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Deriving expressions for angular velocity and acceleration

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Derive the expressions for the [itex]i_r[/itex] and [itex]i_θ[/itex] components of velocity and acceleration.

    2. Relevant equations
    [itex]r=|r|i_r[/itex]

    [itex]\omega=\frac{dr}{dt}[/itex]

    [itex]\alpha=\frac{dω}{dt}=\frac{d^2r}{dt}[/itex]

    3. The attempt at a solution
    [itex]r=|r|i_r[/itex]

    [itex]\omega=\frac{dr}{dt}i_r+r\frac{di_r}{dt}[/itex]

    [itex]\alpha=\frac{d^2r}{dt^2}i_r+\frac{dr}{dt}\frac{di_r}{dt}+\frac{d^2i_r}{dt^2}r+\frac{di_r}{dt}\frac{dr}{dt}[/itex]

    4. The solutions I need to get to
    [itex]\omega=\frac{dr}{dt}i_r+r\frac{d_\theta}{dt}i_θ[/itex]

    [itex]\alpha=\frac{d^2r}{dt^2}i_r+\frac{dr}{dt}\frac{di_\theta}{dt}i_\theta+\frac{dr}{dt} \frac{d\theta}{dt}i_\theta+r\frac{d^2\theta}{dt^2}i_\theta-r\left(\frac{d\theta}{dt}\right)^2i_r[/itex]

    5. Thoughts
    I think that they're substituting in something for parts of my answer. I just don't know what they're substituting in, or how to find it.
     
    Last edited: Nov 18, 2011
  2. jcsd
  3. Nov 19, 2011 #2
    [itex]\omega[/itex] [itex]\neq[/itex] [itex]\frac{dr}{dt}[/itex] but [itex]\omega[/itex] = [itex]\frac{d\theta}{dt}[/itex].

    [itex]\alpha[/itex] [itex]\neq[/itex] [itex]\frac{d^{2}r}{dt^{2}}[/itex] but [itex]\alpha[/itex] = [itex]\frac{d\omega}{dt}[/itex] = [itex]\frac{d^{2}\theta}{dt^{2}}[/itex]
     
  4. Nov 19, 2011 #3
    So basically I'm doing it completely wrong. Would my work be correct for v and a instead of ω and [itex]\alpha[/itex]?

    I'm not quite sure how to solve the problem. Could you point me in the right direction?
     
  5. Nov 19, 2011 #4
    oops!
    let me try again.
     
    Last edited: Nov 19, 2011
  6. Nov 19, 2011 #5
    I think that what you were asked to derive are:

    [itex]v=\frac{dr}{dt}i_r+r\frac{d_\theta}{dt}i_θ[/itex]

    and

    [itex]a=\frac{d^2r}{dt^2}i_r+\frac{dr}{dt}\frac{d\theta}{dt}i_\theta+\frac{dr}{dt} \frac{d\theta}{dt}i_\theta+r\frac{d^2\theta}{dt^2}i_\theta-r\left(\frac{d\theta}{dt}\right)^2i_r[/itex]
     
  7. Nov 20, 2011 #6
    I have solved the problem. Hopefully typing up my work will help someone someday.


    SOLVING FOR KNOWNS

    [itex]i_r=cos\theta+sin\theta[/itex]

    [itex]\frac{di_r}{dt}=-sin\theta+cos\theta[/itex]

    [itex]\frac{di_r}{dt}=\frac{d\theta}{dt}i_\theta[/itex]



    [itex]i_\theta=-sin\theta+cos\theta[/itex]

    [itex]\frac{di_\theta}{dt}=-cos\theta-sin\theta[/itex]

    [itex]\frac{di_\theta}{dt}=-\frac{d\theta}{dt}i_r[/itex]




    POSITION VECTOR

    [itex]r=ri_r[/itex]




    SOLVING FOR VELOCITY

    [itex]v=\frac{dr}{dt}[/itex]

    [itex]v=\frac{dr}{dt}i_r+\frac{di_r}{dt}r[/itex]

    After substituting what we know, we get:

    [itex]v=\frac{dr}{dt}i_r+r\frac{d\theta}{dt}i_\theta[/itex]

    So the solutions for velocity are:

    [itex]v_r=\frac{dr}{dt}[/itex]

    [itex]v_\theta=r\frac{d\theta}{dt}[/itex]




    SOLVING FOR ACCELERATION

    [itex]a=\frac{dv}{dt}[/itex]

    [itex]a=\frac{d^2r}{dt^2}i_r+\frac{di_r}{dt}\frac{dr}{dt}+\frac{dr}{dt}\frac{d\theta}{dt}i_\theta+r \frac{d^2\theta}{dt^2}i_\theta+r\frac{d\theta}{dt}\frac{di_\theta}{dt}[/itex]

    After substituting what we know, we get:

    [itex]a=\frac{d^2r}{dt^2}i_r+\frac{d\theta}{dt}\frac{dr}{dt}i_\theta+\frac{dr}{dt}\frac{d\theta}{dt}i_θ+r \frac{d^2\theta}{dt^2}i_\theta-r\frac{d\theta}{dt}\frac{d\theta}{dt}i_r[/itex]

    And then we simplify to get:

    [itex]a=\frac{d^2r}{dt^2}i_r+2\frac{d\theta}{dt}\frac{dr}{dt}i_\theta+r\frac{d^2\theta}{dt^2}i_\theta-r\left(\frac{d\theta}{dt}\right)^2i_r[/itex]

    So the solutions for acceleration are:

    [itex]a_r=\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2[/itex]

    [itex]a_\theta=2\frac{d\theta}{dt}\frac{dr}{dt}+r\frac{d^2\theta}{dt^2}[/itex]
     
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