On a horizontal ground, a very thin, homogeneous rod stands in a vertical position. The rod starts to fall without an initial velocity. Calculate the angular acceleration of the rod as a function of the angle it makes with the horizontal! With what angular velocity does the rod reach the ground? The friction between the rod and the ground can be neglected.
From the diagram, the change in the potential energy of the rod can be read as: $m g(l / 2)(1-\sin \alpha)$, thus with $v$ denoting the velocity of the center of mass, and $\omega$ the angular velocity, the equation of conservation of mechanical energy is:

$$
m g(l / 2)(1-\sin \alpha)=(1 / 2) m v^{2}+(1 / 2)(1 / 12) m l^{2} \cdot \omega^{2}
$$

If there is no friction, the forces acting on the rod have no horizontal component. Therefore, the center of mass does not move horizontally. Hence, $v$ is in the vertical direction and

$$
v=\mathrm{d} h / \mathrm{d} t=\mathrm{d}[(l / 2) \sin \alpha] / \mathrm{d} t=(l / 2) \cos \alpha \cdot \mathrm{d} \alpha / \mathrm{d} t
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_5e8a71381245402f4363g-1.jpg?height=404&width=512&top_left_y=584&top_left_x=796)

Considering that the directions of $\alpha$ and $\omega$ are opposite in the diagram: $\omega=-(\mathrm{d} \alpha / \mathrm{d} t)$, thus:

$$
v=-(l / 2) \omega \cdot \cos \alpha
$$

Substituting this into (1) we get:

$$
\omega^{2}=\frac{12 g}{l} \cdot \frac{1-\sin \alpha}{1+3 \cos ^{2} \alpha}
$$

The angular acceleration $\beta=\mathrm{d} \omega / \mathrm{d} t$. However, it is more convenient to differentiate the expression (3) with respect to time to calculate $\beta$.

$$
2 \omega \frac{\mathrm{d} \omega}{\mathrm{d} t}=\frac{12 g}{l} \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{1-\sin \alpha}{1+3 \cos ^{2} \alpha}\right)=\frac{12 g}{l} \frac{\cos \alpha}{1+3 \cos ^{2} \alpha}\left[1+\frac{6 \sin \alpha(1-\sin \alpha)}{1+3 \cos ^{2} \alpha}\right] \omega
$$

thus

$$
\beta=\frac{6 g}{l} \frac{\cos \alpha}{1+3 \cos ^{2} \alpha}\left[1+\frac{6 \sin \alpha(1-\sin \alpha)}{1+3 \cos ^{2} \alpha}\right]
$$

At the moment the rod reaches the ground $(\alpha=0)$, the angular velocity of the rod from (3) is

$$
\omega_{1}=\sqrt{\frac{3 g}{l}}
$$

Lajos Czuczor (Budapest, Fazekas M. Gyak. Gimn., III. o. t.)

Remark. Most people made a mistake when they assumed that there is a relationship similar to (2) between the acceleration of the center of mass and the angular acceleration, i.e., $\alpha=-(l / 2) \cos \beta$. The constraint relations between velocities and accelerations are generally of the same form only when there is a linear relationship with a constant coefficient between the velocity and the angular velocity. In our case, however, $\alpha$ depends on time. The condition for the accelerations is obtained by differentiating (2):

$$
\alpha=\mathrm{d} v / \mathrm{d} t=(l / 2)(\mathrm{d} \alpha / \mathrm{d} t) \cdot \omega \sin \alpha-(l / 2)(\mathrm{d} \omega / \mathrm{d} t) \cos \alpha=-(l / 2) \omega^{2} \sin \alpha-(l / 2) \beta \cos \alpha
$$