# Speed and tangential acceleration in pendulum motion

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.

• 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.
– user107153
Commented Apr 23, 2019 at 22:13

## 3 Answers

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:

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

• Times of $a$, $b$ and $c$ are very confusing, especially since $a$ already has another meaning in this question. How about $t_0$, $2 t_0$ and $3 t_0$? Or $t_1$, $t_2$ and $t_3$? Commented Apr 3, 2022 at 13:10
• What do the plus and minus signs on the diagram mean? Commented Apr 3, 2022 at 13:43
• Sign of the gradient Commented Apr 3, 2022 at 22:35
• The answer says: “If you are talking about speed”. There is no “if” here: the OP is definitely talking about speed. The entire question is predicated on there being a local minimum, which can only happen with speed, not velocity. The rest of the answer’s text mentions only speed, not velocity, and certainly does not mention the gradient of velocity. Yet the graph shows velocity too, complete with the sign of the gradient. I find this very confusing. Commented Apr 5, 2022 at 3:21

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.

• The first paragraph’s reasoning is not entirely correct: for example, $x^3$ changes sign at $x = 0$, but $\left| x^3 \right|$ is differentiable everywhere. I do not understand the second paragraph. Why would the acceleration always be in the same direction as the velocity? Why would imagining that it is be helpful? Commented Apr 3, 2022 at 13:46

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

No, it is not. As user107153’s comment says:

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.

For practical purposes (at least in this case), we can probably equate “tangential component of velocity” to “velocity”. But it is still velocity, not just speed; its direction matters.

The question would be correct if it used magnitudes on both sides:

$$$$\left| a_\text{tan} \right| = \left| \frac{d \left| \vec{v} \right|}{dt} \right|,$$$$

With this correction, we still have the same issues about whether the derivative is zero.

When it reaches it highest position, the acceleration is exclusively tangential, since it has zero speed/velocity.

Correct.

the pendulum's speed is at a local minimum

Correct.

the derivative should be zero

Incorrect.

how exactly can we make mathematical sense out of this?

A local extreme satisfies one of the following conditions:

1. It occurs at the “end of the domain”, in a sense: there is no interval with the extreme point in the middle, where the function is defined everywhere on the interval.
2. The derivative is not defined.
3. The derivative is zero.

The first and second cases are the kind of special cases that people tend to gloss over, but it is the second case that applies here. The reason for this is best explained with a diagram, and one is included in Farcher’s answer.

(It is possible that the derivative is indeed zero, and the “exclusively tangential” acceleration is also zero for some time, perhaps just for an instant. But only if some unusual forces are applied. Of course, in the normal situation, the magnitude of tangential acceleration is at a maximum when the pendulum is at its highest position.)