Question

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?

Answer 1

The DVD makes a total of 29 revolutions.

Step-by-step explanation:

To solve this problem, we need to break it into three parts: the acceleration phase, the constant angular speed phase, and the deceleration phase.

In the acceleration phase, the DVD starts from rest and accelerates to an angular speed of 580 rpm in 1.0 s. To find the angular acceleration, we can use the formula α = (w_f - w_i) / t, where α is the angular acceleration, w_f is the final angular speed, w_i is the initial angular speed, and t is the time. Plugging the values into the equation, we get α = (580 rpm - 0 rpm) / 1.0 s = 580 rpm/s.

In the constant angular speed phase, the DVD rotates at 580 rpm for 3.0 s. To find the number of revolutions in this phase, we can use the formula n = w * t / 60, where n is the number of revolutions, w is the angular speed in rpm, and t is the time in seconds.

Plugging the values into the equation, we get n = 580 rpm * 3.0 s / 60 = 29 revolutions.

In the deceleration phase, the DVD steadily decelerates to a halt in 2.0 s.

Since the DVD comes to a halt, its final angular speed is 0 rpm. Using the same formula as in the acceleration phase, we can find the angular acceleration. Plugging the values into the equation, we get 0 rpm - 580 rpm / 2.0 s = -290 rpm/s.

Therefore, the total number of revolutions the DVD makes is the sum of the revolutions in the acceleration phase, constant angular speed phase, and deceleration phase, which is 0 revolutions + 29 revolutions + 0 revolutions = 29 revolutions.

Answer 2

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