Relation between r ,ω and θ for rotation around fixed axis.

  • #1
relation between r ,ω and θ for rotation around fixed axis.

[tex] \frac{d\textbf {r}}{dt} = \textbf {ω} [/tex]


[tex] \frac{dθ}{dt} = ω [/tex]


[tex] \lvert\frac{d\textbf {r}}{dt}\rvert = \frac{dθ}{dt} [/tex]

bold means vector. Is this right?
 
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  • #2
tiny-tim
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Hi AakashPandita! :smile:

(type \left and and \right before two ordinary ||s, and they resize to fit! :wink:)

[tex] \frac{dθ}{dt} = ω = |\textbf{ω}|[/tex]
ω (the angular velocity vector) is along the axis of rotation (ie, out of the page)
[tex] \left|\frac{d\hat{\textbf {r}}}{dt}\right| = r\frac{dθ}{dt} = rω [/tex]
and
[tex] \frac{d\hat{\textbf {r}}}{dt} = rω\hat{\textbf{θ}} [/tex]
where ##\hat{\textbf {r}}## and ##\hat{\textbf {θ}}## are the unit vectors in the radial and tangential directions
 
  • #3
vanhees71
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Let the rotation be around the [itex]z[/itex] axis of a Cartesian coordinate system. Then
[tex]\vec{r}(t)=r(t) \begin{pmatrix}
\cos[\alpha(t)] \\
\sin[\alpha(t)] \\
0
\end{pmatrix}.
[/tex]
This gives
[tex]\vec{v}(t)=\dot{r}(t) = \dot{r}(t) \begin{pmatrix}
\cos[\alpha(t)] \\
\sin[\alpha(t)] \\
0
\end{pmatrix} + r(t) \dot{\alpha}(t) \begin{pmatrix}
-\sin[\alpha(t)] \\
\cos[\alpha(t)] \\
0
\end{pmatrix} = \dot{r} \hat{r} + r \omega \hat{\theta}.
[/tex]
If the mass is fixed to a circle, then you have [itex]\dot{r}=0[/itex] and you can write
[tex]\vec{v}=\vec{\omega} \times \vec{r} \quad \text{with} \quad \vec{\omega}=\omega \vec{e}_z=\dot{\theta} \vec{e}_z.[/tex]
 
  • #4
ZapperZ
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[tex] \frac{d\textbf {r}}{dt} = \textbf {ω} [/tex]


[tex] \frac{dθ}{dt} = ω [/tex]


[tex] \lvert\frac{d\textbf {r}}{dt}\rvert = \frac{dθ}{dt} [/tex]

bold means vector. Is this right?

You should already know that something is not quite right here, because the dimensions are all wrong! Always check that first, because that is your first "line-of-defense"!

Zz.
 
  • #5
ω (the angular velocity vector) is along the axis of rotation (ie, out of the page)
[tex] \left|\frac{d\hat{\textbf {r}}}{dt}\right| = r\frac{dθ}{dt} = rω [/tex]

how did you get this?
 
  • #6
Let the rotation be around the [itex]z[/itex] axis of a Cartesian coordinate system. Then
[tex]\vec{r}(t)=r(t) \begin{pmatrix}
\cos[\alpha(t)] \\
\sin[\alpha(t)] \\
0
\end{pmatrix}.
[/tex]
This gives
[tex]\vec{v}(t)=\dot{r}(t) = \dot{r}(t) \begin{pmatrix}
\cos[\alpha(t)] \\
\sin[\alpha(t)] \\
0
\end{pmatrix} + r(t) \dot{\alpha}(t) \begin{pmatrix}
-\sin[\alpha(t)] \\
\cos[\alpha(t)] \\
0
\end{pmatrix} = \dot{r} \hat{r} + r \omega \hat{\theta}.
[/tex]
If the mass is fixed to a circle, then you have [itex]\dot{r}=0[/itex] and you can write
[tex]\vec{v}=\vec{\omega} \times \vec{r} \quad \text{with} \quad \vec{\omega}=\omega \vec{e}_z=\dot{\theta} \vec{e}_z.[/tex]

i dont understand matrices
 
  • #7
D H
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vanhees71 wasn't using matrices, AakashPandita. He was using vectors.
 
  • #8
i dont understand those brackets
 
  • #9
vanhees71
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How are you supposed to solve this question without vectors? I guess you just use a different notation than I use. In your original posting you used abstract vectors, but the equations are unfortunately not correct. So I thought, it's best to use components wrt. a Cartesian basis to explicitly calculate the derivatives.
 

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