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The instantaneous speed of a point along a circular path is given by $v=\omega r$, where $\omega = \frac{\Delta \theta}{\Delta t}$, $s=\Delta \theta r$, and $v=\frac{s}{t}$.

However, isn’t the displacement measured like this?

enter image description here

The relationship between normal speed ($v$) and angular speed ($\omega$) is related via arc length ($s$). The displacement ($d$) above is a straight line, not the arc length. So why is $v=\frac{s}{t}$ when it should be $v=\frac{d}{t}$? Would this not invalidate the equation $v=\omega r$?

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  • $\begingroup$ I would suggest you revisit the definitions of average and instantaneous velocity. $\endgroup$ – Declan Aug 30 '15 at 3:42
  • $\begingroup$ the distance traveled is $s$ and not $d$. $\endgroup$ – ja72 Aug 30 '15 at 5:06
  • $\begingroup$ instantaneous velocity is taken in a time interval so small that displacement and distance can approximately be taken equal to each other. Hope this clears the confusion. $\endgroup$ – Saad Aug 30 '15 at 7:08

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