How can velocity be expressed as a function of time in polar coordinates?

[Clever_name]

Homework Statement

Here is a picture of the situation http://i48.tinypic.com/vnmi5t.jpg

Homework Equations

polar coordinate system

The Attempt at a Solution

ok so first I'm attempting to find velocity as a function of time,
first I know V=(dR/dt)er +(R)(d∅/dt)e∅ - this is a vector

so, R = 20+15cos(∅) and d∅/dt = ∏ and dR/dt = -15sin(∅)

now this is wher ei get stuck.

so I'm trying to get velocity as a function of time,
the only thing i can think of to get there is my finding the magnitude of the vector, so
I get, ((-15sin(∅))^(2)+(20∏ +125cos(∅))^(2))^(1/2)

simplifying i get 1 + (8/3)∏cos(∅) + (16/9)∏^(2) = V(∅)
but i want v as a function of time not theta, and i have no idea to go about getting there. Any help would be greatly appreciated thanks!

 

[voko]

If $R = 20 + 15 \cos \theta$ and $\theta = \pi t$, then what is $\frac {dR} {dt}$?

 

[Clever_name]

-15*∏*sin(∏*t) = dR/dt

 

[Clever_name]

oh i see, so 1+(8/3)*pi*cos(pi*t)+(16/9)*pi^(2) = V(t) ?

 

[Clever_name]

how did you know theta was pi*t though? My intuition with regard to that statement is lacking.

 

[voko]

So what is the entire velocity vector as a function of time?

 

[Clever_name]

did you see post 4?

 

[voko]

I do not see any vectors in post #4. And I don't really understand what that thing in it really is.

 

[Clever_name]

ok so the velocity vector = -15*pi*sin(pi*t)Er + ((20+15cos(pi*t))*pi)Eo

 

[voko]

That's correct. Can you continue from this on?

 

[Clever_name]

I'll give it a go, thank you for the push in the right direction.

 

[voko]

Do you understand why $\theta = \pi t$? Note the problem specified a constant angular velocity.

 

[Clever_name]

Ah i see! the the integral of d(theta) with respect to time is pi*t

 

[Clever_name]

when calculating the magnitude of the velocity and acceleration vectors at t=0.7s do i need to change the equation pi*t into degrees or does it work just as it is?

 

[voko]

Why would you?

 

[Clever_name]

cos(a) equals a different value depending on a being measured in rads or degrees?

 

[voko]

This is not a correct statement. When you compute a trigonometric function, you - or, rather, the calculator - must know what units are used for the angle measure. Then the result is independent of that.

 

[Clever_name]

so just to makesure i understand you the velcoity at time t=0.7s then would be,

(((-15*pi*sin(pi*0.7))^(2)+(20+15cos(pi*0.7)*pi)^(2))^(1/2) which equals 109.92 m/s?

 

[voko]

I get a different result.
$$
\sqrt {(15 \pi \sin \pi t)^2 + (20 + 15 \cos \pi t)^2 \pi^2}

= 5\pi \sqrt { (3 \sin \pi t)^2 + 4^2 + 24 \cos \pi t + (3 \cos \pi t)^2}

\\

= 5\pi \sqrt { 3^2 + 4^2 + 24 \cos \pi t } = 5 \pi \sqrt { 25 + 24 \cos \pi t }
$$

 

[Clever_name]

when putting in all the values in you're equation i get the same result as when using my equation which is 109.93 m/s :S

 

[voko]

Do you set the calculator to use radians?

 

[Clever_name]

no it's set to degrees, should it be set to radians?

 

[voko]

What did you make out of #17?

 

[Clever_name]

that it should be set to radians lol, my bad its late here.