A bicycle tire is spinning counterclockwise at 3.00 rad/s. During a time period Δt = 2.30 s, the tire is stopped and spun in the opposite (clockwise) direction, also at 3.00 rad/s. Calculate the change in the tire's angular velocity Δω and the tire's averageangular acceleration α. (Indicate the direction with the signs of your answers.)
1 answer:
Answer:
Δω = -6.00 rad/s
α = -2.61 m/s²
Explanation:
Step 1: Data given
A bicycle tire is spinning counterclockwise at 3.00 rad/s
Δt = 2.30 s
In theopposite (clockwise) direction, also at 3.00 rad/s
Step 2: Calculate the change in the tire's angular velocity Δω
Δω = ωf - ωi
ωf = - 3.00 rad/s
ωi = 3.00 rad/s
Δω = ωf - ωi = -3.00 - 3.00 = -6.00 rad/s
Step 3: Calculate the tire's average angular acceleration α
α = Δω / ΔT
α = -6.00 rad/s /2.30s
α = -2.61 m/s²
A negative angular acceleration means a decreasing angular velocity
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