Question

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4680rpm to zero in 1.25s.

A.) what is the angular acceleration of the blade?

B) what is the distance traveled by a point on the tim of the blade during the acceleration? (The answer will be in feet)

C.) what is the magnitude of the net displacement of a point on the rim of the blade during deceleration? (Answer will be in inches)

Answer 1

A. Acceleration = change in velocity / time = 4680 / 1.25

a = 3744 revolution / ms

B. Acceleration is uniform that's why average velocity during this period = ( vf - vi) / 2 = 3744 rpm

Distance travelled during that period = velocity * time

Time in minutes = 1.25 / 60 = 0.0208 min

d = 3744 * 0.0208 = 77.87 revs

Distance in 1 rev = pie * d = 3.14 * 10= 31.4inches

In 77.87 rev = 31. 4 * 77.87 = 2445 inches

C. Net displacement is how the final location is different from the original. The last revolution is 0.9 Complete.

So the net displacement was - 0.1 * distance of the full revolution.

It stopped = - 0.1 * 31.4 inches = - 3.14 inches from the starting location.