Speed is a scalar quantity which corresponds to the magnitude of velocity, so it must always be nonnegative.

In a general 2d motion, the tangential component of acceleration is given by the time derivative of speed: $$a_{tan} = \frac{d|\vec{v}|}{dt}$$

Now, consider the motion of a pendulum. When it reaches it highest position, the acceleration is exclusively tangential, since it has zero speed/velocity.

However, according to our equation for tangential acceleration, the derivative should be zero, since the pendulum's speed is at a local minimum. What is wrong here?

I can kind of make sense of this, if I think that speed "changed direction", so it went from positive to "negative" and there was no minimum at all. But this doesn't make sense if we are to consider speed to be a nonnegative scalar.

You could also think that the speed is still positive, but the tangential direction is now different, because the direction of motion has changed. But this explanation doesn't solve the problem of that derivative being supposedly zero.

So, how exactly can we make mathematical sense out of this? I've searched the internet, but usually this situation is tackled using dynamics and getting the acceleration from the forces acting on the pendulum.

  • $\begingroup$ It's not the case that the tangential component of acceleration is the time derivative of speed: it's the time derivative of the tangential component of velocity. $\endgroup$ – tfb Apr 23 at 22:13

If you are talking about speed, which is the magnitude of the velocity, then speed is always a positive quantity.

Consider the following table:

$\begin{array}{cc}\text{time}&-c&-b&-a& 0 & a & b& c \\\text{speed} &3&2&1&0&1&2&3 \end{array}$

For times which are less than zero (and increasing) the change in speed $(\text{speed}_{\rm final} - \text{speed}_{\rm initial})$ is negative whilst for times which are greater than zero the change in speed is positive.

From the table I hope you can see where your idea that the change of speed is zero when the time is zero has come from.
You are saying that at time $-a$ the speed is $1$ and at time $+a$ the speed is also $1$ so the change in the speed at time equal to zero, (as $a$ tends to zero), is zero but this is not correct as shown below:

enter image description here

The speed vs time graph is not smooth when time is zero and so a gradient cannot be found at this time.


Because the velocity changes sign at the topmost point, there is a kink in $|\vec{v}|$, and it is not differentiable there, leaving $a_\text{tan}$ undefined.

You can see this even more clearly if you visualize the tangential acceleration as always pointing in the same direction as the velocity, but the velocity vanishes at the topmost point. The magnitude of the acceleration, however, does not. This leads to an undefined expression at the moment where the velocity is the zero-vector.


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