Deriving expressions for angular velocity and acceleration

[drofenaz]

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.

 

[grzz]

... $\omega=\frac{dr}{dt}$

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

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

$\alpha$ $\neq$ $\frac{d^{2}r}{dt^{2}}$ but $\alpha$ = $\frac{d\omega}{dt}$ = $\frac{d^{2}\theta}{dt^{2}}$

 

[drofenaz]

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?

 

[grzz]

oops!
let me try again.

 

[grzz]

... 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

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

 

[drofenaz]

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:

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

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