Question

A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length $55.0$m. At the instant it makes an angle of ${35.0}^o$ with the vertical as it falls, what are (a) the radial acceleration of the top, and (b) the tangential acceleration of the top. (Hint : Use energy considerations, not a torque.) (c) At what angle $\theta$ is the tangential acceleration equal to g?

Answer 1

$\Delta u=mg\frac{l}{2}(1-\cos 35)\Delta u=\frac{1}{2}{I\omega }^{2}I=\frac{1}{2}{ml}^2\omega =\sqrt{\frac{2\Delta u}{I}}=\sqrt{\frac{3g(1-\cos 35)}{e}}=0.3rad/sZ=mg\frac{l}{2}\cos 55Z=I\alpha \Rightarrow \alpha =\frac{Z}{I}=\frac{3g\cos 55}{2l}radialacceleration={\omega }^{2}l=0.09\times 55=4.95m/stangentialacceleration=\alpha l=\frac{3g}{2}\cos 55=8.45$

Let $\theta$ be the angle when tangential acceleration is $\theta$

$\frac{3g\cos \theta }{2l}\times l=g\Rightarrow \cos \theta =\frac{2}{3}\Rightarrow \theta ={\cos }^{-1}\frac{2}{3}$