Question

If the CD rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, what is α, the magnitude of the angular acceleration of the CD, as it spins to a stop?

Answer 1

Answer:

Explanation:

Given that,

Initial Angular velocity w=500rpm

Converting from rpm to rad/s

1rev =2πrad

1minutes =60secs

500rpm=500rev/mins

w = 500×2π/60

wi=52.36rad/s

The final angular velocity wf=0rad/s

Time to stop is t=2.6sec

We want to find angular acceleration α

Using the equation of angular motion

wf = wi + αt.

0 = 52.36 + 2.6α

-52.36=2.6α

α = -52.36/2.6

α = -20.14rad/s²

The angular acceleration is negative because it is decelerating.

Then, α=20.14rad/s²

Answer 2

Answer:

Explanation:

Given:

Initial angular velocity, wi = 500 rpm

Converting from rom to rad/s,

500 rev/min × 2pi rad/1 rev × 1 min/60s

= 52.36 rad/s

Final angular velocity, wf = 0 rad/s

Time, t = 2.6 s

Using equations of angular motion,

wf = wi + αt

α = (52.36 - 0)/2.6

= 20.138 rad/s^2

= 20.14 rad/s^2