Answer
$$
a_{r}=5.32 \mathrm{m} / \mathrm{s}^{2}
$$
Work Step by Step
The radial component of the acceleration of the chimney top is given by $$a_{r}=H \omega^{2},$$ so
$$
a_{r}=3 g(1-\cos \theta)=3\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)\left(1-\cos 35.0^{\circ}\right)=5.32 \mathrm{m} / \mathrm{s}^{2}
$$
