Quick question with spherical coordinates and vectors

In summary, in spherical coordinates, the position of a point is written as r = b, \phi = \omegat and \vartheta = \pi / 2 [1 + \frac{1}{4} cos (4\omegat). The velocity is given by \mathbf{v} = \dot{r} \mathbf{\hat r} + r\,\dot\theta\,\boldsymbol{\hat\theta } + r\,\dot\varphi\,\sin\theta \mathbf{\boldsymbol{\hat \varphi}} and the acceleration is given by \mathbf{a} = \left( \ddot{r} - r\,\dot\theta^
  • #1
Johnson
29
0
So here's the question:
An ant crawls on the surface of a ball of radius b in such a manner that the ants motion is given in spherical coordinates by the equations:
r = b, [itex]\phi[/itex] = [itex]\omega[/itex]t and [itex]\vartheta[/itex] = [itex]\pi[/itex] / 2 [1 + [itex]\frac{1}{4}[/itex] cos (4[itex]\omega[/itex]t).

Find the speed as a function at time t and the radial acceleration of the ant.

I found the speed, doing [itex]\left|v\right|[/itex] = b[itex]\omega[/itex][cos[itex]^{2}[/itex]([itex]\frac{\pi}{8}[/itex]cos 4[itex]\omega[/itex]t) + [itex]\frac{\pi^{2}}{4}[/itex] sin[itex]^{2}[/itex] 4[itex]\omega[/itex]t] [itex]^{1/2}[/itex]

Now I don't even know where to begin to take the derivative of that, lol. I know i derive the actual vector v, not the magnitude of it. But how do i derive e[itex]_{\phi}[/itex] and e[itex]_{\vartheta}[/itex]?

I got for velocity

v = [itex]\widehat{e}[/itex][itex]_{\phi}[/itex]b[itex]\omega[/itex]cos [[itex]\frac{\pi}{8}[/itex]cos 4[itex]\omega[/itex]t] - [itex]\widehat{e}[/itex][itex]_{\vartheta}[/itex]b[itex]\omega[/itex] [itex]\frac{\pi}{2}[/itex]sin (4[itex]\omega[/itex]t)

Any help on deriving that to find acceleration would be awesome :s Maybe I'm missing a rule with [itex]\widehat{e}[/itex][itex]_{\phi}[/itex], but I'm getting stuck.

Thanks :)
 
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  • #2
Any help would be greatly appreciated.
 
  • #3
See: http://en.wikipedia.org/wiki/Spherical_coordinate_system,

"Kinematics"
In spherical coordinates the position of a point is written,

[tex] \mathbf{r} = r \mathbf{\hat r} [/tex]

its velocity is then,

[tex] \mathbf{v} = \dot{r} \mathbf{\hat r} + r\,\dot\theta\,\boldsymbol{\hat\theta } + r\,\dot\varphi\,\sin\theta \mathbf{\boldsymbol{\hat \varphi}} [/tex]

and its acceleration is,

[tex] \mathbf{a} = \left( \ddot{r} - r\,\dot\theta^2 - r\,\dot\varphi^2\sin^2\theta \right)\mathbf{\hat r}
+ \left( r\,\ddot\theta + 2\dot{r}\,\dot\theta - r\,\dot\varphi^2\sin\theta\cos\theta \right) \boldsymbol{\hat\theta }
+ \left( r\ddot\varphi\,\sin\theta + 2\dot{r}\,\dot\varphi\,\sin\theta + 2 r\,\dot\theta\,\dot\varphi\,\cos\theta \right) \mathbf{\boldsymbol{\hat \varphi}} [/tex]


ehild
 

FAQ: Quick question with spherical coordinates and vectors

What are spherical coordinates and how are they different from other coordinate systems?

Spherical coordinates are a way of representing points in three-dimensional space using three parameters: radius, inclination, and azimuth. They are different from other coordinate systems, such as Cartesian or cylindrical coordinates, because they are based on the distance from a fixed point (the origin) and two angles instead of just x, y, and z coordinates.

How do you convert between spherical and Cartesian coordinates?

To convert from spherical to Cartesian coordinates, you can use the following formulas:
x = r * sin(θ) * cos(φ)
y = r * sin(θ) * sin(φ)
z = r * cos(θ)
Where r is the radius, θ is the inclination angle, and φ is the azimuth angle. To convert from Cartesian to spherical coordinates, you can use the following formulas:
r = √(x^2 + y^2 + z^2)
θ = arccos(z / r)
φ = arctan(y / x)
Note that the arccos and arctan functions may need to be adjusted depending on the quadrant of the point.

How do you represent vectors using spherical coordinates?

Vectors in spherical coordinates are represented by three components: a magnitude, an inclination angle, and an azimuth angle. The magnitude is the length of the vector, and the inclination and azimuth angles describe the direction in which the vector is pointing. To convert from Cartesian to spherical coordinates, you can use the following formulas:
r = √(x^2 + y^2 + z^2)
θ = arccos(z / r)
φ = arctan(y / x)
Note that the arccos and arctan functions may need to be adjusted depending on the quadrant of the vector.

How are spherical coordinates used in real-world applications?

Spherical coordinates are commonly used in physics, astronomy, and engineering to describe the position and orientation of objects in three-dimensional space. They are also used in navigation and mapping systems, such as GPS, to determine the location of objects on the Earth's surface. Additionally, spherical coordinates are useful for solving certain types of differential equations and for representing wave functions in quantum mechanics.

What are the limitations of using spherical coordinates?

One limitation of spherical coordinates is that they can only be used to represent points in three-dimensional space. They are not suitable for describing higher-dimensional spaces. Additionally, the inclination angle in spherical coordinates is limited to a range of 0 to 180 degrees, which can make it difficult to represent certain types of rotations or orientations. Finally, converting between spherical and Cartesian coordinates can be complex and may require special tools or software.

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