The work energy theorem in polar coordinates

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Homework Statement

Mass m whirls on a frictionless table, held to circular motion by a string which passes through a hole in the table. The string is slowly pulled through the hole so that the radius of the circle changes from l₁ to l₂. Show that the work done in pulling the string equals the increase in kinetic energy of the mass.

Homework Equations

\int \boldsymbol{F}\cdot d\boldsymbol{r}
a_{r}=r\omega^{2}+\ddot{r}

The Attempt at a Solution

I'm sorry that I don't have much to say about the solution. I want a hint because I have tried so hard. The work can anyway be evaluated to
\int \boldsymbol{F}\cdot d\boldsymbol{r}=\int_{l_1}^{l_2}m(r\omega^{2}+ \ddot{r} )dr=m\int_{l_1}^{l_2}r\omega^{2}+m\int_{l_1}^{l_2}\ddot{r}dr

The second term is easily evaluated using the substition dr=dr/dt*dt this is because you can derivate
\frac{d}{dt}(m\frac{v^2}{2})=mv\frac{dv}{dt}
Now you know that the result of the first term should be
m\frac{r^2\omega^{2}}{2}
I try to make a similar approach as with the second term and therefore evaluate
\frac{d}{dt}(m\frac{r^2\omega^{2}}{2})=mr\omega( \frac{dr}{dt} \omega+\frac{d\omega}{dt}r)
The result is two terms, so how can I find a variable substitution? Now it is easy to find a solution once you know that
\omega=\omega_0 \sqrt{\frac{r_0}{r}}
This might be found by setting up a force diagram and finding a differential equation. But there are two problems here. First we don't know anything about the force (except that it is weak) and secondly the result will be in time t not in position r. So therefore my pessimism is still.

 

[ideasrule]

I'm sorry that I don't have much to say about the solution. I want a hint because I have tried so hard. The work can anyway be evaluated to
\int \boldsymbol{F}\cdot d\boldsymbol{r}=\int_{l_1}^{l_2}m(r\omega^{2}+ \ddot{r} )dr=m\int_{l_1}^{l_2}r\omega^{2}+m\int_{l_1}^{l_2}\ddot{r}dr
The second term is easily evaluated using the substition dr=dr/dt*dt this is because you can derivate
\frac{d}{dt}(m\frac{v^2}{2})=mv\frac{dv}{dt}

Actually, the second term can be taken as 0. The string is slowly being pulled in, so acceleration in r is negligible.

The result is two terms, so how can I find a variable substitution? Now it is easy to find a solution once you know that
\omega=\omega_0 \sqrt{\frac{r_0}{r}}

Try the conservation of angular momentum.

 

[Order]

Try the conservation of angular momentum.

Ok, I use the differential m(r + dr)(\omega + d \omega )^{2} - mr\omega^{2}=0 to arrive at the differential equation 2 \frac{d \omega}{\omega}=-\frac{dr}{r} which gives the desired result.

Now I can integrate m \int_{l_{1}}^{l_{2}} r \omega^{2}_{0}\frac{r_{0}}{r}dr=m \omega^{2}_{0}r_{0}(l_{2}-l_{1})=m(l_{2}^2\omega^{2}_{2}-l_{1}^2\omega^{2}_{1})=m(v_{\theta}^2(l_{2})-v_{\theta}^2(l_{1}))
One problem here is that a factor 1/2 is missing.

But another problem is that the problem is in the chapter of Newtons laws, before angular momentum is discussed (that is why I don't know about it). Is there no way to solve this problem using Newtons laws?!