Question

As you finish listening to your favorite compact disc (cd), the cd in the player slows down to a stop. assume that the cd spins down with a constant angular acceleration. part a 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

The angular acceleration of the CD as it spins to a stop is 20.14 rad/s², and it goes through approximately 10.72 revolutions during the process.

Step-by-step explanation:

To calculate the angular acceleration (α) of the compact disc (CD) as it spins to a stop, we can use the kinematic equation for rotational motion, which is similar to the linear kinematic equations. First, we need to convert the initial angular speed from revolutions per minute (rpm) to radians per second (rad/s).

The initial angular speed (ω_i) is 500 rpm, which can be converted by multiplying by 2π rad/rev and dividing by 60 s/min,

ω_i = (500 rev/min) × π/30 rad/rev ≈ 52.36 rad/s.

Since the final angular speed (ω_f) is 0 (because the CD comes to a stop), and the time (t) taken to stop is 2.60 s, we can use the angular kinematic equation:

ω_f = ω_i + αt

Now solve for the angular acceleration α:

0 = 52.36 rad/s + (α)(2.60 s)

α = -52.36 rad/s / 2.60 s ≈ -20.14 rad/s²

The magnitude of α is simply the absolute value, which is | α | = 20.14 rad/s².

To find the number of revolutions the CD goes through as it stops, we can use another kinematic equation:

θ = ω_it + 0.5αt²

Plugging in our values:

θ = (52.36 rad/s)(2.60 s) + 0.5(-20.14 rad/s²)(2.60 s)² = 135.736 - 68.3712 = 67.3648 rad.

To convert radians to revolutions, divide by 2π rad/rev:

67.3648 rad / (2π rad/rev) ≈ 10.72 revolutions.

Answer 2

Remember that the angular acceleration is the rate at which the angular velocity changes. Given that the angular acceleration is constant then we have tho following formula relating the angular acceleration
`\alpha` and the angular velocity
`\omega`:


`\alpha =(\omega_f-\omega_i)/(\Delta t)`

Where
`\omega_f` is the final angular velocity and
`\omega_i` is the initial angular velocity.
`\Delta t` is the time the rotation took from the initial angular velocity to reach the final angular velocity.
Let's convert
`\omega_i` to rotations per second.
If we have 500 rotations per minute and a minute has 60 seconds we can apply a simple rule of three to get the answer:


`500 rotations\to60sec\\?rotation\to1sec`
We get the following for
`\omega_i`.

`\omega_i=(50)/(6)`
Finally the angular acceleration is:

`image`
The result is a negative value but this is expected as the rotating motion is deceletating. In the case of an increasing angular velocity we would have a positive angular acceleration.