Answer to Question #248192 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

Solution;

Taking counterclockwise as +ve direction;

Initial velocity,$w_i$ =+2.60$rad/s$

Final velocity,$w_f$ =-2.60 $rad/s$

$\Delta t=1.75s$

(a)change in the tire's angular velocity ;

$\Delta w=(w_f-w_i)$

$\Delta w=-2.60-2.60=-5.2rad/s$

(b)tire's average angular acceleration;

$\alpha_{av}=\frac{\Delta w}{\Delta t}=\frac{-5.20rad/s}{1.75s}$

$\alpha_{av}=2.97rad/s^2$