General Mechanics/Motion in Two and Three Dimensions

Motion in 2 and 3 Directions

Previously, we discussed Newtonian dynamics in one dimension. Now that we are familiar with both vectors and partial differentiation, we can extend that discussion to two or three dimensions.

Work becomes a dot product

$W=\mathbf {F} \cdot \Delta x$

and likewise power

$P=\mathbf {F} \cdot \mathbf {v}$

If the force is at right angles to the direction of motion, no work will be done.

In one dimension, we said a force was conservative if it was a function of position alone, or equivalently, the negative slope of a potential energy.

The second definition extends to

$\mathbf {F} =-\nabla V$

In two or more dimensions, these are not equivalent statements. To see this, consider

$D_{y}\mathbf {F} _{x}-D_{x}\mathbf {F} _{y}=V_{xy}-V_{yx}$

Since it doesn't matter which order derivatives are taken in, the left hand side of this equation must be zero for any force which can be written as a gradient, but for an arbitrary force, depending only on position, such as F=(y, -x, 0), the left hand side isn't zero.

Conservative forces are useful because the total work done by them depends only on the difference in potential energy at the endpoints, not on the path taken, from which the conservation of energy immediately follows.

If this is the case, the work done by an infinitesimal displacement dx must be

$W=V(\mathbf {x} )-V(\mathbf {x} +d\mathbf {x} )=-(\nabla V)\cdot d\mathbf {x}$

Comparing this with the first equation above, we see that if we have a potential energy then we must have

$\mathbf {F} =-\nabla V$

Any such F is a conservative force.

Circular Motion

An important example of motion in two dimensions is circular motion.

Consider a mass, m, moving in a circle, radius r.

The angular velocity, ω is the rate of change of angle with time. In time Δt the mass moves through an angle Δθ= ωΔt. The distance the mass moves is then r sin Δθ, but this is approximately rΔθ for small angles.

Thus, the distance moved in a small time Δt is rωΔt, and divided by Δt gives us the speed, v.

$v=\omega r\,$

This is the speed not the velocity because it is not a vector. The velocity is a vector, with magnitude ωr which points tangentially to the circle.

The magnitude of the velocity is constant but its direction changes so the mass is being accelerated.

By a similar argument to that above it can be shown that the magnitude of the acceleration is

$a=\omega v\,$

and that it is pointed inwards, along the radius vector. This is called centripetal acceleration.

By eliminating v or ω from these two equations we can write

$\mathbf {a} =-\omega ^{2}\mathbf {r} =-{\frac {v^{2}}{r}}\mathbf {\hat {r}}$