Question

Blue-ray disc accelerates from rest to a constant rotational speed of 466 rpm, it rotates through an angular displacement of 0.250 rev. What is the angular acceleration of the CD?

Answer 1

`\alpha =434312\ rev/s^2`

Step-by-step explanation:

Given that,

Initial angular speed,
`\omega_i=0`

Final angular speed,
`\omega_f=466\ rpm`

Angular displacement,
`\theta=0.25\ rev`

We need to find the angular acceleration of the CD. Using third equation of rotational kinematics as follows :

`\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =(\omega_f^2-\omega_i^2)/(2\theta)\\\\\alpha =(466^2-0^2)/(2* 0.25)\\\\\alpha =434312\ rev/s^2`

So, the angular acceleration of the CD is equal to
`434312\ rev/s^2`.