Steam engine's and revolutions it takes to slow down...?

Question

The flywheel of a steam engine runs with a constant rotational velocity of 150 rev/min. When the steam is shut off, the friction of the bearings and of the air stops the wheel in 2.2 h. (a) What is the constant rotational acceleration, in revolutions per minute-squared, of the wheel during the slowdown?

(b) How many rotations does the wheel make during the slowdown?

(c) At the instant the flywheel is turning at 75 rev/min, what is the tangential component of the translational acceleration of a flywheel particle that is 50 cm from the axis of rotation? (d) What is the magnitude of the net translational acceleration of the particle in (c)?

Explain your work. Thanks! :)

Answer 1

No conversions needed for answers A, B.
α = ang. accel in rev/min^2, ω = ang. rate in rev/min, θ = angle in rev
t = time in min = 132
A. α = ω/t = 150/132 = 1.1364 rev/min^2
B. θ = ωt/2 = 9900 rev
C. 1st answer confused tangential acceleration (=αr) and centripetal acceleration (=ω^2r).
now α must be in rad/s^2
atang = αr = 1.1364*2pi/3600*0.5 m/s^2
D. Radial acceleration is centripetal.
now ω must be in rad/s
ω = 2*pi*75/60
acent = ω^2r
atot = sqrt(acent^2+atang^2)

Answer 2

covert units

w0 = 150 rev/min * 1 min /60 sec * 2pi rad/rev = 3.93 rad/sec

t = 2.2 hrs * 3600 sec/1hr = 7920 sec

a) use wf = w0 + alpha * t
wf =0
alpha = -w0/t

b) theta = theta0 + w0*t + 1/2*alpha*t^2
theta0 = 0
plug in your numbers to get theta.
take theta and divide it by 2pi and you'll get your number of revs.

c) at = alpha*r (remember to use the correct units)

d) ar = w^2*r
a = sqrt(at^2 + ar^2)