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Consider a system of particles where the kinetic energy of the system is varying with time. I'd like to know the significance (or meaning) of the time derivative of the kinetic energy being zero at a point. What is the significance of time instances where the kinetic energy has maxima and minima ?

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It depends what you mean by system. A simple system of atoms composing a ball, for example, thrown in the air against gravity, has a minimum in kinetic energy at the top of the parabola and a maximum of the potential energy. The significance is that the motion is against a gravitational field and once the aggregate kinetic energy becomes minimum the ball will start falling, in other words "what goes up must come down". Note that the kinetic energy of the atoms is just a constant in this instance. –  anna v Jun 18 '11 at 18:35
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As Anna pointed out in her comment, it really depends on the system. In general, there's not much you can say except that the minimum of kinetic energy corresponds to the minimum speed and the maximum of kinetic energy corresponds to the maximum speed.

There are many systems in which the minimum speed (and thus the minimum kinetic energy) is zero, and in those systems, finding the times at which the kinetic energy is a minimum tells you when the object stops.

Mathematically, if you actually take the derivative of kinetic energy with respect to time, you get

$$\frac{\mathrm{d}K}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\biggl[\frac{1}{2}mv^2\biggr] = m\vec{v}\cdot\vec{a}$$

(using the nonrelativistic expression for $K$). There are three ways this can be equal to zero:

  • $v = 0$: the object is not moving. This corresponds to a minimum of kinetic energy.
  • $a = 0$: the object is not accelerating (and by $F = ma$ is also experiencing no net force). This corresponds to either constant velocity motion, or a maximum of kinetic energy, in a simple harmonic oscillator for example.
  • $\vec{v}\perp\vec{a}$: the object is experiencing pure centripetal acceleration, which means it's moving with a momentarily constant radius of curvature. This can be either a minimum or a maximum of kinetic energy, in an orbit for example.

Using the relativistic expression, you get

$$\frac{\mathrm{d}K}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\bigl[(\gamma - 1)mc^2\bigr] = \frac{m\vec{v}\cdot\vec{a}}{\bigl(1-\frac{v^2}{c^2}\bigr)^{\frac{3}{2}}}$$

from which the same conclusions follow.

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