# Polar coordinates question

1. Apr 17, 2013

### Clever_name

1. The problem statement, all variables and given/known data

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

2. Relevant equations

polar coordinate system

3. 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!

2. Apr 17, 2013

### voko

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

3. Apr 17, 2013

### Clever_name

-15*∏*sin(∏*t) = dR/dt

4. Apr 17, 2013

### Clever_name

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

5. Apr 17, 2013

### Clever_name

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

6. Apr 17, 2013

### voko

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

7. Apr 17, 2013

### Clever_name

did you see post 4?

8. Apr 17, 2013

### voko

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

9. Apr 17, 2013

### Clever_name

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

10. Apr 17, 2013

### voko

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

11. Apr 17, 2013

### Clever_name

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

12. Apr 17, 2013

### voko

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

13. Apr 17, 2013

### Clever_name

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

14. Apr 18, 2013

### 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?

15. Apr 18, 2013

### voko

Why would you?

16. Apr 18, 2013

### Clever_name

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

17. Apr 18, 2013

### 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.

18. Apr 18, 2013

### 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?

19. Apr 18, 2013

### voko

I get a different result.
$$\sqrt {(15 \pi \sin \pi t)^2 + (20 + 15 \cos \pi t)^2 \pi^2} = 5\pi \sqrt { (3 \sin \pi t)^2 + 4^2 + 24 \cos \pi t + (3 \cos \pi t)^2} \\ = 5\pi \sqrt { 3^2 + 4^2 + 24 \cos \pi t } = 5 \pi \sqrt { 25 + 24 \cos \pi t }$$

20. Apr 18, 2013

### 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