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

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...

• Since you mention a tangent, the path must be contionuously differentiable, right? Can it be argued that any contionuously differentiable path locally can be approximated as a circular arc? (a straight line is a circular arc for the circle with infinite radius). I think this might be done by a Taylor expantion. Something to investigate at least... Best regards Christoffer Jan 7 at 19:38
• I was wondering if this could be proved using a description of the motion through polar coordinates... That is, letting the position vector be R(theta) in the R hat direction, deriving that to get the velocity vector (and then once again for the acceleration vector), and setting the condition of constant speed through the length of the velocity vector having to be constant. I tried this approach but it didn't seem to lead to anything, could it work? Jan 7 at 22:23

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).

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: $$|\vec v| = C \to a_T = 0$$

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.

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.

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

• The time derivative of v is not the tangential acceleration. Consider an object moving in the purely radial direction. Jan 7 at 20:48
• @jensenpaull tangential acceleration is quite literally defined as the time-derivative of speed. Jan 8 at 7:39
• tutorial.math.lamar.edu/Classes/CalcIII/… Jan 8 at 7:40
• You're right, I confused this term to mean perpendicular to the centripetal acceleration in circular motion. Jan 8 at 12:23

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.