Mechanics |07-06-2021|

At a general time $t$ let the angle the rod makes with the vertical be $\theta$ . At this instant, computing the net torque about the point O leads to the equation:

$\left(\frac{mL^2}{3}\right)\ddot{\theta} = FL\cos{\theta} - \frac{mgL\sin{\theta}}{2}$ $\implies \ddot{\theta} =\frac{3F\cos{\theta}}{mL} - \frac{3g\sin{\theta}}{2L}$

The initial conditions are: $\theta(0) = \dot{\theta}(0) =0$ . Since the rod hangs vertically, it is initially assumed to be at rest.

Now, applying the following manipulation:

$\frac{d^2\theta}{dt^2} = \frac{d \dot{\theta}}{d\theta}\frac{d\theta}{dt} = \dot{\theta}\frac{d \dot{\theta}}{d\theta}$

Plugging this into the equation of motion:

$\ddot{\theta}=\dot{\theta}\frac{d \dot{\theta}}{d\theta}=\frac{3F\cos{\theta}}{mL} - \frac{3g\sin{\theta}}{2L}$ ]

Separating variables and integrating:

$\int \dot{\theta} \ d \dot{\theta} = \int \left( \frac{3F\cos{\theta}}{mL} - \frac{3g\sin{\theta}}{2L} \right)d\theta$ $\implies \frac{\dot{\theta}^2}{2} =\frac{3F\sin{\theta}}{mL} + \frac{3g\cos{\theta}}{2L} + c$

Plugging in initial conditions and solving for $c$ leads to the final relation:

$\frac{\dot{\theta}^2}{2} = \frac{3F\sin{\theta}}{mL} - \frac{3g}{2L} \left(1 - \cos{\theta}\right)$

The maximum angular displacement is achieved when the rod again comes to rest. This means $\dot{\theta}=0$ . This means:

$\frac{3g}{2L} \left(1 - \cos{\theta}\right)=\frac{3F\sin{\theta}}{mL}$ $\frac{3g}{2L} 2\sin^2\left(\frac{\theta}{2}\right) = \frac{3F}{mL}\left(2 \sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)\right)$ $\implies \tan\left(\frac{\theta}{2}\right)= \frac{2F}{mg}$ $\theta_{max} = 2\arctan\left(\frac{2F}{mg}\right)$

$\implies \alpha = \beta = 2$