(a)What is the angular acceleration of the blade? solved
(b)What is the distance traveled by a point on the rim of the blade during the deceleration, in feet? need help
(c)What is the magnitude of the net displacement of a point on the rim of the blade during the deceleration? in inches need help.
I used rotational kinematics equations but the answers kept coming back wrong for the last two? any help?
1 Answer

ωi = 4310rev/min 2πrad/rev 1min/60s = 451 rad/s
(a) π = (ωf  ωi) / t = 451rad/s / 3s = 150 rad/s²
(b) A number of ways to go about this. Radius = 5 in, so
tangential vi = 451rad/s 5in 1ft/12in = 188 ft/s
and distance = Vavg t = ½188ft/s * 3s = 282 ft
(c) Angular displacement Θ = ωavg t = ½451rad/s * 3s = 677 rads
which divided by 2π is 107.75 revolutions.
Now, the 107 revolutions have a net displacement of zero, leaving 0.75 revolutions.
Since the wheel has radius 5in, the distance between the starting point and the ending point for any point P on the rim is d = 5in*√2 = 7.1 in.