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Polar coordinates theta

  • Thread starter ma18
  • Start date
93
1
1. Homework Statement

A particle moves with const speed v along the curve r(θ) = a(1+cos θ).

Starting with the general expression for the velocity vector v in polar coordinates solve for θ_dot in terms of v, k, and θ. What does the sign of θ_dot signify?


2. Homework Equations

v = r_dot*r_hat + r*θ_dot*θ_hat

v = v * v_hat

r(θ) = a(1+cos θ)

r_hat = x_hat cos Θ + y_hat sin θ

v

3. The Attempt at a Solution

r_dot = -a sin θ

solving for θ_dot:

θ_dot = (v - r_dot * r_hat)/(r*Θ_hat)

I am lost here, I think I am missing some important relation and something that comes out of the fact that the speed in constant but I don't know what.

Any help would be greatly appreciated thanks.


Edit:

I think I've got it, since the magnitude of unit vectors are one and θ_dot is a scalar I can just stop at

θ_dot = (v - r_dot * r_hat)/(r*Θ_hat)
= (v*v_hat - r_dot * r_hat)/(r*Θ_hat)
= (v-r_dot)/r
= (v+a sin θ)/(a + a cos θ)

Is this right? What does the sign mean then, just the value?
 
Last edited:

verty

Homework Helper
2,157
198
1. Homework Statement

A particle moves with const speed v along the curve r(θ) = a(1+cos θ).

Starting with the general expression for the velocity vector v in polar coordinates solve for θ_dot in terms of v, k, and θ. What does the sign of θ_dot signify?


2. Homework Equations

v = r_dot*r_hat + r*θ_dot*θ_hat

v = v * v_hat

r(θ) = a(1+cos θ)

r_hat = x_hat cos Θ + y_hat sin θ

v

3. The Attempt at a Solution

r_dot = -a sin θ

solving for θ_dot:

θ_dot = (v - r_dot * r_hat)/(r*Θ_hat)

I am lost here, I think I am missing some important relation and something that comes out of the fact that the speed in constant but I don't know what.

Any help would be greatly appreciated thanks.


Edit:

I think I've got it, since the magnitude of unit vectors are one and θ_dot is a scalar I can just stop at

θ_dot = (v - r_dot * r_hat)/(r*Θ_hat)
= (v*v_hat - r_dot * r_hat)/(r*Θ_hat)
= (v-r_dot)/r
= (v+a sin θ)/(a + a cos θ)

Is this right? What does the sign mean then, just the value?
Remember that you can't divide by a vector, that operation is not defined. So dividing by r*θ_hat is not allowed. What you want to do is start by getting the squared magnitude of ##\vec{v}##.

Then also, this is wrong: ##\dot{r} = -a \; sinθ##. Remember that θ is a function of time, you are doing implicit differentiation to get ##dr \over dt## which is ##\dot{r}##.
 
93
1
Okay I've got the answer to this as

θ_dot2 = v2/(2ar)

but I'm still not sure if I've got the hang of it, what would θ_dotdot and r_dotdot be?

Would Θ_dotdot just be the

v/(k√2) * 0.5 sin θ / (1+cos Θ)^1.5

and then r_dotdot be

r_dotdot = -k θ_dot ^2 cos θ

? I think this is right
 

rude man

Homework Helper
Insights Author
Gold Member
7,512
667
You started with v = dr/dt r + r dθ/dt θ. Good.
(Note: I use bold for vectors. θ and r are the unit vectors in polar coordinates.)

So, what is v2 ? Remember the bit about differentiating r that you were alerted to in post 2. Get v2 as a function of a and θ only.

BTW I have no idea what k is supposed to be. Ignore it. Your only variables are v, r, a and θ, obviously. a is assumed a positive constant.

Then, solving for dθ/dt is easy.

Finally, you can ponder the idea that dθ/dt can be negative as well as positive.
 

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