Answer :
Answer:
The angular acceleration of the tire is 0.454 rad/s²
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 = [tex]\frac{260}{2 \pi r} *\frac{2 \pi \ rad}{rev}[/tex]
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;
[tex]v^2 = u^2 + 2 \alpha \theta\\\\2 \alpha \theta = v^2 - u^2\\\\\alpha = \frac{v^2-u^2}{2 \theta} \\\\\alpha = \frac{(34.562)^2-(21.3656)^2}{2 (812.5)}\\\\\alpha = \frac{738.043}{1625} \\\\\alpha = 0.454 \ rad/s^2[/tex]
Therefore, the angular acceleration of the tire is 0.454 rad/s²