Question Detail
Question
A grindstone rotates at constant angular acceleration α = 0.35 rad/s². At time t = 0, it has an angular velocity of ω = 4.6 rad/s and a reference line on it is horizontal, at the angular position θ = 0. At what time after t = 0 is the reference line at the angular position θ = 5.0 rad?
Correct Answer
Option D is the correct answer.
Detailed Explanation
Using θ = ω₀t + ½αt²: 5.0 = 4.6t + ½(0.35)t². 5.0 = 4.6t + 0.175t². Rearranging: 0.175t² + 4.6t - 5.0 = 0. Multiply by 1000: 175t² + 4600t - 5000 = 0. Divide by 25: 7t² + 184t - 200 = 0. Using quadratic formula: t = [-184 ± √(184² + 4×7×200)]/(14) = [-184 ± √(33856 + 5600)]/14 = [-184 ± √39456]/14 = [-184 ± 198.6]/14. Positive root: (14.6)/14 ≈ 1.04 s.
Hint
Use rotational kinematics equation θ = ω₀t + ½αt².