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The particle is moving towards the center as the disc is spinning, the position of the particle is described by the following expression:

$r = r(t) cos(ωt)i + r(t) sin(ωt)j$

enter image description here

How do I calculate the particles velocity?(not angular velocity)

Andrews
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noname197
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  • Is $r$ the radius? – Kandinskij Apr 24 '19 at 14:50
  • You should note that you have two different $r$'s here. On the left it is a position vector, on the right it is the radius in polar coordinates. The two terms on the right need unit vectors multiplying them. The correct expression is $\vec r = r(t) \cos(ωt)\hat x + r(t) \sin(ωt)\hat y$ – Ross Millikan Apr 24 '19 at 14:53
  • The position as you wrote it is not a vector. I think you wanted $\mathbf r = r(t) \cos(\omega t) \mathbf i + r(t) \sin(\omega t)\mathbf k$ or something like that where you substitute your notation for the $x$ unit vector for $\mathbf i$ and your notation for the $z$ unit vector for $\mathbf k$. – David K Apr 24 '19 at 14:54
  • MathJax hint: if you put a backslash before common functions you get the proper font and spacing, so \cos (\omega t) gives $\cos (\omega t)$ compared to cos (\omega t) which gives $cos (\omega t)$ – Ross Millikan Apr 24 '19 at 14:55

2 Answers2

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It is easier to work in polar coordinates, where the position is $(r(t),\omega t)$ because $\theta=\omega t$. Then use the equation for velocity in polar coordinates $$\vec v=\dot r \hat r+r\dot \theta \hat \theta$$ Now if you want the result in Cartesian coordinates you can transform back.

Ross Millikan
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you have the displacement vector $$\underline{r}=\left(\begin{matrix}r\cos\omega t\\r\sin\omega t\end{matrix}\right)$$

so the velocity vector is $$\frac{d\underline{r}}{dt}=\left(\begin{matrix}\dot{r}\cos\omega t-r\omega\sin\omega t\\ \dot{r}\sin\omega t+r\omega\cos\omega t\end{matrix}\right)$$

David Quinn
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