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3 years ago

12

If the CD rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angu

lar 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?

2 answers:

gavmur [86]3 years ago

8 0

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

Tems11 [23]3 years ago

8 0

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

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