Atall, 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° 

The answer is 12, for the points

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j