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A

Anonymous

Feb 14, 2025

When a carpenter shuts off his circular saw, the 10.0-inch diameter blade slows from 4310rpm to zero in 3.00s?

When a carpenter shuts off his circular saw, the 10.0-inch diameter blade slows from 4310 RPM to zero in 3.00 seconds.

(a) What is the angular acceleration of the blade?

(b) What is the distance traveled by a point on the rim of the blade during the deceleration, in feet?

(c) What is the magnitude of the net displacement of a point on the rim of the blade during the deceleration, in inches?

I have been using rotational kinematics equations, but the answers for the last two questions keep coming back incorrect. Any assistance would be appreciated.

1 Answers

A
Adeline Price

Feb 19, 2025

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.

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