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Angular Velocity Differential Equation

  1. Oct 8, 2013 #1
    Problem:

    Newton's 2nd Law for rotational motion states that the product of the moment of Inertia, I, and the angular acceleration alpha(t) is equal to the net Torque acting on a rotating body. Consider a wheel that is being turned by a motor that exerts a constant torque T. Friction provides an opposing torque proportional to the square root of the angular velocity w(t). The moment of inertia of the wheel is I.

    A) Write out a differential equation that represents Newton's 2nd law for rotational motion in this case. Note that alpha(t)=w'(t)

    B) If the initial angular velocity is w_o, determine the angular velocity as a function of time, w(t). Can you find an expression for w(t)

    Attempt at Solution:

    I was able to find the answer for A. I found that it was: I(dw/dt)=T-k(sqrt(w))

    As for part B, I am really confused. I know I have to seperate the equation and integrate it. I know that dw=w'(t)dt.Other than that I'm lost. First of all, should I be plugging in Wo now or at the end? I'm trying to isolate the w's but I keep getting:
    I *w'(t)+k(sqrt(w))=T

    I think I need to find a way to get the w's as a product or quotient of each other, so I can use dw=w'(t)dt.
     
  2. jcsd
  3. Oct 8, 2013 #2

    Dick

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    It's a separable differential equation. I(dw/dt)=T-k(sqrt(w)) so dw/(T-k(sqrt(w))=dt/I. Integrate both sides. It's a differential equations thing. The w0 part is after you've got the general solution.
     
  4. Oct 8, 2013 #3
    Ok, now I see it! However, I get: int (dw/T-Ksqrt(w)) on one side. I tried subbing for the whole denominator, but it doesn't seem to work..
     
  5. Oct 8, 2013 #4
    I [itex]\frac{dω}{dt}[/itex] = τ-k[itex]\sqrt{w}[/itex]
    dt(τ-k[itex]\sqrt{w}[/itex])=(I)dω
    [itex]\frac{dt}{I}[/itex] = [itex]\frac{dω}{τ-\sqrt{w}}[/itex]
    [itex]\int \frac{dt}{I} [/itex]= [itex]\int \frac{dω}{τ-k\sqrt{w}} [/itex]
    [itex]\frac{1}{I}[/itex][itex]\int dt [/itex]= [itex]\int \frac{dω}{τ-k\sqrt{w}} [/itex] (I is constant so you can do this I believe?)
    [itex]\frac{t}{I}[/itex]=[itex]\int \frac{dω}{τ-k\sqrt{w}} [/itex]
    [itex]\int \frac{dω}{τ-k\sqrt{w}} [/itex] u=τ-k[itex]\sqrt{ω}[/itex], du=-[itex]\frac{k}{2\sqrt{ω}}[/itex] dω
    [itex]\frac{1}{k}[/itex]2[itex]\sqrt{w}[/itex]du=dω, ([itex]\frac{1}{k}[/itex])(-u+τ)=[itex]\sqrt{ω}[/itex] so dω=([itex]\frac{1}{k}[/itex])2(-u+τ)du=[itex]\frac{-2u+2τ}{k}[/itex]du
    [itex]\int \frac{dω}{τ-\sqrt{w}} [/itex] = [itex]\int \frac{-2u+2τ}{ku} [/itex] du
    = [itex]\frac{1}{k}[/itex][itex]\int\frac{-2u}{u} [/itex]du + [itex]\frac{1}{k}[/itex][itex]\int \frac{2τ}{u} [/itex]du
    = [itex]\frac{-2}{k}[/itex][itex]\int du [/itex] + [itex]\frac{2τ}{k}[/itex][itex]\int \frac{1}{u} [/itex] = -2[itex]\frac{u}{k}[/itex]+[itex]\frac{2τ}{k}[/itex]*ln|u|+c
    = -2(τ-[itex]k\sqrt{w}[/itex])[itex]\frac{1}{k}[/itex] + 2[itex]\frac{τ}{k}[/itex] *ln|τ-[itex]k\sqrt{ω|}[/itex]+c
    [itex]\frac{-2}{k}[/itex] (τ-[itex]k\sqrt{ω}[/itex]+τln|τ-k[itex]\sqrt{ω}[/itex]|)+c = [itex]\frac{t}{I}[/itex]

    I'm lost at this point. I think everything is OK up to here? How would I plug ωo in? Would I plug it in so that [itex]\sqrt{ω}[/itex] becomes [itex]\sqrt{w-wo}[/itex] which is like [itex]\sqrt{\Deltaω}[/itex] ? or ω(0)=ωo ?
    My solutions page says it should be k(√ω-√ωo)+τ ln [itex]\frac{τ-k\sqrt{w}}{τ-k\sqrt{ωo}}[/itex]= -k2t/2I
     
    Last edited: Oct 8, 2013
  6. Oct 8, 2013 #5

    Dick

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    You are almost right. I think you dropped a factor of k and a sign. Check it. To get the constant c put t=0, and ##\omega(0)=\omega_0##. Once you've solve for c put it back into the solution and simplify.
     
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