Why the work of the weight of a pendulum is $mg\ell\cos \theta $? Consider the coordinate system $(O,e_r,e_\theta )$. Then $P=(mg\cos\theta ,mg\sin \theta )$, and thus $$\delta W_P= P\cdot dr.$$ Since the mouvement is a circle, then $dr=(-\sin\theta ,\cos\theta )$ and thus $$\delta W_P=-mg\cos\theta \sin\theta +mg\cos\theta \sin\theta =0.$$ Therefore $W_P=0$. What's wrong in my argument?

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  • $\begingroup$ That's a very complicated way of calculating that work! Simply calculate the change in height of the bob: $l[1 - \cos\theta]$. $\endgroup$ – Gert Dec 15 '18 at 17:19
  • $\begingroup$ @Gert Yes, but it is also a instructive way when you are studying classical mechanics, which is what I assume OP is doing. $\endgroup$ – Pygmalion Dec 15 '18 at 17:21

Welcome to Physics.SE!

You wrote infinitesimal vector $d\vec{r}$ wrongly. It has only component in $\vec{e}_\theta$ direction (see your own picture!) that is

$$d\vec{r} = l \: d\theta \: \vec{e}_\theta.$$

Here $l \: d\theta$ is the infinitesimal move or arc length for an infinitesimal angle $d\theta$.

So infinitesimal work is

$$dW = \vec{P} \cdot d\vec{r} = (mg \cos \theta \: \vec{e}_r + mg \sin \theta \: \vec{e}_\theta) \cdot l \: d\theta \: \vec{e}_\theta = mg \: l \sin \theta \: d\theta.$$

You integrate that for $\theta$ from $\theta$ to $0$ and you get wanted expression

$$W =\int dW = \int_\theta^0 m g l \sin\theta d\theta = m g \: l [1 - \cos\theta].$$

Note that $l[1 - \cos\theta]$ is actually height difference.


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