Downward Acceleration of a Falling Chimney's Tip

Question

I was checking out this explanation on Harvard’s site about why the top of a falling chimney seems to fall faster than gravity (https://sciencedemonstrations.fas.harvard.edu/presentations/falling-faster-g), and it got me thinking.

If we consider the downward acceleration of the chimney's tip, shouldn't we include not just the vertical component of the tangential acceleration but also the vertical component of the radial acceleration? Am I missing something here? Any insights or clarifications would be greatly appreciated thanks the derivation use these equations to get the angular acceleration $\alpha$ $$\begin{gathered}I \alpha=\frac{L}{2} m g \cos \theta \\ I=1 / 3 m L^2 \\ \end{gathered}$$

and then to get the vertical acceleration (direction of g,Y) use the downward component of the tangential acceleration $$a_t=L \alpha$$ which is $$a_Y=a_t \cos \theta=\ldots=\frac{3}{2} \cos ^2(\theta)$$

why not also add the radial acceleration's contribution to the Y-Component?

Answer 1

The radial acceleration depends on the angular velocity $\omega$ by $a_r=r\omega^2$. Therefore, when the rod is first released from rest, $a_r=0$. The analysis is considering the instant in time when the rod is first released, and they have set the initial angle so that the downward component of the acceleration (the acceleration is completely tangential at this instant) already starts off at g and will increase from there. This is sufficient to have the ball fall into the cup, so there is no need to look at the radial acceleration after the release.

You are correct though, to analyze the overall downward acceleration of the tip across the whole motion, one would need to consider the radial acceleration as well.

Answer 2

If the vertical component of acceleration of the chimney tip is > g, the ball will be left above the tip. The chimney will exert no horizontal or vertical force on the ball. The ball will drop straight down.

Answer 3

The position vector to the tip is: \begin{align*} &\begin{bmatrix} X \\ Y \\ \end{bmatrix}= L\,\begin{bmatrix} \cos(\theta) \\ \sin(\theta) \\ \end{bmatrix} \end{align*} form here you obtain the velocity and the acceleration

\begin{align*} &\begin{bmatrix} \dot X \\ \dot Y \\ \end{bmatrix}= L\,\dot{\theta}\begin{bmatrix} -\sin(\theta) \\ \cos(\theta) \\ \end{bmatrix}\\ &\begin{bmatrix} \ddot X \\ \ddot Y \\ \end{bmatrix}= L\,\ddot{\theta}\begin{bmatrix} -\sin(\theta) \\ \cos(\theta) \\ \end{bmatrix}+ L\,\dot{\theta}^2\begin{bmatrix} -\cos(\theta) \\ -\sin(\theta) \\ \end{bmatrix} \end{align*}

thus $~a_Y=\ddot{Y}~$ \begin{align*} &a_Y=L\,\ddot\theta\,\cos(\theta)\underbrace{-L\,\dot{\theta}^2\sin(\theta)}_{\text{ radial component}}\\ &\text{with }\quad,I\,\ddot{\theta}=-\frac{L}{2}\,m\,g\,\cos(\theta) \end{align*}

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