Question Detail
Question
A grindstone rotates at constant angular acceleration α = 0.35 rad/s². At t = 0, angular velocity ω0 = 4.6 rad/s and angular position θ0 = 0. At what time after t = 0 is the reference line at θ = 5.0 rev?
Correct Answer
Option B is the correct answer.
Detailed Explanation
For rotational motion with constant angular acceleration, the angular position is given by:
theta = theta₀ + ω₀t + ½αt²
Step 1: Convert revolutions to radians
1 revolution = 2π radians
theta = 5.0 × 2π
theta = 10π ≈ 31.42 rad
Step 2: Substitute the given values
theta₀ = 0
ω₀ = 4.6 rad/s
α = 0.35 rad/s²
31.42 = 4.6t + ½(0.35)t²
31.42 = 4.6t + 0.175t²
Step 3: Rearrange into quadratic form
0.175t² + 4.6t − 31.42 = 0
Step 4: Solve using the quadratic formula
t = [-4.6 ± √(4.6² − 4(0.175)(−31.42))] / (2 × 0.175)
4.6² = 21.16
4 × 0.175 × 31.42 = 21.99
√(21.16 + 21.99) = √43.15 ≈ 6.57
t = (-4.6 + 6.57) / 0.35
t = 1.97 / 0.35
t ≈ 5.63 s
(The negative root is discarded because time cannot be negative.)
Final Answer:
t ≈ 5.63 seconds.
Hint
Convert revolutions to radians and use rotational kinematic equations.