Question

When a carpenter shuts off his circular saw, the 10.0-inch diameter blade slows from 4452 rpm to zero in 2.50 s .

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?

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

Answer 1

A.)

From first rotational kinematics law,

wf - wi = *t

here, wf = final angular speed = 0

wi = initial angular speed = 4452 rpm= 4452*(1/60) = 74.2 rev/s

t = time = 2.50 s

= angular acceleration = (wf - wi)/t

= (0 - 74.2)/2.50

= -29.68 rev/s^2

B.)

From second rotational kinematics law,

= wi*t + 0.5**t^2

using known values:

= 74.2*2.50 + 0.5*(-29.68)*2.50^2

= 92.75 rev = 92.75*2*pi rad

Since, total distance(S) = r*

here, r = radius of saw = 10.0/2 inch = 5.00 inch = 5.00/12 ft

So, S = (5/12)*(92.75*2*pi)

S = 232.38 ft

C.)

Since blades makes total 92.75 revolutions.

So, after 92 revolution, it will come back to the same initial point.

then, the net angular displacement of a point = 0.75 rev

net displacement(d) = sqrt(R^2 + R^2)

d = R*sqrt(2) = 5.00*sqrt(2) inch

d = 7.07 inch