 29, 2015 #6
  8. Jun 30, 2015 #7
    So if the tangential acceleration increases the speed of the particle, what does the centripetal acceleration do?
     
  9. Jun 30, 2015 #8

    A.T.

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    Changing the direction.
     
  10. Jun 30, 2015 #9
    And how do you know that? Can you prove it?
     
  11. Jun 30, 2015 #10

    Drakkith

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    I believe that's what the the wiki page linked earlier does.
     
  12. Jun 30, 2015 #11
    No, it doesn't. It just computes the 2 components of acceleration, viz tangential and centripetal with the tangential one increasing the speed of the particle. But it nowhere says that the centripetal component changes the direction. The expression of tangential acceleration itself i.e. at = d|v|/dt states that it changes the speed, but the expression for centripetal acceleration, i.e. at = -ω2r, does not state that it changes the direction of the particle so as to keep it going on a tangent at all times.
     
  13. Jun 30, 2015 #12

    Nathanael

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    Centripetal acceleration is like a "perpetually sideways push"

    If you push something sideways (for an instant), it can't change the speed, it can only change the direction.
    (Granted, you can only change the direction infinitesimally before you have to start pushing in another direction; the "new sideways" direction. This is what centripetal acceleration does.)

    Work it out yourself. What acceleration is necessary to keep something moving in a circle at a constant speed? If you assume that it has to be pointed towards the center at all times, then it's not complicated to work out. (It's a purely mathematical problem.) Use an arbitrary "v" and "R"
     
  14. Jul 1, 2015 #13
    If it is that mathematical then show me how is it done? You're stating that it would keep it gong in a circle because of intuitiveness, but I want to see the actual proof behind it, which I am not able to figure out myself.

    What math are you talking about? Once we have the expression for centripetal acceleration, what further math needs to be done?
     
    Last edited: Jul 1, 2015
  15. Jul 1, 2015 #14

    Drakkith

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    Of course it does. Acceleration is a vector, and centripetal acceleration points towards the center of the circle, which means that the object is continually accelerating towards the center. When you do the math, as provided in the link, you will find how the position changes over time given a tangential velocity and a centripetal acceleration.

    In addition, it flat out states: While objects naturally follow a straight path (due to inertia), this centripetal acceleration describes the circular motion path caused by a centripetal force.
     
  16. Jul 1, 2015 #15

    Nathanael

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    My point is that, if you derive an expression for the acceleration of a particle traveling in a circle at constant speed and it turns out to be the centripetal acceleration formula, then does that not mean that centripetal acceleration causes particles to move tangent to a circle at a constant speed? Isn't this what you're looking for?

    The derivation of centripetal acceleration based on the assumption of constant speed implies that it does not change the speed. The math I was talking about was simply to derive the formula from the constraint of constant speed.
     
  17. Jul 1, 2015 #16
    What you are saying is that we get 2 accelerations, one radial and and one tangential, when we take the time derivative of velocity v. The tangential component changes the speed as its expression itself clearly states that. And now we are left with the centripetal/radial acceleration. Now what you are saying is that since the accelerations we got were derived from a v that is always tangential in nature (because the v we chose was tangential from the very beginning even before we had computed centripetal acceleration), so that is why the radial component helps in keeping the v along tangent. But I don't think you can relate the two so simply without having any formal-proof to back up your claims.
     
  18. Jul 1, 2015 #17

    Nathanael

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    I'm not sure exactly what you are trying to prove? That the radial component of acceleration can't change the magnitude of velocity?
     
  19. Jul 1, 2015 #18
    No. That how does radial component of acceleration change the particle's direction so as to keep it moving in a circle.
     
  20. Jul 1, 2015 #19

    Drakkith

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    I'm not sure what to tell you. I've programmed an orbital simulation where I split the velocity, force, and acceleration of the Earth into X and Y components that works just fine. It you look at those components after an infinitesimal amount of time dT you will find that the force components change the velocity components in exactly the right manner to produce circular motion. I don't even need to use angular units. If you'd like, I can copy my equations and post them here.
     
  21. Jul 1, 2015 #20

    Nathanael

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    Okay so if it's specifically about a circle then we can parameterize it's motion:
    [itex]\vec R=R(\cos(\theta)\hat i+\sin(\theta)\hat j)[/itex]
    [itex]\vec V=R\dot \theta(-\sin(\theta)\hat i+\cos(\theta)\hat j)[/itex]
    [itex]\vec a=R\ddot \theta(-\sin(\theta)\hat i+\cos(\theta)\hat j)+R\dot \theta^2(-\cos(\theta)\hat i-\sin(\theta)\hat j)[/itex]

    If [itex]\ddot \theta[/itex] is zero, then it traveling at a constant speed and the only component left of acceleration is that which is centripetal. It is traveling in a circle because that is from where we derived this formula for acceleration (we differentiated ##\vec R## which traces out a circle).

    If it is moving then ##\dot \theta## is not zero therefore it must have centripetal acceleration given by ##R\dot \theta^2## and if ##\ddot \theta## is not zero then there is an extra component of acceleration (which is obvious to prove that it's tangential) that corresponds to changing speed (nonzero ##\ddot \theta## implies the ##|\vec V|## is changing).

    This is pretty much all there is to say about circular motion, I'm not really sure what else you're looking for.
     
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