Question

When a carpenter shuts off his circular saw, the 10.0-inch-diameter blade slows from 4865 rpm to zero in 2.75 s .

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

Answer 1

here,

the diameter of saw, d = 10 inch

radius , r = d/2 = 5 inch = 0.127 m

the initial angular speed , w0 = 4865 rpm = 509.2 rad/s

the final angular speed , w = 0 rad/s

the time take , t = 2.75 s

let the angular acceleration be alpha

using first equation of motion

w = w0 + alpha * t

0 = 509.2 + alpha * 2.75

alpha = - 185.2 rad/s^2

the total angle covered , theta = w0 * t + 0.5 * alpha * t^2

theta = 509.2 * 2.75 - 0.5 * 185.2 * 2.75^2 rad

theta = 700 rad

the number of revolution , N = theta/2pi = 111.5 rev

the net angle , phi = 0.5 * 2pi = 3.14 rad

the magnitude of the net displacement of a point on the rim of the blade during the deceleration , s = r *2

s = d = 10 inches