Vector differentiation of velocity (polar coord.)

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SUMMARY

The discussion focuses on the differentiation of velocity in polar coordinates, specifically addressing the formula for acceleration derived from the velocity vector. The velocity of a particle is expressed as {\bf{v}} = v_r {\bf{\hat r}} + r\omega \hat \theta, where v_r = \frac{{dr}}{{dt}} and \omega = \frac{{d\theta}}{{dt}}. The acceleration is calculated as {\bf{a}} = \left( {\frac{{dv_r }}{{dt}} - \omega ^2 r} \right){\bf{\hat r}} + \left( {2\omega v_r + r\frac{{d\omega }}{{dt}}} \right)\hat \theta. A key point of confusion is the term \omega ^2 r in the derivative, which arises from the product rule applied to the radial component of the velocity.

PREREQUISITES
  • Understanding of polar coordinates and their applications in physics.
  • Familiarity with vector calculus, particularly differentiation of vector functions.
  • Knowledge of angular velocity and its relationship to linear velocity.
  • Proficiency in using dot notation for time derivatives in physics.
NEXT STEPS
  • Study the application of the product rule in vector differentiation.
  • Learn about the relationship between angular velocity and linear velocity in polar coordinates.
  • Explore the concept of unit vectors and their derivatives in vector calculus.
  • Investigate more complex motion in polar coordinates, including forces and energy considerations.
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Students and professionals in physics, particularly those focusing on mechanics and dynamics, as well as educators teaching vector calculus and polar coordinates.

quozzy
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The velocity of a particle moving in a plane in polar coordinates is

[tex]{\bf{v}} = v_r {\bf{\hat r}} + r\omega \hat \theta[/tex]

where [tex]v_r = \frac{{dr}}{{dt}}[/tex] and [tex]\omega = \frac{{d\theta }}{{dt}}[/tex].

By differentiating w.r.t. time, show that the acceleration of the particle is

[tex]{\bf{a}} = \left( {\frac{{dv_r }}{{dt}} - \omega ^2 r} \right){\bf{\hat r}} + \left( {2\omega v_r + r\frac{{d\omega }}{{dt}}} \right)\hat \theta[/tex]

(The no-subscript v should be bold, as should the a and the r's with hats.)

Okay, I'm confident I can work this one out, except for one thing: how does that [tex]\omega ^2 r[/tex] get into the derivative? I assume the [tex]\bf{\hat r}[/tex] is a unit vector, so the derivative of [tex]v_r[/tex] should just be [tex]{\frac{{dv_r }}{{dt}}[/tex] right? Anyway, if anyone could just explain that detail to me, I'll be on my way.

Thanks!
 
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I like writing in the dot notation because it helps me to remember the dependencies of each variable.

[tex]\dot{\mathbf{r}}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta}\hat{\mathbf{\theta}}[/tex]

So r depends on time and theta depends on time, what are all the variables that now have r in them and theta in them? (Hint: unit vectors might count!)
 
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