# A simple pendulum

The idealized simple pendulum model (see the following figure) assumes that at every point of time the string to which the bob is attached exerts an equal and opposite force on the bob as does the component of the force of gravity acting on the bob along the line determined by the string. Is this assumption the mathematical way of modeling the string, or can it be derived from the (in my opinion more intuitive) assumption that the bob's movement is constrained to take place along the circumference of the circle centered at the string's axis whose radius is the string's length?

Here's how I have attempted to solve this problem. Unfortunately, this solution makes no sense, but I can't see where my mistake is.

Let $\mathbf{p} : \mathbb{R} \rightarrow \mathbb{R}_{2, 1}$ be a twice differentiable function from $\mathbb{R}$ to the space of real $2 \times 1$ column vectors. $\mathbf{p}$ will represent the location of the bob in a two-dimensional space. Denoting the length of the string by $L$ ($L > 0$) and assuming the string's axis coincides with the origin, the model's assumptions reduce to the following equation: $$\|\mathbf{p}\|^2 = L^2$$ Using this notation, we wish to show that the force acting on the point-mass is entirely perpendicular to the line determined by the string. Since this force is proportional to the mass's acceleration, this reduces to the following equation: $$\mathbf{p}^T\ddot{\mathbf{p}} = 0$$

By differentiating both sides of the assumption, we get $$\mathbf{p}^T \dot{\mathbf{p}} = 0$$

By differentiating both side of the last equation, we get $$\|\dot{\mathbf{p}}\|^2 + \mathbf{p}^T\ddot{\mathbf{p}} = 0$$

Therefore, if the proposition that we wish to prove is true, the last equality implies $\dot{\mathbf{p}} = \mathbf{0}$, i.e. the pendulum is motionless. However, this is evidently false in general. So what did I do wrong?

• The first statement can give rise to the second, and the second to the first. Both of them are the same thing, really. – Hritik Narayan Feb 14 '15 at 11:47
• @HritikNarayan: So how can the first be mathematically derived from the second? – Evan Aad Feb 14 '15 at 11:48
• The force applied by the string is equal and opposite to the pull generated by the bob, this implies the bob has no acceleration along the string! (only tangential acceleration.) This implies that the bob always stays at a fixed distance from the pivot point of the pendulum. The locus of all points at the same distance from any given point is a circle, and hence the second statement arises. For the opposite, the bob is constrained along a circle. So it has no acceleration in the direction of the string, which implies a zero net force along the string. And hence we get the first statement. – Hritik Narayan Feb 14 '15 at 11:52

## 2 Answers

The string is "modelled" as a stretched out, unflexible piece of string. The premise in the motion is that the string is at all times stretched (the lenght, and therefor the distance to the rotational center is kept constant).

If you believe that this premise is true then the movement has to be circular, and the stone at the end of the string can only move at the circumference of the circle with a radius equal to the string lenght.

The assumption you are talking about is:

at every point of time the string to which the bob is attached exerts an equal and opposite force on the bob as does the component of the force of gravity acting on the bob along the line determined by the string.

Since the "bob" at the end of the string is fixed to the string, and the string is not-breaking and stretched, then this has to follow the circular path. It can move to the sides around the path, but cannot get closer or further away - that is, there must be no movement perpendicular to the path. For this to be true, the string must counteract any other forces acting perpendicular to the path on the bob.

Other forces are only the perpendicular component of gravity, and you get:

$$T=mg\cos{\theta}$$

This is an idealized physical model of the situation mathematically described with components of forces. To answer your question: It is a mathematical way of modelling the situation, but I would say that it is derived physically from the knowledge of Newtons 1st law.

It is not true that "the string to which the bob is attached exerts an equal and opposite force on the bob as does the component of the force of gravity acting on the bob along the line determined by the string". The string's tension is not equal, in general, to the component of gravity along the line determined by the string. If you look closely at the picture you linked to, you'll see that the two arrows are not the same length. Also consult the following gif.