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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!

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  • $\begingroup$ @Forbenius Cheers for the correction. Do you have any idea about the question? $\endgroup$ – hellowurf May 11 '16 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 '16 at 8:51
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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:

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

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  • $\begingroup$ 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 '16 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 $ $\endgroup$ – GodotMisogi May 11 '16 at 9:03

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