Angular Velocity and Acceleration Problem

[VitaX]

Homework Statement

The flywheel of a steam engine runs with a constant angular velocity of 180 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.5 h. (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 90.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 33 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Homework Equations

5 equations of constant acceleration (α is constant)

The Attempt at a Solution

a) ωo = 180 rev/min (2π/1 rev)(1 min/60 s) = 18.8496 rad/s
ω = 0
t = 2.5 h (60 min/1 h)(60 s/1 min) = 9000 s
ω = ωo + αt
0 = 18.8496 + 9000α
α = -.0021 rad/s (1 rev/2π)(60 s/1 min) = -1.032 rev/min^2

b) ϴ = ωot + .5αt^2
ϴ = 18.8496*9000 + .5(-.0021)(9000)^2
ϴ = 84,596.4 rad (1 rev/2π) = 13,463.935 rev

Parts c and d go together and I'm not quite sure how to find part c all I know is this equation should apply: at = rα

And if someone can confirm my work and answers for parts a and b that would be great since I'm not 100% sure they are correct, but I believe my method is right.

Edit: I figured it out.

 

[jbsteele]

c) at = rα (33 cm)(.01 m/1 cm)(.0021 rad/s) = .006987 rad/s^2 (1 rev/2π)(60 s/1 min) = .1182 rev/min^2d) a = √(ar^2 + at^2) = √(.1182^2 + .006987^2) = 11.8382 rev/min^2