Deriving expressions for angular velocity and acceleration

AI Thread Summary
The discussion focuses on deriving the expressions for the radial (i_r) and angular (i_θ) components of velocity and acceleration in polar coordinates. The key equations include the relationships for angular velocity (ω) and angular acceleration (α), which are expressed in terms of the derivatives of the position vector. The participant realizes that their initial approach was incorrect, particularly in confusing ω with dr/dt and α with d²r/dt². After revising their work, they successfully derive the correct expressions for both velocity and acceleration components, highlighting the importance of substituting the correct terms. The final solutions for velocity and acceleration are presented clearly, demonstrating a thorough understanding of the topic.
drofenaz
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Homework Statement


Derive the expressions for the i_r and i_θ components of velocity and acceleration.

Homework Equations


r=|r|i_r

\omega=\frac{dr}{dt}

\alpha=\frac{dω}{dt}=\frac{d^2r}{dt}

The Attempt at a Solution


r=|r|i_r

\omega=\frac{dr}{dt}i_r+r\frac{di_r}{dt}

\alpha=\frac{d^2r}{dt^2}i_r+\frac{dr}{dt}\frac{di_r}{dt}+\frac{d^2i_r}{dt^2}r+\frac{di_r}{dt}\frac{dr}{dt}

4. The solutions I need to get to
\omega=\frac{dr}{dt}i_r+r\frac{d_\theta}{dt}i_θ

\alpha=\frac{d^2r}{dt^2}i_r+\frac{dr}{dt}\frac{di_\theta}{dt}i_\theta+\frac{dr}{dt} \frac{d\theta}{dt}i_\theta+r\frac{d^2\theta}{dt^2}i_\theta-r\left(\frac{d\theta}{dt}\right)^2i_r

5. Thoughts
I think that they're substituting in something for parts of my answer. I just don't know what they're substituting in, or how to find it.
 
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drofenaz said:
... \omega=\frac{dr}{dt}

\alpha=\frac{dω}{dt}=\frac{d^2r}{dt}...

\omega \neq \frac{dr}{dt} but \omega = \frac{d\theta}{dt}.

\alpha \neq \frac{d^{2}r}{dt^{2}} but \alpha = \frac{d\omega}{dt} = \frac{d^{2}\theta}{dt^{2}}
 
So basically I'm doing it completely wrong. Would my work be correct for v and a instead of ω and \alpha?

I'm not quite sure how to solve the problem. Could you point me in the right direction?
 
oops!
let me try again.
 
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drofenaz said:
... The solutions I need to get ...
\omega=\frac{dr}{dt}i_r+r\frac{d_\theta}{dt}i_θ

\alpha=\frac{d^2r}{dt^2}i_r+\frac{dr}{dt}\frac{di_\theta}{dt}i_\theta+\frac{dr}{dt} \frac{d\theta}{dt}i_\theta+r\frac{d^2\theta}{dt^2}i_\theta-r\left(\frac{d\theta}{dt}\right)^2i_r...

I think that what you were asked to derive are:

v=\frac{dr}{dt}i_r+r\frac{d_\theta}{dt}i_θ

and

a=\frac{d^2r}{dt^2}i_r+\frac{dr}{dt}\frac{d\theta}{dt}i_\theta+\frac{dr}{dt} \frac{d\theta}{dt}i_\theta+r\frac{d^2\theta}{dt^2}i_\theta-r\left(\frac{d\theta}{dt}\right)^2i_r
 
I have solved the problem. Hopefully typing up my work will help someone someday.SOLVING FOR KNOWNS

i_r=cos\theta+sin\theta

\frac{di_r}{dt}=-sin\theta+cos\theta

\frac{di_r}{dt}=\frac{d\theta}{dt}i_\theta
i_\theta=-sin\theta+cos\theta

\frac{di_\theta}{dt}=-cos\theta-sin\theta

\frac{di_\theta}{dt}=-\frac{d\theta}{dt}i_r

POSITION VECTOR

r=ri_r

SOLVING FOR VELOCITY

v=\frac{dr}{dt}

v=\frac{dr}{dt}i_r+\frac{di_r}{dt}r

After substituting what we know, we get:

v=\frac{dr}{dt}i_r+r\frac{d\theta}{dt}i_\theta

So the solutions for velocity are:

v_r=\frac{dr}{dt}

v_\theta=r\frac{d\theta}{dt}

SOLVING FOR ACCELERATION

a=\frac{dv}{dt}

a=\frac{d^2r}{dt^2}i_r+\frac{di_r}{dt}\frac{dr}{dt}+\frac{dr}{dt}\frac{d\theta}{dt}i_\theta+r \frac{d^2\theta}{dt^2}i_\theta+r\frac{d\theta}{dt}\frac{di_\theta}{dt}

After substituting what we know, we get:

a=\frac{d^2r}{dt^2}i_r+\frac{d\theta}{dt}\frac{dr}{dt}i_\theta+\frac{dr}{dt}\frac{d\theta}{dt}i_θ+r \frac{d^2\theta}{dt^2}i_\theta-r\frac{d\theta}{dt}\frac{d\theta}{dt}i_r

And then we simplify to get:

a=\frac{d^2r}{dt^2}i_r+2\frac{d\theta}{dt}\frac{dr}{dt}i_\theta+r\frac{d^2\theta}{dt^2}i_\theta-r\left(\frac{d\theta}{dt}\right)^2i_r

So the solutions for acceleration are:

a_r=\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2

a_\theta=2\frac{d\theta}{dt}\frac{dr}{dt}+r\frac{d^2\theta}{dt^2}
 
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