During the time a compact disc (CD) accelerates from rest to a constant rotational speed of 477 rev/min, it rotates through an angular displacement of 0.250 rev. What is the angular acceleration of the CD?
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The angular acceleration can be found using the equation of motion for rotational motion:
ω^2 = ω0^2 + 2αθ
where ω is the final angular speed, ω0 is the initial angular speed, α is the angular acceleration, and θ is the angular displacement.
In this case, the CD starts from rest, so ω0 = 0. The final angular speed is 477 rev/min, which needs to be converted to rad/s (since 1 rev = 2π rad and 1 min = 60 s):
ω = 477 rev/min * 2π rad/rev * 1 min/60 s = 50 rad/s
The angular displacement is 0.250 rev, which also needs to be converted to rad:
θ = 0.250 rev * 2π rad/rev = 0.5π rad
Substituting these values into the equation of motion gives:
(50 rad/s)^2 = 0 + 2α * 0.5π rad
Solving for α gives:
α = (50 rad/s)^2 / (2 * 0.5π rad) = 2500 rad^2/s^2 / π rad = 795.77 rad/s^2
So the angular acceleration of the CD is approximately 796 rad/s^2.
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