A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 53.2 m. At the instant it makes an angle of 34.1° 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

From an energy balance, we can use this formula to solve for the angular speed of the chimney

ω^2 = 3g / h sin θ

Substituting the given values:

ω^2 = 3 (9.81) / 53.2 sin 34.1

ω^2 = 0.987 / s

The formula for radial acceleration is:

a = rω^2

So,

a = 53.2 (0.987) = 52.494 / s^2

The linear velocity is:

v^2 = ar

v^2 = 52.949 (53.2) = 2816.887

The tangential acceleration is:

a = r v^2

a = 53.2 (2816.887)

a = 149858.378 m/s^2

If the tangential acceleration is equal to g:

g = r^2 3g / sin θ

Solving for θ

θ = 67°