Question

9. A bicycle tire is spinning clockwise at 3.80 rad/s. During a time period of t = 2.35 s, the tire is stopped and spun in the opposite (counterclockwise) direction, also at 3.80 rad/s. Calculate the tire's average angular acceleration. a. 7.63 rad/s2 b. 5.43 rad/s2 c. 2.53 rad/s2 d. 4.03 rad/s2 e. 3.43 rad/s2

Answer 1

Answer:

option E

Explanation:

given,

angular speed of bicycle tire ()= 3.80 rad/s

time period = 2.35 s

tire stops and spun in opposite direction with angular velocity =  3.80 rad/s

average angular acceleration

initial angular speed = ω₀ = - 3.80 rad/s

final angular speed = ω₁ = 3.80 rad/s

change in angular velocity

$\Delta \omega = \omega - \omega_0$

$\Delta \omega = 3.80 - (-3.80)$

$\Delta \omega = 7.6\ rad/s$

average angular acceleration

$\alpha = \dfrac{\Delta \omega}{t}$

$\alpha = \dfrac{7.6}{2.35}$

$\alpha = 3.23\ rad/s^2$

approximately equal to 3.4 rad/s²

the correct answer is option E