- #1

PhizKid

- 477

- 1

## Homework Statement

## Homework Equations

Energy

## The Attempt at a Solution

So I'm not exactly sure what this question even wants considering the statement is true by the work - energy theorem. Does it want me to show it explicitly for this case? If so, then, since there are no forces present along [itex]\hat{\theta }[/itex], [itex]W = m\int_{a}^{b}<\ddot{r} - r\dot{\theta }^{2}, 0>\cdot <dr,d\theta > = m\int_{l_{1}}^{l_{2}}\ddot{r} - r\dot{\theta }^{2}dr[/itex] so [itex]W = \frac{1}{2}m\dot{r}^{2}|_{l_1}^{l_2} - \int_{l_{1}}^{l_{2}}mr\dot{\theta }^{2}dr[/itex]. The change in kinetic energy is given by [itex]\Delta T = \frac{1}{2}m\dot{r}^{2}|_{l_1}^{l_2} + \frac{1}{2}mr^{2}\dot{\theta }^{2}|_{l_1}^{l_2}[/itex]. If we choose a coordinate system centered around the hole, then the position vector to the mass is parallel to the net force so the torque is identically zero thus angular momentum is conserved so [itex]r^{2}\dot{\theta }[/itex] is a constant of motion which implies [itex]r^{4}\dot{\theta }^{2}[/itex] is also a constant so [itex]-\int_{l_1}^{l_2}mr\dot{\theta }^{2}dr = -\int_{l_1}^{l_2}m\frac{r^{4}\dot{\theta }^{2}}{r^{3}}dr =-mr^{4}\dot{\theta }^{2}\int_{l_1}^{l_2}\frac{dr}{r^{3}} = \frac{1}{2}mr^{2}\dot{\theta }^{2}|_{l_1}^{l_2}[/itex] so [itex]W = \Delta T[/itex] but I'm not sure if my solution is valid and I'm also not satisfied with this long winded solution, if it is correct, because the chapter this problem is from was before angular momentum was introduced so there must be a solution that doesn't use the concept but I can't think of it so if anyone could guide me towards that it would be great. Thanks!