Determine the angular speed theta_dot

[CGI]

Homework Statement

Homework Equations

Vθ = r x θdot
Vr = rdot

Accelθ = r ⋅ θdot^2 + 2(r dot)(theta dot)
Accelr = r double dot - (r ⋅(θdot)^2)

The Attempt at a Solution

I know that the answer is supposed to be

But I'm not quite sure on how I'm supposed to get to an answer like that.
I'm still relatively new to polar coordinates, but I was hoping someone
could get me to start thinking in the right direction?

I thought that my position would be R = bθ and if I took the derivative with
respect to θ, I would get R_dot = b, which I think would be equal to π/2.

I'm not quite sure that's right though. I'd really appreciate it if someone could
help me out!

 

[Simon Bridge]

I thought that my position would be R = bθ and if I took the derivative with
respect to θ, I would get R_dot = b, which I think would be equal to π/2.

The position is not ##R=b\theta## ... ##\dot R \neq dR/d\theta## ... and there is no reason b should be any particular value.
You need to start again - what is the definition of "position"? What is the definition of "velocity" and how does that relate to "speed"? What does the dot notation tell you?

 

[CGI]

Okay, I'm sorry it took so long to respond. I actually waited to talk to my professor to understand polar coordinates a little better.
I can't come up with a good definition of position except that they are coordinates to a particular point on a graph. When I think of velocity I often think
of something with a magnitude and direction. The magnitude of the velocity would also give you the speed. The dot notation was tripping me up for
a while, but from what I understand it signals the 1st or 2nd derivatives of something, depending on how many dots there are.

 

[haruspex]

The position is not R=bθ

Just to be clear, what are you objecting to there? I agree that it would be more accurate to say "the position in polar coordinates is (r,θ) where r=bθ."

Vθ = r x θdot
Vr = rdot

Writing it that way is open to misinterpretation. Better:
##V_{\theta}=r\dot \theta##
##V_r=\dot r##
I.e., these are the tangential and radial components of the velocity. Since you are given the speed, what equation does that allow you to write?
Yes, the dots indicate differentiation with respect to time. If you differentiate both side of r=bθ with respect to time, what do you get?

 

[CGI]

I'm sorry, I'd like to write the proper terms, but I'm just not sure how. I'll try to be as clear as possible. If I differentiate r=bθ, would I just get (r dot) = (b ⋅ θ dot)?

 

[Simon Bridge]

I can't come up with a good definition of position except that they are coordinates to a particular point on a graph. When I think of velocity I often think
of something with a magnitude and direction. The magnitude of the velocity would also give you the speed. The dot notation was tripping me up for
a while, but from what I understand it signals the 1st or 2nd derivatives of something, depending on how many dots there are.

In this case, the position would be given in polar coordinates ... so position is ##\overrightarrow{OB} = \vec r = R\hat r + \theta \hat \theta## or you can write it as an ordered pair like haruspex did.
The velocity is the rate of change of position... so you differentiate the position with respect to time.
The dot represents the derivative with respect to time. The "prime" notation refers to the derivative with length.

If you differentiate r = b\theta wrt time, you get ##\dot r = b\dot\theta## ... yes... which gives you the rate of change of the radial distance from the origin.
That is: the component of the velocity that points along ##\overrightarrow{OB}##.

The velocity vector will have another component that points perpendicular to ##\overrightarrow{OB}## which harispex gives as ##V_\theta##.

 

[CGI]

Okay great, so wouldn't theta_dot just be r_dot/b where r_dot equals Initial velocity?

 

[haruspex]

Okay great, so wouldn't theta_dot just be r_dot/b where r_dot equals Initial velocity?

Yes, ##\dot \theta=\dot r/b##, but ##\dot r## is only the radial component of velocity, v_r. As I posted, and as shown in your own relevant equations, there is also the tangential component ##v_\theta=r\dot\theta##. These two components are at right angles. So what is the overall speed?

 

[CGI]

Could I say the overall velocity would be Vr^2 + V theta^2? And square root everything?

 

[haruspex]

Could I say the overall velocity would be Vr^2 + V theta^2? And square root everything?

Yes, but please try to use clearer notation, either with LaTeX or using the subscript and superscript featuresin the toolbar (X₂, X²). You mean V_r² +V_theta².

 

[CGI]

Okay I'll try the best I can with LaTeX. So could I set the whole velocity equation equal to that velocity and solve for θdot?

##sqrt{V_r^2 + V_(theta)^2}## = r_dot + r ⋅ θ_dot

And would my V_o just equal my r_dot?

 

[haruspex]

##sqrt{V_r^2 + V_(theta)^2}## = r_dot + r ⋅ θ_dot

You need to put backslashes in front of the keywords sqrt, theta, and the closing double hash should be right at the end. Use the prefix \dot to get the overhead dot. Enclose superscripts and subscripts of more than one character in curly braces {}.
So your equation above should be written:
\sqrt{V_r^2 + V_{\theta}^2} = \dot r + r \dot\theta
Which comes out as:
##\sqrt{V_r^2 + V_{\theta}^2} = \dot r + r \dot\theta##
(To see the latex someone else has used, right click on the algebraic expression in their post and select Show Math As... You can copy and paste it out of there, then edit it to your own needs.)
But that's wrong because ##V_r=\dot r##, and ##V_{\theta}=r\dot\theta##, so what have you left out on the right hand side?

And would my V_o just equal my r_dot?

No, it is the overall speed, as in your root-sum-square expression above.

 

[CGI]

I feel like it should have something to do with b, but I'm not sure quite sure how I would put it on the right hand side..

 

[haruspex]

I feel like it should have something to do with b, but I'm not sure quite sure how I would put it on the right hand side..

Let's take this in simple steps. Try to answer each in turn.
1. The velocity has two components, the radial velocity and the tangential velocity. Call these ##V_r## and ##V_{\theta}##.
These are at right angles.
In terms of those two variables, what is the overall speed? (You already got this in post #11.)
2. What are you told the overall speed is?
3. What equation can you write down from the answers to 1 and 2?
4. What expression do you have for ##V_r## in terms (some of) of r, b and theta?
5. What expression do you have for ##V_{\theta}##in terms (some of) of r, b and theta?
6. Using 4. and 5., substitute for ##V_r## and ##V_{\theta}## in your equation from 3.
That's enough for now.

 

[CGI]

1. The overall speed would equal
##\sqrt{V_r^2 + V_{\theta}^2}##
2. V_o is the constant speed of the collar.
3. I could say that ##\sqrt{V_r^2 + V_{\theta}^2}## = V_o
4. V_r is the same as ##\dot r## which is the same as b##\dot θ##
5. V_θ would be r##\dot θ##.. I think
6. ##\sqrt{b\dot θ + r\dot θ}## Would be the final equation after everything is plugged in.

Is this correct?

 

[haruspex]

1. The overall speed would equal
##\sqrt{V_r^2 + V_{\theta}^2}##
2. V_o is the constant speed of the collar.
3. I could say that ##\sqrt{V_r^2 + V_{\theta}^2}## = V_o
4. V_r is the same as ##\dot r## which is the same as b##\dot θ##
5. V_θ would be r##\dot θ##.. I think
6. ##\sqrt{b\dot θ + r\dot θ}## Would be the final equation after everything is plugged in.

Is this correct?

Close, but you forgot to square the terms.
Next step is to use the equation of the curve so that you can replace r.

 

[CGI]

Wow! I got it! Thank you so much for all the help. I know it was a loooong process, but I fully understand now and I have the right answer.
Appreciate all the help!

 

[Simon Bridge]

Well done.