The pendulum of mass m and length l is released from rest at $\theta=0$. Using only the principle of angular impulse and momentum, determine the expression for $\ddot \theta $ in terms of $\theta$, and find the velocity v, of the pendulum at $\theta=90\unicode{xb0}$ . Compare this approach with a solution by the work-energy principle.

I can easily solve using the work- energy principle. But the other approach wasn't clearly explained in the answers:

$\sum M=\dot H : mg\ell cos(\theta)=\frac{d(m\ell^2 \dot\theta)}{dt}\Rightarrow\ddot \theta=\frac{gcos(\theta)}{\ell}$.

What I don't get is this :$\int \dot \theta d\dot \theta=\int \ddot \theta d \theta$

Note that $\theta$ starts from the horizontal.

Please explain how they got this. Thanks!

enter image description here

  • $\begingroup$ @Forbenius Cheers for the correction. Do you have any idea about the question? $\endgroup$
    – hellowurf
    May 11, 2016 at 6:38
  • $\begingroup$ $$ dW=\mathbf{F}\circ d\mathbf{r}=F_{\theta}\cdot ds=\underbrace{\left(mg\cos\theta\right)}_{m\ell \ddot \theta}\cdot\left(\ell d\theta\right)=m \ell^{2} \ddot \theta d \theta \tag{01} $$ $$ dE_{kin}=d\left(\dfrac{1}{2}mv^{2}\right)=d\left(\dfrac{1}{2}m\ell^{2} \dot\theta^{2}\right)=m \ell^{2} \dot \theta d \dot \theta \tag{02} $$ $$ dW=dE_{kin} \longrightarrow \quad \textbf{???} \tag{03} $$ $\endgroup$
    – Frobenius
    May 11, 2016 at 8:51

1 Answer 1


The torque derivation is as follows:

$$ \sum \vec \tau = \frac{\mathrm{d}L}{\mathrm{d}t} $$

The magnitude of the torque (measuring $\theta$ with respect to the horizontal) is:

$$ |\vec{\tau}| = mgl\cos\theta $$

The angular momentum $L$ is given by:

$$ L = mr^2\omega = ml^2\frac{\mathrm{d}\theta}{\mathrm{d}t} $$

So the rate of change of angular momentum is:

$$ \frac{\mathrm{d}L}{\mathrm{d}t} = ml^2\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}$$

Which gives the differential equation:

$$ mgl\cos\theta = ml^2\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} $$

$$ \frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} - \frac{g}{l}\cos\theta = 0 $$

Without going into elliptic integrals, the most you can find out of this differential equation is the velocity:

Multiply the differential equation by $ 2\frac{d\theta}{dt} $ and inspect: $$ 2\frac{\mathrm{d}\theta}{\mathrm{d}t}\cdot\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} - \frac{2g}{l}\cos\theta\cdot\frac{\mathrm{d}\theta}{\mathrm{d}t} = 0 $$

$$ \left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 - \frac{2g}{l}\sin\theta = 0$$

Note that this is pretty much a trick, so it's fair to use. Now you can find the velocity and compare!

EDIT: For clarity, you can use the chain rule to find the angular velocity by using information from the angular displacement:

$$ \dot{\theta} = \frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{\mathrm{d}\theta}{\mathrm{d}\dot{\theta}}\frac{\mathrm{d}\dot{\theta}}{\mathrm{d}t} = \frac{\mathrm{d}\theta}{\mathrm{d}\dot{\theta}}\ddot{\theta}$$

Which gives the expression you're looking for by rearranging the differentials and integrating:

$$ \int\dot{\theta}\,\mathrm{d}\dot{\theta} = \int \ddot{\theta}\,\mathrm{d}\theta $$

  • $\begingroup$ For the last part, how did you get from 2*d(theta)/dt to (d(theta)/dt)^2? Did you implicit integrate? $\endgroup$
    – hellowurf
    May 11, 2016 at 8:23
  • $\begingroup$ Well, it's just looking for the anti-derivative of the expression. It's basically $ \frac{d}{dt}\left(\frac{d\theta}{dt}\right)^2 $ $\endgroup$ May 11, 2016 at 9:03

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.