Question

When a carpenter shuts off his circular saw, the 10.0-inch-diameter blade slows from 4217 rpm to zero in 1.75 s .

What is the angular acceleration of the blade?

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

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

Answer 1

ωi = 4217 rpm = 4217 *2pi /60 = 441.603207 rad/sec

ωf = 0 , ∝ = ? & t= 1.75

1)

ωf =  ωi + ∝*t

∝ = - ωi /t = - 441.603207 /1.75 = 252.344 rad/s²  

angular acceleration of the blade = - 252.344 rad/s² answer

2)

θ =  ωi^2 /( 2*∝) =441.603207^2 / ( 2*252.344)

=386.40386225 rad

no of revolution = = 386.40386225 / (2*pi ) = 61.4980847

radius = 5 inch

Total Distance = 2*pi*r* n = 2*pi* 5* 61.4980847

= 1932.01931 inch =161.001609166667 feet answer

3)

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

total displacement =d

d =  √2* r =  √2* 5 = 7.07106781 inch answer

Goodluck for exams Comment in case any doubt, will reply for sure..