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 0.250 rev. what is the angular acceleration of the cd? 358 rad/s² 126 rad/s² 866 rad/s² 901 rad/s² 794 rad/s²

Explanation:

Step1: Convert final angular velocity to rad/s

The final rotational speed is \( \omega = 477\ \text{rev/min} \). To convert to rad/s, we use the conversion factors: \( 1\ \text{rev} = 2\pi\ \text{rad} \) and \( 1\ \text{min} = 60\ \text{s} \). So, \( \omega = 477\times\frac{2\pi}{60}\ \text{rad/s} \approx 49.9\ \text{rad/s} \). The initial angular velocity \( \omega_0 = 0\ \text{rad/s} \) (starts from rest). The angular displacement \( \theta = 0.250\ \text{rev} \), converting to rad: \( \theta = 0.250\times 2\pi\ \text{rad} \approx 1.571\ \text{rad} \).

Step2: Use rotational kinematic equation

The rotational kinematic equation is \( \omega^2 = \omega_0^2 + 2\alpha\theta \). Since \( \omega_0 = 0 \), the equation simplifies to \( \omega^2 = 2\alpha\theta \). We can solve for \( \alpha \): \( \alpha = \frac{\omega^2}{2\theta} \).

Step3: Substitute values

Substitute \( \omega \approx 49.9\ \text{rad/s} \) and \( \theta \approx 1.571\ \text{rad} \) into the formula: \( \alpha = \frac{(49.9)^2}{2\times 1.571} \approx \frac{2490.01}{3.142} \approx 792\ \text{rad/s}^2 \), which is approximately \( 794\ \text{rad/s}^2 \) (due to rounding differences in intermediate steps).

Answer:

794 rad/s² (the option with 794 rad/s²)