find the cylindrical , rectangular, spherical coordinates

• Jan 9th 2007, 10:53 AM
Jenny20
find the cylindrical , rectangular, spherical coordinates
A right circular cylinder of radius 10cm spinning at 3 revolutions per minute about the z-axis. At time t=0s, a bug at the point (0,10,0) begins walking straight up the face of the cylinder at the rate of 0.5cm/min.

a) find the cylindrical coordinates of the bug after 2 min.

b) Find the rectangular coordinates of the bug after 2 min.

c) find the spherical coordinates of the bug after 2 min.
• Jan 9th 2007, 11:16 AM
topsquark
Quote:

Originally Posted by Jenny20
A right circular cylinder of radius 10cm spinning at 3 revolutions per minute about the z-axis. At time t=0s, a bug at the point (0,10,0) begins walking straight up the face of the cylinder at the rate of 0.5cm/min.

a) find the cylindrical coordinates of the bug after 2 min.

b) Find the rectangular coordinates of the bug after 2 min.

c) find the spherical coordinates of the bug after 2 min.

The linear and rotational motions are independent, so after 2 min the bug has a z coordinate of $z = (0.5 \, cm/min)(2 \, min) = 1 cm$.

Rotationally speaking, the bug starts on the y-axis at 10 cm ( $\theta_0 = \pi/2 \, rad$) and spins for 2 min. The angular speed is constant, so
$\theta = \theta _0 + \omega t$

$\theta = \pi/2 + (3 \cdot 2\pi \, rad/min)(2 \, min) = 25 \pi /2 \, rad$ ==> $\theta = \pi /2 \, rad$

So in cylindrical coordinates the bug is at
$(r, \theta, z) = (10, \pi /2, 1)$
after 2 minutes.

In rectangular coordinates we have
$x = r cos( \theta )$
and
$y = r sin( \theta )$

So
$x = 10 cos( \pi /2 ) = 0 \, cm$
$y = 10 sin( \pi / 2 ) = 10 \, cm$

So
$(x, y, z) = (0, 10, 1)$

Using rectangular coordinates as the starting point, we have for spherical coordinates:
$x = \rho cos( \phi ) sin( \theta )$ <-- Not the same $\theta$ as before!
$y = \rho sin( \phi ) sin( \theta )$
$z = \rho cos( \theta )$

So
$\rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{101}$

Now, $\phi$ would normally be $tan^{-1} \left ( \frac{y}{x} \right )$ but since x is 0 this cannot be calculated directly. Suffice it to say that $\phi$ needs to be such that $tan( \phi )$ is positive and infinite. Thus $\phi = \pi /2 \, rad$. (The other way to see this is that $\phi$ in spherical coordinates is the same as $\theta$ in cylindrical coordinates.)

$\theta = cos^{-1} \left ( \frac{z}{\rho} \right ) = cos^{-1} \left ( \frac{1}{\sqrt{101}} \right ) \approx 1.47113 \, rad$ (As far as I know there is no exact expression of this.)

So in spherical coordinates we have:
$(\rho, \phi, \theta ) = (\sqrt{101}, \pi /2, 1.47113 )$

-Dan
• Jan 9th 2007, 11:32 AM
Jenny20
Hi topsquark,

for part a) why r = 10?
• Jan 9th 2007, 11:41 AM
topsquark
Quote:

Originally Posted by Jenny20
Hi topsquark,

for part a) why r = 10?

Because the bug is on a spinning cylinder of radius 10 cm. I'm assuming it stays on the cylinder.

-Dan
• Jan 9th 2007, 11:53 AM
Jenny20
Hi topsquark,

i see. Also , in part a) why thetasub0 = pi/2?
• Jan 9th 2007, 12:16 PM
topsquark
Quote:

Originally Posted by Jenny20
Hi topsquark,

i see. Also , in part a) why thetasub0 = pi/2?

The bug starts at (0, 10, 0) in Cartesian coordinates. That means the bug is on the y-axis, which corresponds to $\theta_0 = \pi /2 \, rad$.

-Dan
• Jan 9th 2007, 12:42 PM
Jenny20
Hi topsquark,

Thank you very much. I got the question now.

You have a typo in part c):

cos phi = z/rho = 1/sqrt(101)
so phi approximately equal to 1.47 radians.
• Jan 9th 2007, 02:15 PM
topsquark
Quote:

Originally Posted by Jenny20
Hi topsquark,

Thank you very much. I got the question now.

You have a typo in part c):

cos phi = z/rho = 1/sqrt(101)
so phi approximately equal to 1.47 radians.

Heh. :o Sorry about that. I've fixed it in my original post.

-Dan