Work from the bottom of Unit circle to its top in Polar Coordinates

[soopo]

Homework Statement

Calculate the work $$W_{A B}$$ done by the force $$F$$ using Newton's laws ($$F=ma$$, etc), when a particle moves from the point $$A$$ to the point $$B$$ along the unit circle. The angle is $$\theta$$. No friction. How do you define kinetic energy in polar coordinates?

Homework Equations

Acceleration in polar coordinates is:

$$\bar{a} = ( \ddot{r} - r ( \dot{ \theta } )^2 ) \hat{r} + ( r \ddot{ \theta } + 2 \dot{ r } \dot{ \theta } ) \hat{ e_{\theta} }$$

The Attempt at a Solution

I know from cartesian coordinates that $$PE=KE <=> 1/2 * mv^2 = mg*h$$. I should verify it in polar coordinates. So integrating $$\bar{a}*m$$, with respect to the radius $$r$$ and the angle $$\theta$$, probably give me the energy, like $$W=F*distance$$ in carteesian coordinates.

 

[jdwood983]

The kinetic energy is given by $$T=\frac{1}{2}mv^2$$ and in polar coordinates, the velocity is given by

$$v(r,\theta)=\frac{dr}{dt}\hat{\mathbf{r}}+r\frac{d\theta}{dt}\hat{\boldsymbol{\theta}}$$

(you can check that taking the time-derivative of this does give the acceleration you have) so that takes care of the first part.

For the second part, you will need to integrate $$m\mathbf{a}$$ over $$r,\,\theta$$ to get the work:

$$\begin{array}{ll}W&=\int \mathbf{F}\cdot d\mathbf{r} \\ &=\int F_r\,dr+\int F_\theta\,rd\theta \\ &=\int ma_r\,dr+\int ma_\theta\,rd\theta \end{array}$$

 

[soopo]

$$\begin{array}{ll}W&=\int \mathbf{F}\cdot d\mathbf{r} \\ &=\int F_r\,dr+\int F_\theta\,rd\theta$$

Why do you have $$r$$ in the last integral?

It seems that you have used the relation
$$dr = r d\theta$$

 

[jdwood983]

Why do you have $$r$$ in the last integral?

It seems that you have used the relation
$$dr = r d\theta$$

It comes from the change of coordinate systems. If anything, the relation should be

$$d\mathbf{r}=dr\hat{\mathbf{r}}+rd\theta\hat{\mathbf{\theta}}$$

And when taking the dot-product with $$\mathbf{F}(r,\theta)$$, you get the integral relation I gave.

 

[soopo]

The kinetic energy is given by $$T=\frac{1}{2}mv^2$$ and in polar coordinates, the velocity is given by
$$\begin{array}{ll}W&=\int \mathbf{F}\cdot d\mathbf{r} \\ &=\int F_r\,dr+\int F_\theta\,rd\theta \\ &=\int ma_r\,dr+\int ma_\theta\,rd\theta \end{array}$$

How do you define $$KE$$ and $$PE$$ with respect to a conservative force $$\bar{D} = - m g \hat{k}$$ ?

Is it just a dot product? So I get the work, and then I can calculate the $$PE$$ with the new work:

$$\begin{array}{ll}W_{ n e w }&=\int \mathbf{D}\cdot d\mathbf{r} \\ &=0 \end{array}$$

Zero? I assumed $$\hat{k}$$ is perdendicular to the plane by $$\theta$$ and $$r$$. So $$PE$$ is zero with respect to the force $$\bar{D}$$.

 

[jdwood983]

How do you define $$KE$$ and $$PE$$ with respect to a conservative force $$\bar{D} = - m g \hat{k}$$ ?

Is it just a dot product? So I get the work, and then I can calculate the $$PE$$ with the new work:

$$\begin{array}{ll}W_{ n e w }&=\int \mathbf{D}\cdot d\mathbf{r} \\ &=0 \end{array}$$

Zero? I assumed $$\hat{k}$$ is perdendicular to the plane by $$\theta$$ and $$r$$. So $$PE$$ is zero with respect to the force $$\bar{D}$$.

I think you are combining two separate problems into one.

Work is defined as the force over a distance, and neither kinetic energy nor potential energy is a force, so you can't say that W=0 as you did above.

Kinetic energy is defined as the mass times the square of the velocity (really the square of the time derivative of the position). And potential energy is defined as the mass times the relative height times the acceleration due to gravity. Neither of these is work.

 

[jdwood983]

Your problem asks two questions:
(1) What is the work done by a force moving from point $$A$$ to $$B$$ in polar coordinates?
(2) What is the kinetic energy in polar coordinates?

For the first problem,

$$W=\int\mathbf{F}\cdot d\mathbf{r}=\int_A^BF_r\,dr+\int_0^{\phi}F_\theta\,rdr$$

where $$\phi$$ is the final angle, sweeping from $$A$$ to $$B$$. For the second problem,

$$KE=\frac{1}{2}mv^2=\frac{1}{2}m\left(\frac{dr}{dt}\hat{e}_r+r\frac{d\theta}{dt}\hat{e}_\theta}\right)$$