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 Reshma Feb23-06 10:46 AM

Energy of a particle

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?

 topsquark Feb23-06 03:18 PM

Quote:
 Quote by 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?
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

 Reshma Feb24-06 12:29 AM

Quote:
 Quote by 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
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?

 topsquark Feb24-06 08:06 AM

Quote:
 Quote by 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?
$$T=(1/2)mv^2$$
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.

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