Vo=4401rev/min * 6.28rad/rev * 1min/60s=
460.6 m/s
V = 430 * 6.28/60 = 45 m/s.
a=(V-Vo)/t = (45-460.6)/4.93=-84.3 m/s^2
The blades of a motor rotate at a rate of 4401 rpm. When the motor is turned off, the blades slow to a new angular speed of 430 rpm in 4.93 seconds. What is the angular acceleration of the blades (in rads/sec^2)?
2 answers
Correction:
Vo = 460.6 rad/s
V = 45 rad/s
a = -84.3 rad/s^2
Vo = 460.6 rad/s
V = 45 rad/s
a = -84.3 rad/s^2