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Find Angular Acceleration

  1. Aug 7, 2008 #1

    IBY

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    1. The problem statement, all variables and given/known data
    Find angular acceleration.


    2. Relevant equations
    [tex]\omega=\frac{2\pi}{t}[/tex]
    [tex]\frac{\delta*\omega}{\delta*t}=a_{angular}[/tex]

    3. The attempt at a solution
    [tex]\frac{\delta*\omega}{\delta*t}=-1*2\pi*t^{-1-1}[/tex]
    [tex]-2\pi*t^{-2}=\frac{-2\pi}{t^2}[/tex]
    [tex]\frac{-2\pi}{t^2}=a_{angular}[/tex]

    Is this right?
     
  2. jcsd
  3. Aug 7, 2008 #2

    alphysicist

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    Hi IBY,

    It looks to me like the [itex]t[/itex] in the first equation is the period (time to complete a revolution or [itex]2 \pi[/itex] radians), while the [itex]t[/itex] in the second equation is the time, and the left side of that is a derivative with respect to time.

    Since time and period are two diffrent things here, it's best to give them separate symbols. That's what's leading to errors in the work that follows--to determine the acceleration you want to take the derivative with respect to time, but in your original post you are finding the derivative with respect to the period.
     
  4. Aug 7, 2008 #3

    IBY

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    Okay, but how can I do it in respect to time if omega doesn't have it. I assumed T for period was the one to use because frequency is 1/T, and T is measured in seconds. So how can I do it in respect to time itself instead of the period?
     
  5. Aug 7, 2008 #4

    alphysicist

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    What if I said, an object rotates through 1.3 radians in 5.9 seconds. What is the angular speed? What general formula do you use to find that?

    The point is that [itex]\omega= 2 \pi / T[/itex] is not the general formula; it has had values of [itex]2 \pi[/itex] radians plugged in so that it only applies to a specific case.

    But I was just making a point about the formulas in the original post. The problem is to find angular acceleration? Is it asking for a general relationship, or a formula for constant acceleration, or what? Depending on what it's asking for, it may be worthwhile to consider an analogous problem for linear acceleration--the problem often are solved in quite similar ways.
     
  6. Aug 7, 2008 #5

    IBY

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    Let's just say for one cycle, which is 2pi, and 1/T can be any frequency. Omega is the angular velocity, so I thought I could use that to derive angular acceleration. Or you could just do a general relationship, it wouldn't matter.
     
  7. Aug 8, 2008 #6

    alphysicist

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    I don't think you would be keeping track of the problem that way. The general formula for omega is:

    [tex]
    \omega=\frac{d\theta}{dt}
    [/tex]
    (or for average angular velocity:[itex]
    \omega=\frac{\Delta\theta}{\Delta t}
    [/itex]). The form in your original post loses the time dependence of [itex]\theta[/itex]. Do you see where that makes a difference in the expression for alpha?

    Looking again at [itex]\omega=2\pi/T[/itex] in your original post, this specific case could describe a case where the [itex]\omega[/itex] is constant, which would mean that angular acceleration is zero. But the formula you derived in your original post indicates that there is always a negative angular acceleration.
     
  8. Aug 8, 2008 #7

    IBY

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    Would the next step, then be this?:
    [tex]\omega=\frac{d\theta}{dt}[/tex]
    [tex]\omega*dt=d\theta[/tex]
    This gave my acceleration omega. So acceleration is the same as the velocity of the motion?
     
    Last edited: Aug 8, 2008
  9. Aug 8, 2008 #8

    alphysicist

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    It depends on what form you're trying to get. That looks to be how you would derive a formula relating change in [itex]\theta[/itex], once you know how [itex]\omega[/itex] changes with time.

    I'm still not clear as to what you are trying to derive. Are you trying to find a formula for [itex]\alpha[/itex] that's true in all cases? Or are you deriving the constant-[itex]\alpha[/itex] kinematic equations for rotations? Or something else?

    But you already had that

    [tex]
    \alpha=\frac{d\omega}{dt}
    [/tex]
    Depending on what you are trying to find, that would probably be a good starting point.
     
  10. Aug 8, 2008 #9

    IBY

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    That still makes this:

    [tex]\theta*t^{-1}[/tex]

    [tex]-\theta*t^{-1-1}[/tex]

    [tex]\frac{-\theta}{t^2}[/tex]
     
  11. Aug 8, 2008 #10

    IBY

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    Oh, and I am trying to derive alpha that is true for all cases when things are rotating..
     
  12. Aug 8, 2008 #11

    alphysicist

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    No, you are treating [itex]\theta[/itex] as a constant, but it's the thing you're supposed to be differentiating. You are not differentiating the time factor, because the [itex]t[/itex] is not an explicit factor of time; the time [itex]t[/itex] shows up as a derivative with respect to time, and it occurs inside the function [itex]\theta[/itex].

    Look at this notation:

    [tex]
    \begin{align}
    \omega(t) &= \frac{d}{dt}\ (\theta(t) )\nonumber\\
    \alpha(t) &= \frac{d}{dt}\ (\omega(t) )\nonumber
    \end{align}
    [/tex]

    When you plug the first into the second, what do you get? Until you specify something like the time dependence for [itex]\theta[/itex], [itex]\omega[/itex], or [itex]\alpha[/itex], I think that's about all you can do.


    And think back to the linear case. You have:

    [tex]
    v = dx/dt
    [/tex]
    [tex]a=dv/dt[/tex].
    Does that mean that [itex]a=\, -\, x/t^2[/itex] always? What is the expression for linear acceleration that is always true, for all one-dimensional motion? That will be essentially the same expression that you'll find for the rotational case.

    (It's when you specify something, like for example that the acceleration is constant, or the explicit time dependence of the speed, etc., that you'll be able to get expressions that are more explicit.)
     
  13. Aug 8, 2008 #12

    IBY

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    What is omega(t), though? What I know of omega is [tex]\frac{\theta}{T}[/tex], but I don't know how to do it with respect to time. That means there is [tex]\theta(t)[/tex] What is the function of theta, then?
     
  14. Aug 8, 2008 #13

    IBY

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    Okay, I got somethign else working. I found that theta is arclenght/radius. That makes theta purely radians. There is no variable of time because time=t^0. So, I did:
    [tex]s*r^{-1}*T^{-1}*t^0[/tex]
    [tex]s*r^{-1}*T^{-1}*t^0-1[/tex]
    [tex]s*r^{-1}*T^{-1}*t^-1[/tex]
    [tex]\frac{s}{r*T*t}[/tex]
    t is in the bottom, so it becomes frequency.
    [tex]\frac{s}{r*T*T}=\frac{\theta}{T^2}=\alpha[/tex]
    Wait..., argh, that would mean I would have had to multiply the whole thing by 0 because of the exponent 0! At least tell I must be somewhere in the right track, right?
     
  15. Aug 8, 2008 #14

    alphysicist

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    That's the point; they're unknown. I think you're searching for results that just aren't there until you move away from the most general case.

    In the general case, [itex]\omega(t) [/itex] , [itex]\theta(t)[/itex], and [itex]\alpha(t)[/itex] are unknown functions. You can find relationships between these, and you already have. For example, [itex]\omega(t) = d \theta(t)/dt[/itex].


    It's just like the linear case I keep bringing up--for what you are searching for they are the same. What is [itex]x(t)[/itex],[itex] v(t)[/itex], or [itex]a(t)[/itex]? You don't know what they are in general. But you know how they are related: [itex]v = dx/dt, a=dv/dt, \mbox{ and } a=d^2 x/dt^2[/itex].

    Once you have those, let's say you specify that [itex]a(t)=\mbox{constant}[/itex]. Once you specify that, you get all of the kinematic results for constant acceleration.

    Or you can specify that [itex]a= (- c x)[/itex] for some constant [itex]c[/itex]. Then you get the results for a spring type motion.


    But those specific results only follow once you've narrowed the problem down.
     
  16. Aug 8, 2008 #15

    IBY

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    Oh, I get it, thanks. :)
    Though, I did think of a geometric way to do it when theta is arclenght/radius (as you said, now I have specified it).
    Basically, velocity is tangen to angular position, right? And in this case, the position is the angle of the circle. Well, if you put all vectors of velocity in circular motion together, you will find that it forms a circle, with the tips at the circumference. Therefore, acceleration is tangent to the circle of velocity. Therefore: [tex]\frac{\theta}{T}=\omega[/tex],
    so [tex]\frac{\omega}{T}=\frac{\theta}{T^2}=\alpha[/tex].
    That feels right. Hey! It is the same result I got previously!
     
    Last edited: Aug 8, 2008
  17. Aug 8, 2008 #16

    alphysicist

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    I think you're confusing some properties of the linear velocity and position vector for uniform circular motion. The angular acceleration you're looking for is related to the tangential acceleration, but it looks like you're thinking about the centripetal acceleration.

    If you're looking for general expressions for the angular acceleration, they have to be valid for all possible angular accelerations. Consider the simplest case: [itex]\alpha=0[/itex]. That's a perfectly valid angular acceleration, but your formula would not allow that. If something is rotating uniformly, it has an angle, a change in angle, and a period, but if you plug some possible numbers in your formula you don't get zero.

    The form of your formula is similar to the form of the kinematic equation for [itex]\Delta\theta[/itex] assuming constant [itex]\alpha[/itex]:
    [tex]
    \Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2
    [/tex]
    if you make some assumptions about the initial angular velocity; but it is missing a factor of (1/2); and that is far from being a general case.

    But you can also have situations with [itex]\alpha=c t[/itex], [itex]\alpha= c t^2[/itex],[itex]\alpha= c \sin( d t)[/itex], etc. (with d and c constants, for example), or just about anything else. A general expression for [itex]\alpha[/itex] will need to be able to encompass all of these, which is why from your posts I think you are looking for something that isn't there.
     
  18. Aug 9, 2008 #17

    IBY

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    Thanks, that was really helpful. :) I think this is the end of the line, so, yeah, thanks.
     
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