Pendulum question- Angular momentum-impulse

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! • @Forbenius Cheers for the correction. Do you have any idea about the question? – hellowurf May 11 '16 at 6:38
• $$dW=\mathbf{F}\circ d\mathbf{r}=F_{\theta}\cdot ds=\underbrace{\left(mg\cos\theta\right)}_{m\ell \ddot \theta}\cdot\left(\ell d\theta\right)=m \ell^{2} \ddot \theta d \theta \tag{01}$$ $$dE_{kin}=d\left(\dfrac{1}{2}mv^{2}\right)=d\left(\dfrac{1}{2}m\ell^{2} \dot\theta^{2}\right)=m \ell^{2} \dot \theta d \dot \theta \tag{02}$$ $$dW=dE_{kin} \longrightarrow \quad \textbf{???} \tag{03}$$ – Frobenius May 11 '16 at 8:51

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:

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

• For the last part, how did you get from 2*d(theta)/dt to (d(theta)/dt)^2? Did you implicit integrate? – hellowurf May 11 '16 at 8:23
• Well, it's just looking for the anti-derivative of the expression. It's basically $\frac{d}{dt}\left(\frac{d\theta}{dt}\right)^2$ – GodotMisogi May 11 '16 at 9:03