When a carpenter shuts off his circular saw, the 10.0-inch- diameter blade slows from 3984 rpm to 0 in 1.25 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?

a) -333.8 rad/s² . b) 33.1 m c) 10.0 inch = 0.254 m

Explanation:

a) We can apply the definition of angular acceleration, as the rate of change of angular velocity over a time interval, as follows:

aθ = ωf - ωi / t

As ωf=0, we get the following value for aθ:

aθ = - ωi /t

Now, it is advisable to convert the ωi units from rpm to rad/sec, as follows:

3984 rev/min = 3984 rev. (2π /rev) . (60 s/1 min) = 417.2 rad/s

Replacing in the expression for aθ, we get:

aθ = -333.8 rad /s²

b) Assuming a constant angular acceleration, we can find the total angle travelled, Δθ, as follows:

ωf² - ωi² = 2 aθ Δθ ⇒ Δθ = -ωi² / 2 aθ = -(417.2)² /2 (-333.8) = 260.71 rad

By definition, an angle is the relationship between the arc subtended and the radius, so we can write the following expression:

Δs = r Δθ ⇒ Δs = 0.127 m. 260. 71 rad = 33.1 m

c) In order to get the net displacement, we need to know how many revolutions are equivalent to the distance travelled, in terms of circumferences, so we can write the following:

Nbr. of circumferences: 33.1 m / 2π r = 33.1 m / 2 π. 0.127 m = 41.5

So, taking any point on the rim of the blade, it travelled an entire number of circumferences during deceleration, plus half a circumference, which is equal to one diameter in net displacement.

So, the net displacement is equal to one diameter, i.e., 10 inch or 0.254 m.

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