Question

When a carpenter shuts off his circular saw, the 10.0-inch diameter blade slows from 4830 rpm to zero in 3.00 s .

1. What is the distance traveled by a point on the rim of the blade during the deceleration?

2. What is the magnitude of the net displacement of a point on the rim of the blade during the deceleration?

Answer 1

ωi = 4830rev/min * 2πrad/rev * 1min/60s = 505.6 rad/s
π = (ωf - ωi) / t = -505.6 rad/s / 3s = -168 rad/s²

(1) A number of ways to go about this. Radius = 5 in, so
tangential vi = 505.6rad/s * 5in = 2528 inch/s
and distance = Vavg * t = ½*2528 inch/s * 3s = 3792 inch

(2) Angular displacement Θ = ωavg * t = ½*505.6 rad/s * 3s = 758.4 rads
which divided by 2π is 120.77 revolutions.
Now, the 120 revolutions have a net displacement of zero, leaving 0.77 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 = 7.07 in.