If a body moves along a path (any path, not just circular) with constant speed, is it's tangential acceleration necessarily zero?

Question

If a body moves along a path (any path, not just circular) with constant speed, is it's tangential acceleration necessarily zero?

I could only find general proofs for the case of circular motion and not some more generalised form...

Answer 1

The most general relation between speed and acceleration can be written like this: $$\frac{d(|\vec{v}|^2)}{dt} = \frac{d(\vec{v}\cdot\vec{v})}{dt} = 2\vec{v}\cdot\frac{d\vec{v}}{dt} = 2\vec{v}\cdot\vec{a}$$ I'm taking advantage of the fact that if $|v|$ is constant, then $|v|^2$ is constant and vice versa. From this vector equation, we can see that if acceleration is perpendicular to velocity, then the speed doesn't change. Also, if speed is constant, then acceleration is perpendicular to velocity (or zero).

Answer 2

If a body moves along a path (any path, not just circular) with constant speed, is it's tangential acceleration necessarily zero?

Yes.

First, recall that the velocity vector is pointing in the same direction as the tangent to the path. Therefore the tangential acceleration is: $$ a_T = \hat v \cdot \vec a\;, $$ where $\hat v = \vec v/|\vec v|$.

Next, we know that the speed is defined as the magnitude of the velocity, so we have $$ |\vec v| = \sqrt{\vec v \cdot \vec v}= C\;, $$ where $C$ is constant, and therefore: $$ \frac{d|\vec v|}{dt} = 0 = \frac{1}{|\vec v|}\vec v \cdot \vec a = \hat v\cdot \vec a = a_T\;, $$ where we used $\frac{d(\vec v \cdot \vec v)^{1/2}}{dt} = \frac{1}{2}(\vec v \cdot \vec v)^{-1/2}(\vec v \cdot \frac{d\vec v}{dt}+\frac{d\vec v}{dt} \cdot \vec v)=(\vec v \cdot \vec v)^{-1/2}(\vec v \cdot \frac{d\vec v}{dt}) = \frac{1}{|\vec v|}\vec v \cdot \vec a$.

In other words, we see directly that, yes: $$ |\vec v| = C \to a_T = 0 $$

Answer 3

Yes. Several other answers have already answered this, but here is another viewpoint.

The kinetic energy is $$T=\frac{1}{2}mv^2.$$ The work done is equal to the change in kinetic energy: $$W=\Delta T$$ And it is also equal to: $$W = F \cdot \vec{v}.$$ If it has a constant speed, then the kinetic energy does not change. Hence, $$F \cdot \vec{v}=0.$$ This means that the force is zero, or perpendicular to the velocity. Therefore, it has no tangential force, and hence no tangential acceleration.

Answer 4

Constant speed means no work is done.

$$\vec{F} \cdot \vec{dr} = 0$$

$$ [\vec{F} \cdot \vec{v}]dt = 0$$

The component of force in the direction of velocity (tangential direction) must therefore be zero, and hence the tangential acceleration is zero.

Answer 5

Since the tangential acceleration is defined as $a_T = \dot v$, then necessarily if $v$ is not changing, $a_T$ must be zero.

Answer 6

Too lazy to write down formulas, but my first instinct is to say Yes if the path is smooth. All smooth curves are locally similar to a segment of some circle (up to the second order). So, locally you have a motion with a constant velocity along some circle.