# Energy of a particle

1. Feb 23, 2006

### Reshma

A particle of mass 'm' is moving in a circular orbit under the influence of the potential $V(x) = \frac{ar^4}{4}$ where 'a' is a constant. Given that the allowed orbits are those whose circumference is $n\lambda$, where 'n' is an integer and $\lambda$ is the de-Broglie wavelength of the particle. Obtain the energy of the particle as a function of 'n' and $\lambda$.

So,
$$2\pi r_n = n\lambda$$
I don't understand how the potential of the orbit comes into the picture here. Isn't PE = 2KE for Bohr orbits? How is the energy calculated?

2. Feb 23, 2006

### topsquark

Total energy is always potential plus kinetic. From general central motion techniques, how do you find the total energy of an orbit when the central potential goes as r^4?

-Dan

3. Feb 24, 2006

### Reshma

I don't have a clue. The force can be shown as $\vec F = -\vec \nabla V$. The work done will be: $W = \int \vec F \cdot dr$. Will this equal the energy?

Last edited: Feb 24, 2006
4. Feb 24, 2006

### topsquark

$$T=(1/2)mv^2$$
Now, what is this is polar coordinates? Well, $$v^2=\dot{r^2}+r^2 \dot{\theta^2}$$

And, of course, E = T + V(r)...

-Dan

BTW, you can get rid of the $$\dot{\theta}$$ by using angular momentum conservation.