Vector differentiation of velocity (polar coord.)

In summary, the velocity of a particle moving in a plane in polar coordinates is represented by the vector {\bf{v}} = v_r {\bf{\hat r}} + r\omega \hat \theta, where v_r = \frac{{dr}}{{dt}} and \omega = \frac{{d\theta }}{{dt}}. By differentiating w.r.t. time, the acceleration of the particle is given by {\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
  • #1
quozzy
15
0
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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  • #2
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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FAQ: Vector differentiation of velocity (polar coord.)

1. What is vector differentiation of velocity in polar coordinates?

Vector differentiation of velocity in polar coordinates is a mathematical process used to find the derivative of a vector (velocity) in terms of its components (speed and direction) in a polar coordinate system. It allows us to determine the rate of change of velocity in a specified direction.

2. How is vector differentiation of velocity in polar coordinates different from Cartesian coordinates?

In Cartesian coordinates, velocity is represented by two components (x and y) in a horizontal and vertical direction. In polar coordinates, velocity is represented by two components (radial and tangential) in a radial and tangential direction. Therefore, the process of finding the derivative is different for each coordinate system.

3. What is the formula for finding the derivative of velocity in polar coordinates?

The formula for finding the derivative of velocity in polar coordinates is:
dV/dt = (dr/dt)ur + (rdθ/dt)uθ
where dV/dt is the derivative of velocity, dr/dt is the derivative of the radial component, rdθ/dt is the derivative of the tangential component, and ur and uθ are unit vectors in the radial and tangential directions, respectively.

4. What is the physical significance of vector differentiation of velocity in polar coordinates?

The physical significance of vector differentiation of velocity in polar coordinates is that it helps us understand the direction and magnitude of an object's velocity at a given point in time. It also allows us to determine the acceleration of an object in a specific direction, which is useful in analyzing the motion of objects in circular or curved paths.

5. In what real-life situations is vector differentiation of velocity in polar coordinates applicable?

Vector differentiation of velocity in polar coordinates is applicable in many real-life situations, such as analyzing the motion of planets in our solar system, determining the trajectory of satellites, and understanding the movement of objects in circular or curved paths (e.g. roller coasters, ferris wheels, etc.). It is also useful in engineering and physics for designing and analyzing complex motion systems.

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