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}