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From the pictures below, it has been said that the tangential component (represented by the tangent at $B$) the linear velocity of a particle moving from $A$ to $B$ is $r \omega$, which is equal to the linear velocity itself. My first question is if this would imply that the tangetntial velocity here is the linear velocity? Now, it has also been said that the radial component of the linear velocity is $0$, implying that the radial component is represented by the tangent at $A$. However, I'm not sure if I am correct here either.

Pictures:

enter image description here enter image description here

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  • $\begingroup$ Sounds correct to me. $\endgroup$
    – Floris
    Jun 12, 2016 at 12:01
  • $\begingroup$ @Floris: okay, I either didn't get the question or didn't get what you nodded on (or maybe both). $\endgroup$
    – user36790
    Jun 12, 2016 at 12:08
  • $\begingroup$ @MAFIA36790 I'm trying to see what represents the tangential component and radial component of the linear velocity from the diagram above. The first thing I wanted to ask was if the tangential component of the linear velocity is represented by the tangent at B, and the second was what represented the radial velocity. $\endgroup$ Jun 12, 2016 at 12:11
  • $\begingroup$ @ReinhildVanRosenú: To your first question; yes. To your second question, radial component must point along the instantaneous radius unit vector; so what do you think? Would it exist for circular motion? $\endgroup$
    – user36790
    Jun 12, 2016 at 12:14
  • $\begingroup$ @MAFIA36790 I'm guessing no. But regarding your answer to my first question, would that means the linear velocity is represented by the tangential component? Because my source tells me (in the 2nd picture) that the tangenial component is $r \delta$ which is equal to the linear velocity. $\endgroup$ Jun 12, 2016 at 12:19

1 Answer 1

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The position vector of a particle in polar coordinates is given by

$$\mathbf r ~=~ r\mathbf e_\mathrm r$$

Velocity $\bf v$ is

\begin{align}\mathbf v&=\frac{\mathrm dr}{\mathrm dt}~\mathbf e_\mathrm r+ r\frac{\mathrm d\mathbf e_\mathrm r}{\mathrm dt}\\ &= \frac{\mathrm dr}{\mathrm dt}~\mathbf e_\mathrm r+ r\frac{\mathrm d\theta}{\mathrm dt}~\mathbf e_\theta \;_;\end{align}

\begin{align}\frac{\mathrm dr}{\mathrm dt}&= \textrm{radial component of velocity}\;,\\ r\frac{\mathrm d\theta}{\mathrm dt}&=\textrm{transverse component of velocity}\;. \end{align}

In circular motion the length of $r$ doesn't change and so $\dot r$ and hence the radial component of $\bf v$ for circular motion is always zero.

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