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Sever21 [200]
3 years ago
9

During the time a compact disc (CD) accelerates from rest to a constant rotational speed of 477 rev/min, it rotates through an a

ngular displacement of 1.5758 rev. What is the angular acceleration of the CD
Physics
1 answer:
atroni [7]3 years ago
8 0

Answer:

<em>126.01 rad/s^2</em>

<em></em>

Explanation:

since it starts from rest, initial angular speed ω' = 0 rad/s

angular speed N = 477 rev/min

angular speed in rad/s ω = \frac{2\pi N}{60} =  \frac{2*3.142* 477}{60} = 49.95 rad/s

angular displacement ∅ = 1.5758 rev

angular displacement in rad/s = 2\pi N = 2 x 3.142 x 1.5758 = 9.9 rad

angular acceleration \alpha = ?

using the equation of angular motion

ω^2 = ω'^2 + 2\alpha∅

imputing values, we have

49.95^{2}  = 0^{2}  + (2 *\alpha*9.9 )

2495 = 19.8\alpha

\alpha = 2495/19.8 = <em>126.01 rad/s^2</em>

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