0
$\begingroup$

What is the net force affecting a simple gravity pendulum?

I know the restoring force equals to:

$$ \vec{F}_s= -mg\cdot\sin{\varphi(t)}\cdot\hat{\theta} $$

And then there is the centripetal force (?):

$$ \vec{F}_c = \frac{mv^2}{L}\cdot-\hat{r} $$

$$ = \hat{r}(mg\cdot\cos{\varphi(t)}-T) $$

I am guessing that the net force equals to the vectoral sum of the forces I have highlighted:

$$ \vec{F}_{net} = \vec{F}_s+\vec{F}_c $$

But I honestly am confused about the centripetal force and don't quite understand how it equals to $\frac{mv^2}{L}$.

Improve this question
$\endgroup$
2
  • $\begingroup$ think about circular motion where the radius is L $\dfrac{mv^{2}}{L}=m\omega ^{2}L=m\left( \dfrac{d\varphi }{dt}\right) ^{2}L$ $\endgroup$
    – Eli
    Commented Jun 7, 2021 at 19:15
  • $\begingroup$ explorer, note that "centripetal force" is a catch-all term to describe a force that causes circular motion. In the case of a pendulum, that force is the tension in a string. $\endgroup$
    – David White
    Commented Jun 7, 2021 at 19:38

1 Answer 1

Reset to default
1
$\begingroup$

Since we know our pendulum is not getting any longer, we know that the net force in the radial direction must be zero. And so the net force acting on the pendulum is in the direction of $\hat{\theta}$ as you mentioned. This force is obviously going to be $( -mg \sin{\theta} )\hat{\theta} $ , which of course gives us:

$$m\vec{a}=( -mg \sin{\theta} )\hat{\theta} $$

And since $a$ is in th rotational direction, we can say $\vec{a}=(l \frac{\mathrm{d}^2}{\mathrm{d} t^2}\theta) \hat{\theta}$

Plugging that back in we obtain:

$$\frac{\mathrm{d}^2}{\mathrm{d} t^2}\theta = -\frac{g}{l}\sin{\theta}$$

This differential equation tells you everything about the pendulum, and you can use a small angle approximation to obtain a Harmonic Oscillator, but if you are interested in centripetal force it will be a little more subtle since we are now dealing with non-uniform circular motion, and so the velocity vector is not so nice to deal with. This is because in uniform circular motion $v$ is just a number, however in the case of a pendulum its a time varying vector.

But to understand why in general $a_c=\frac{v^2}{r} \rightarrow F_c=ma_c=\frac{mv^2}{r}$ consider the following:

enter image description here

Triangles ABC and PQR are similar and therefore:

$$\frac{\Delta v}{v}=\frac{\Delta s}{r}\rightarrow a:=\frac{\Delta v}{\Delta t}=\frac{v}{r}\frac{\Delta s}{\Delta t}= \frac{v^2}{r}$$

And so we obtain $a_c=\frac{v^2}{r}$

Improve this answer
$\endgroup$
3
  • $\begingroup$ This differential equation tells you everything about the pendulum Not entirely true: w/o an initial condition you won't know the pendulum's amplitude. $\endgroup$
    – Gert
    Commented Jun 7, 2021 at 20:56
  • $\begingroup$ If the radial acceleration is 0, then how come the object is maintaining its circular trajectory? $\endgroup$
    – explorer
    Commented Jun 7, 2021 at 21:19
  • $\begingroup$ en.wikipedia.org/wiki/Pendulum_(mathematics)#/media/… This is an animation from Wikipedia. When $\vec{F}_s = 0$ (the object is at the equilibrium position) there is still acceleration towards the center, pivot point. If the radial acceleration is indeed zero and the net force is: $-mg\cdot\sin{\varphi}\cdot\hat{\theta}$, at the equilibrium position, the object should have no net force, therefore acceleration, but according to this animation, Wikipedia; there is as I mentioned. Please address these issues regarding your answer. $\endgroup$
    – explorer
    Commented Jun 8, 2021 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.