Answer to Question #246135 in Molecular Physics | Thermodynamics for Kelani

Question

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 𝛼_av. (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 𝛼_av (in rad/s²)

___rad/s²

Answer 1

(a) The change in angular velosity is,

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

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

(b) The average angular acceleration is,

$α_{av} = \frac{Δω}{Δt} \\ = \frac{5.20 \;rad/s}{1.75 \;s} \\ = 2.97 \;rad/s^2$