Question #246135

A bicycle tire is spinning counterclockwise at 2.60 rad/s. During a time period Δt = 1.75 s, the tire is stopped and spun in the opposite (clockwise) direction, also at 2.60 rad/s. Calculate the change in the tire's angular velocity Δ𝜔 and the tire's average angular acceleration 𝛼av. (Indicate the direction with the signs of your answers.)


(a) the change in the tire's angular velocity Δ𝜔 (in rad/s)

 ___rad/s


(b) the tire's average angular acceleration 𝛼av (in rad/s2)

___rad/s2

1
Expert's answer
2021-10-03T13:21:58-0400

(a) The change in angular velosity is,

Δω=ωfωi=(2.60  rad/s)(2.60  rad/s)=5.20  rad/sΔω = ω_f-ω_i \\ = (2.60 \;rad/s) -(-2.60 \;rad/s) \\ = 5.20 \;rad/s

Here, ω_i is negative because it is in the clockwise direction.

(b) The average angular acceleration is,

αav=ΔωΔt=5.20  rad/s1.75  s=2.97  rad/s2α_{av} = \frac{Δω}{Δt} \\ = \frac{5.20 \;rad/s}{1.75 \;s} \\ = 2.97 \;rad/s^2


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