3) A bicycle wheel has an initial angular velocity of 1.50rad/s. If its angular acceleration is constant and equal to 0.200rad/s^2 what is its angular velocity after it has turned through 3.50 revolutions. 3.32rad/s

Here's how to solve this problem:<br /><br />**1. Convert revolutions to radians:**<br /><br />* One revolution is equal to 2π radians.<br />* 3.50 revolutions * 2π radians/revolution = 7π radians<br /><br />**2. Use the following rotational kinematics equation:**<br /><br />ω_f² = ω_i² + 2αθ<br /><br />Where:<br /><br />* ω_f is the final angular velocity (what we want to find)<br />* ω_i is the initial angular velocity (1.50 rad/s)<br />* α is the angular acceleration (0.200 rad/s²)<br />* θ is the angular displacement (7π radians)<br /><br />**3. Substitute the known values and solve for ω_f:**<br /><br />ω_f² = (1.50 rad/s)² + 2 * (0.200 rad/s²) * (7π radians)<br />ω_f² = 2.25 rad²/s² + 8.796 rad²/s²<br />ω_f² = 11.046 rad²/s²<br />ω_f = √(11.046 rad²/s²)<br />ω_f ≈ 3.32 rad/s<br /><br />Therefore, the final angular velocity after the bicycle wheel has turned through 3.50 revolutions is approximately $\boxed{3.32 \, rad/s}$. The provided answer is correct.<br />