Forces on simple pendulum bob at turning point?

Question

A simple pendulum consists of a massless string and an oscillating bob. The resultant force on the bob is pointing out the direction of acceleration, and varying with time. My question concerns the turning point of the bob, i.e where the speed is zero (at $0<\theta<90^\circ$). Why is the resultant force in the tangential direction in this moment, and not just $mg$ pointing downwards?

I like to see inertia as a "speed memory". But does inertia also acts like an "acceleration memory", so the acceleration at some time $t$ depends on the acceleration of the moment before $t$?

Answer 1

Short answer

You forgot the force $\mathbf{N}$ on the bob coming from the wire, assumed to be aligned with the wire itself (for mass-less wire this is true), pointing towards the point $P$. Once you take into account the reaction $\mathbf{N}$, you get the expected results at the turning point (see the last line of the answer).

Derivation of the equation of motion and the reaction

The forces acting on the bob are its weight $m\mathbf{g}$ (constant magnitude and direction, always pointing downwards) and the reaction of the wire $\mathbf{N}$ (variable magnitude and direction, pointing towards $P$). The $2^{nd}$ principle of the dynamics for the bob reads:

$m \ddot{\mathbf{r}} = \mathbf{F}^{tot} = m \mathbf{g} + \mathbf{N}$,

and it can be projected along the radial and azimutal directions as

$r: m L \dot{\theta}^2 = N - m g \cos \theta$
$\theta: m L \ddot{\theta} = - m g \sin \theta$

Our problem has only one degree of freedom, $\theta$, so we could solve the problem solving just one equation if it doesn't contain reactions. This is the case of the projection along $\theta$ above, that only contains $\theta(t)$ (unknown), but not $N$. The very same equation can be obtained from

• the balance equation of angular momentum around the point $P$: $ m L \dot{v}_{\theta} = - m g L \sin \theta$.
• the conservation of energy: $\text{const.} = E^{mec} = K - U = \dfrac{1}{2}m|\mathbf{v}|^2 + m g h$, where the kinematics tells us that $\mathbf{v} = v_{\theta} \boldsymbol{\hat{\theta}}$ and $h = - \cos \theta$, $v_{\theta} = L \dot{\theta}$

Now, provided the initial conditions (as an example $\theta(0) = \theta_0$ and $\dot\theta(0) = \Omega_0$, we can solve (numerically, or for small-amplitude oscillations analitically if we assume $\sin \theta \sim \theta$) the ODE and find $\theta(t)$.

Once we have solve for $\theta(t)$, we can find the magnitude of the wire reaction

$N(t) = m L \dot\theta^2(t) + m g \cos \theta(t)$.

Highest point of the trajectory

From the last formula, we can easily see that in correspondence of the highest point of the trajectory, when the motion reverses and $\dot\theta = 0$, the reaction in the wire is $N = m g \cos \theta_{\max}$.

In this point, the dynamical equations tell us that

$ 0 = N_{\theta} - m g \theta_{max}$
$ m L \ddot\theta = - m g \sin \theta_{max} $

Thus, doing the vector sum of the weight and the reaction $\mathbf{N}$ you get the total force acting in azimutal direction with magnitude $m g \sin \theta_{max}$, as you expected.

Answer 2

There are two forces acting on the pendulum bob - its weight $mg$ and the tension in the string $T$. If the string is taut and its length does not vary then the acceleration of the bob has no radial component, so the net radial force on the bob must be zero. This tells us that

$T = mg \cos \theta$

So the net force on the bob must act in a tangential direction, and so the bob's acceleration is also tangential (which we also know because the distance between the bob and the pivot point $P$ does not change). And the magnitude of the tangential force is the tangential component of the bob's weight, which is $mg \sin \theta$.