Answer
The tangential element of the acceleration of the chimney top is given by $a_{t}=H \alpha,$ where $\alpha$ is the angular acceleration. We are unable to use Table $10-1$ since the acceleration is not (Suit)uniform. Hence, we differentiate
$$
\omega^{2}=(3 g / H)(1-\cos \theta)
$$
with respect to time, replacing $d \omega / d t$ with $\alpha,$ and $d \theta / d t$ with $\omega,$ and obtain
$$
\frac{d \omega^{2}}{d t}=2 \omega \alpha=(3 g / H) \omega \sin \theta \Rightarrow \alpha=(3 g / 2 H) \sin \theta
$$
Consequently,
$$
a_{t}=H \alpha=\frac{3 g}{2} \sin \theta=\frac{3\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)}{2} \sin 35.0^{\circ}=8.43 \mathrm{m} / \mathrm{s}^{2} .
$$
Work Step by Step
The tangential element of the acceleration of the chimney top is given by $a_{t}=H \alpha,$ where $\alpha$ is the angular acceleration. We are unable to use Table $10-1$ since the acceleration is not (Suit)uniform. Hence, we differentiate
$$
\omega^{2}=(3 g / H)(1-\cos \theta)
$$
with respect to time, replacing $d \omega / d t$ with $\alpha,$ and $d \theta / d t$ with $\omega,$ and obtain
$$
\frac{d \omega^{2}}{d t}=2 \omega \alpha=(3 g / H) \omega \sin \theta \Rightarrow \alpha=(3 g / 2 H) \sin \theta
$$
Consequently,
$$
a_{t}=H \alpha=\frac{3 g}{2} \sin \theta=\frac{3\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)}{2} \sin 35.0^{\circ}=8.43 \mathrm{m} / \mathrm{s}^{2} .
$$
