- #1
Order
- 97
- 3
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
[tex]\int \boldsymbol{F}\cdot d\boldsymbol{r}[/tex]
[tex]a_{r}=r\omega^{2}+\ddot{r}[/tex]
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
[tex]\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[/tex]
The second term is easily evaluated using the substition dr=dr/dt*dt this is because you can derivate
[tex]\frac{d}{dt}(m\frac{v^2}{2})=mv\frac{dv}{dt}[/tex]
Now you know that the result of the first term should be
[tex]m\frac{r^2\omega^{2}}{2}[/tex]
I try to make a similar approach as with the second term and therefore evaluate
[tex]\frac{d}{dt}(m\frac{r^2\omega^{2}}{2})=mr\omega( \frac{dr}{dt} \omega+\frac{d\omega}{dt}r)[/tex]
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
[tex]\omega=\omega_0 \sqrt{\frac{r_0}{r}}[/tex]
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.