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Derivatives in relation to physics

  1. Oct 27, 2015 #1
    OK, I have never had physics till this semester and I am in calculus based physics and it is kicking my butt.

    I don't understand how derivatives are properly used in the formulas, and I have an example, my question is the image attached.

    @=theta
    A=alpha

    SO,
    A) I need omega (angular velocity). w= d@/dt right? so if @=0.11(4.9)^2, plugged into that equation...
    we end up with:
    w=d(0.11*4.9^2)/d(4.9)
    w=0.11(2)(4.9)/1=1.08 rad/s, <------this is right I think.

    B) got it: v=rw, no sweat, v=10.8m/s

    C) Here is where I trip up. According to my understanding, A=dw/dt, which means d(1.08)/d(4.9)....

    but thats not right, derivatives of constants=0....I don't understand. It's been a while since I took calculus, so am I just messing something dumb up???
     

    Attached Files:

  2. jcsd
  3. Oct 27, 2015 #2
    and to boot, the formula for omega(w) is just listed as w=@/t, which gives me 0.539 rad/s.

    and neither 0.539rad/s OR 1.08rad/s is correct. grrrrrrrrrrr
     
  4. Oct 27, 2015 #3
    is it because they are differential change of @ and t? sooooo, its actually d((0.11*t^2)-0)/d(4.9-0) so the differential just takes the 0 away?

    ...no that isn't right.
     
  5. Oct 27, 2015 #4

    SteamKing

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    Maybe you should refresh your knowledge of calculus before trying to apply it to physics.

    Back to the problem at hand:

    θ = 0.11 t2

    If you want to find the angular velocity ω = dθ / dt, then differentiate the expression for θ with respect to t first, before plugging in any values for t.

    https://www.mathsisfun.com/calculus/derivatives-rules.html

    What you wrote, w=d(0.11*4.9^2)/d(4.9), made no sense whatsoever. The derivative, dθ / dt, does not mean substitute the expression for θ, the value for t, and cancel out the d's. It means something altogether different.
     
  6. Oct 27, 2015 #5
    So it is in leibniz notation, that explains a lot of the issues I am having.

    So Id I write it this way, I end up with d/dt (0.112t) = 1.08 @ t=4.9s


    but regardless, I still don't understand when I find angular acceleration why it is the dw/dt when to get the answer I don't take a derivative anywhere. I just divide w by t. Is it because it is actually the second derivative of d@/dt so I end up with 0.11(2) =0.22 which is the answer....yuuup

    Please excuse my inabilities with leibniz, My professor stuck to newton notation, with y dots or f'(x). He touched on leibniz notation, but I never quite understood it. Since I hadn't planned on going further I didn't bother to pick it up. I will have to remedy that as it evidently starting to cause me problems.
     
  7. Oct 27, 2015 #6
    I guess my issue here is, how do I know when I am using an equation what level of derivative I am actually at? Is it just something I have to know?
     
  8. Oct 27, 2015 #7
    You need to go back and review calculus. These are all math issues that you're asking about, not Physics. These are typical homework problems covered in a calculus course.

    Chet
     
  9. Oct 27, 2015 #8

    SteamKing

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    No, that's still not correct.

    If θ =0.11 t2, then dθ / dt = 2 * 0.11 * t = 0.22 t, for all values of t.

    If you want to find the value of the angular velocity at a particular time t0, then dθ / dt = 0.22 t is evaluated by setting t = t0, thus

    At t0 =4.9 s, dθ / dt = 0.22 * 4.9 = 1.08 rad/sec = angular velocity @ t = 4.9 sec

    The angular acceleration is defined as the change of angular velocity with respect to time. Since angular velocity = dθ / dt = ω, then angular acceleration α = dω / dt.
    For this particular example, ω = 0.22 t, so α = 0.22 rad / sec2, which is a constant value for any time t = t0.
    The two forms of derivative notation are slightly different, but the concepts are the same:

    ## dθ / dt = \dot{θ} ##
    ## dω / dt = \dot{ω} = \ddot{θ} ##

    Each dot above a variable indicates differentiation with respect to time. More than one dot indicates a higher-order time derivative.

    As Chet indicated above, your problem is with calculus, rather than physics. Spend some time learning (or re-learning) the calculus, and then come back to the physics.
     
  10. Oct 27, 2015 #9
    Thank you for your help.
     
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