Starting from rest, a dvd steadily acceleration to 580 rpm in 1.0 s, rotates at this angular speed for 3.0 s, then steadily decelerates to a halt in 2.0 s. how many revolutions does it make?

it makes N= 43.493≈ 43 revolutions

Explanation

there are 3 stages

1) steady acceleration from rest to 580 rpm in 1.0 seg

2) constant angular speed for 3.0 s

3) steady acceleration to halt in 2.0 s

knowing that

θ final =θ initial  +ω initial * t + 1/2 α * t²

where θ angle respect with the horizontal axis , ω= angular velocity α= angular acceleration

since ω initial = 0 , θ initial=0 , α =(ω final - ω initial)/t = ω final /t

θ final = 1/2 ω final * t

since f = ω/(2π) and  N=(θ final- θ initial)/(2π) , where f=frequency and N= number of revolutions

N1= 1/2 f final * t

N1= 1/2 * 580 revolutions/minute * 1 s * 1 min/60 sec= 4. 833 rev

for 2) θ final - θ initial = ω initial + ω final * t

θ final = ω final * t

N2= f final * t = 580 revolutions/minute * 3 s * 1 min/60 sec= 29 rev

for 3)

θ final =θ initial  +ω initial * t + 1/2 α * t²

since ω final = 0 ,  α =(ω final - ω initial)/t = -ω initial /t

θ final- θ initial  = ω initial * t + 1/2 α * t² = ω initial * t - 1/2 ω initial * t  = 1/2 ω initial * t

θ final- θ initial  = 1/2 ω initial * t

therefore

N3= 1/2 f initial * t = 1/2 * 580 revolutions/minute * 2 s * 1 min/60 sec= 9.66 rev

therefore

N total = N1 +N2 +N3 =  4. 833 rev + 29 rev + 9.66 rev =43.493 revolutions

higher potential

explanation:

because it is in a high speed so it must move into a of high potential