Answer:
The angular acceleration of the tire is 0.454 rad/s²
Step-by-step explanation:
Given;
initial velocity, u = 3.4 rev/s = 3.4 rev/s x 2π rad/rev
u = 21.3656 rad/sec
final velocity, v = 5.5 rev/s = 5.5 rev/s x 2π rad/rev
v = 34.562 rad/sec
Calculate the value of angular rotation, θ, of the tire
θ = Number of revolutions x 2π rad/rev
θ =
![(260)/(2 \pi r) *(2 \pi \ rad)/(rev)](https://img.qammunity.org/2021/formulas/physics/college/6jn6knzdl96eun0n02rmf0e3c1yhowtjle.png)
θ = (260 / r)
r is the radius of the tire = 64 / 2 = 32cm = 0.32 m
θ = (260 / 0.32)
θ = 812.5 rad
Apply the following kinematic equation, to determine angular acceleration of the tire;
![v^2 = u^2 + 2 \alpha \theta\\\\2 \alpha \theta = v^2 - u^2\\\\\alpha = (v^2-u^2)/(2 \theta) \\\\\alpha = ((34.562)^2-(21.3656)^2)/(2 (812.5))\\\\\alpha = (738.043)/(1625) \\\\\alpha = 0.454 \ rad/s^2](https://img.qammunity.org/2021/formulas/physics/college/aygchnyy9ptne3vk09wyz348owp3ve7a2m.png)
Therefore, the angular acceleration of the tire is 0.454 rad/s²