Question

When a carpenter shuts off his circular saw, the 10.0-inch-diameter blade slows from 4488 rpm to zero in 1.25 s .

Part A What is the angular acceleration of the blade? (Express your answer to three significant figures.)

(answer in rad/s2)

Part B What is the distance traveled by a point on the rim of the blade during the deceleration?(Express your answer to three significant figures.)

(answer in ft.)

Part C What is the magnitude of the net displacement of a point on the rim of the blade during the deceleration?(Express your answer to three significant figures.)

(answer in in.)

Answer 1

here,

diameter , d = 10 inch

radius , r = d/2 = 5 in

w0 = 4488 rpm = 469.7 rad/s

time taken , t = 1.25 s

a)

let the angular accelration of the blade be alpha

w = w0 +alpha * t

0 = 469.7 + alpha * 1.25

solving for alpha

alpha = 375.8 rad/s^2

b)

theta = w^2 /(2*alpha)

theta = 469.7^2 /( 2 * 375.8)

theta = 293.5 rad

the distance travelled , s = r * theta = 1467.7 in = 122.4 ft

c)

the net angle displaced , theta' = 4.62 rad

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

s' = 5 * 4.62 in = 23.1 in