A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 58.8 m. At the instant it makes an angle of 30.2° 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 θ is the tangential acceleration equal to g? Assume free-fall acceleration to be equal to 9.81 m/s2.

Using conservation of mechanical energy for rotational motion of chimney

gain of rotational KE = loss of potential energy

1/2 Iω² = mgl/2 (1 - cosθ); I is moment of inertia, ω is angular velocity, m is mass of chilney, l is its length.

I = 1/3 m l²

1/6 m l²ω² = mgl/2 (1 - cosθ)

1/6 m v² = mgl/2 (1 - cosθ)

v² = 3gl (1 - cosθ)

radial acceleration at the top = v² / r

=3gl (1 - cosθ) / l

= 3g (1 - cosθ)

= 3 x 9.8 (1 - cos 30.2)

= 4 m / s²

b) Let angular acceleration be α.

Torque acting at that time

mgsinθ x l / 2 = I α

mgsinθ x l / 2 = 1/3 m l² α

α = 3g sinθ / 2l

tangential acceleration

α x l = 3gsinθ / 2

= 3 x 9.8 sin 30.2 / 2

= 7.4 m / s²

c) tangential acceleration equal to g

3gsinθ / 2 = g

sinθ = 2 / 3

42 degree.