Saturday
May 25, 2013

# Homework Help: physics

Posted by charlie on Saturday, April 11, 2009 at 9:56am.

I've just read that the centripetal force on an electron is mv^2/r; and that this is also equal to the electrostatic force pulling the electron toward the nucleus.

Does mv^2/r hold good in all Newtonian siuations as a formula for centripetal force acting on a body?

Thanks

• physics - Count Iblis, Saturday, April 11, 2009 at 10:19am

No. Suppose the electron is moving in some arbitrary way. Then the compinent of the acceleration in the radial direction (i.e. the direction away from the nucleus) is given by:

a = d^2r/dt^2 - (V_t)^2/r

Here V_t is the velocity of theelectron in the "tangential direction", this is the component of the velocity orthogonal to the distance vector of the electron to the nucleus.

This means that the second derivative of the radius has to be a constant for the formula for the acceleration in the radial direction to reduce to:

V_t^2/r

For V_t to be equal to V ,m requires that there is no component of the velocity in the radial direction, so r must be constant.

So, you obtain the result that if the radius is constant, then the acceleration in the radial direction is given by the V^2/r. Since F = m a, it then follows that the force in the radial dorection equals m V^2/r.

• physics - charlie, Saturday, April 11, 2009 at 12:29pm

If I had made it conditional that the electron was orbiting at a constant radius, would it then be correct; and would it also be accurate in 'ordinary' situations, eg spinning a stone around on a string?

• physics - Count Iblis, Saturday, April 11, 2009 at 1:52pm

Yes, if the radius is constant, then it is correct.

• physics - charlie, Saturday, April 11, 2009 at 3:52pm

Thanks, CI

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