Mechanical Energy of Particle in Horizontal Circle

[ritwik06]

Homework Statement

A particle of mass "m" is revolving in a horizontal circle of radius "r" under a centripetal force -k/(r*r), where k is a constant. What is the mechanical energy of the particle.

The Attempt at a Solution

There are two solutions to the problem I have got. One is mine, the other was published in my book.
My own approach:
I take the potential energy reference level at the horizontal plane at which the particle is.
Therefore its potential energy is 0
Now by equation for circular motion:
\frac{mv^{2}}{r}=\frac{-k}{r^{2}}mv^{2}=\frac{-k}{r}

Kinetic Energy=0.5*mv^{2}=0.5\frac{-k}{r}Now my books solution:
U=-\int\frac{-k * dr}{r^{2}}=-k/r
I know this step is done because magnitude of conservative force acting is negative of the rate of change of potential energy.

\frac{mv^{2}}{r}=\frac{k}{r^{2}}
K=0.5 mv^{2}=\frac{k}{2r}

Now they add these two to get the same result I did. Where is the anomaly? Was my solution incorrect?

 

[Irid]

Your choice of zero potential energy was arbitrary. Of course, one can set potential equal to zero anywhere he/she pleases, but in a case such as this one, the potential energy is usually the energy at infinity. It's because at infinity the attractive force is zero, so the particles do not interact, therefore, they share no potential energy. As they approach to a distance r, the attraction results in increase of kinetic energy, which is also, by energy conservation, the decrease of potential energy.

Your result has an obvious fault. Look here:

E = \frac{mv^2}{2} = \frac{-k}{2r}

If k is positive and r is positive, the velocity v can't be real! It's just a matter of dimensions that you have the same dependence k/r and pure luck that there's the same factor 0.5. If the force was 1/r^3 or something like that, you wouldn't get the same factor.

 

[ritwik06]

solved[/color]
Thanks a lot.