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? express your answer in radians per second squared

The solution for this problem is:

500 revolution per minute = 8.33rev /s = 2π*8.33 rad /s = 52.36 rad /s 

Angular velocity ω = 2π N

Angular acceleration α= (ω2 - ω1) /t

ω2 = 0

α = - ω1/t = -2π N /t

N = 500 rpm = 8.33 r p s.

α = -2π 8.33 /2.6 =- 20 rad/s^2

overcharge is the right answer system

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//