When the play button is pressed, a CD accelerates uniformly from rest to 495 rev/min in 3.1 revolutions. If the CD has a radius of 6 cm, and a mass of 17 g, what is the torque exerted on it?
To find the torque exerted on the CD, we need to use the formula:
Torque (τ) = Moment of Inertia (I) × Angular Acceleration (α)
The moment of inertia of a disc can be calculated using the formula:
Moment of Inertia (I) = (1/2) × Mass (m) × Radius^2
First, let's convert the mass from grams to kilograms since the SI unit is kg:
Mass (m) = 17 g = 0.017 kg
Next, we can calculate the moment of inertia of the CD:
I = (1/2) × 0.017 kg × (0.06 m)^2
I = 5.1 × 10^-5 kg m^2
Now, we need to find the angular acceleration (α) of the CD. We can use the following formula:
Angular Acceleration (α) = (Final Angular Velocity - Initial Angular Velocity) / Time
The final angular velocity is given as 495 rev/min. To convert this to radians per second, we use the conversion factor:
1 revolution = 2π radians
So, the final angular velocity in radians per second is:
Final Angular Velocity = (495 rev/min) × (2π rad/rev) × (1 min/60 s)
Final Angular Velocity = 51.840 rad/s
The initial angular velocity is 0 since the CD starts from rest.
We also have the information that the CD accelerates uniformly in 3.1 revolutions. This gives us the time it takes for the CD to reach the final angular velocity:
Time = (3.1 revolutions) / (Final Angular Velocity)
Time = (3.1 rev) / (51.840 rad/s)
Time ≈ 0.060 s
Now that we have all the necessary values, we can calculate the torque:
Torque (τ) = 5.1 × 10^-5 kg m^2 × (51.840 rad/s) / 0.060 s
Torque ≈ 4.42 N m
Therefore, the torque exerted on the CD is approximately 4.42 N m.