Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Step-by-step explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:
![2\alpha \theta = \omega_f^2 -\omega_I^2](https://img.qammunity.org/2022/formulas/physics/college/4xexll2dqfjkphl9p853n3bm5ipzzej8uh.png)
where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,
![2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = √(77.65\ rad/s^2)](https://img.qammunity.org/2022/formulas/physics/college/zwp9fc6l5hnznkevuwfprayz2j4uzzr7y1.png)
ωf = 8.8 rad/s
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
v = 2.2 m/s