A shaft is turning at 71.6rad/s at time zero. Thereafter, its angular acceleration is given by

Question

α = –10.8rad/s2 – 4.42 t rad/s3
where t is the elapsed time. Calculate its angular speed at t = 2.73s.
and
How far does it turn in these 2.73s

Answer 1

dw/dt = -10 - 4.42 t
so
w = -10 t - 2.21 t^2 + constant

find constant
at t = 0
71.6 = -10(0) - 2.21(0)^2 + c
so c = 71.6

w = 71.6 - 10 t - 2.21 t^2

put in t = 2.73

Now the angle

dTheta/dt = w = 71.6-10 t -2.21 t^2
so
Theta = 71.6 t - 5 t^2 - (2.21/3)t^3