Question

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 1.5758 rev. What is the angular acceleration of the CD

Answer 1

Answer:

126.01 rad/s^2

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 = 126.01 rad/s^2