Question

How much work is done by the motor in a CD player to make a CD spin, starting from rest? The CD has a diameter of 12.70 cm and a mass of 16.30 g. The laser scans at a constant tangential velocity of 1.150 m/s. Assume that the music is first detected at a radius of 20.90 mm from the center of the disk. Ignore the small circular hole at the CD's center.

Answer 1

The work done by the motor in a CD player to make the CD spin can be calculated using the formula: Work = Torque * Angle. The final angular velocity can be calculated using the formula: v = r * ω. The moment of inertia can be calculated using the formula: Moment of Inertia = (1/2) * Mass * Radius^2.

Step-by-step explanation:

The work done by the motor in a CD player to make the CD spin can be calculated using the formula:

Work = Torque * Angle

In this case, since the CD starts from rest, the initial angular velocity is zero. The final angular velocity can be calculated using the formula:

v = r * ω

where r is the radius of the CD and ω is the angular velocity.

The torque can be calculated using the formula:

Torque = Moment of Inertia * Angular Acceleration

Since we have the mass and diameter of the CD, the moment of inertia can be calculated using the formula:

Moment of Inertia = (1/2) * Mass * Radius^2

Once we have the torque and the angle, we can calculate the work done by the motor.

Answer 2

Step-by-step explanation:

Given

Diameter of CD
`d=12.70\ cm`

radius
`r=6.35\ cm`

mass of CD
`m=16.30\ gm`

Tangential velocity
`v_t=1.150\ m/s`

music detected at
`r'=20.90\ mm=2.090\ cm`

Moment of Inertia of disc
`I=(mr^2)/(2)`

`I=(16.30* 10^(-3)* (6.35* 10^(-2))^2)/(2)`

`I=6.572* 10^(-5) kg-m^2`

Work done is equal to change in kinetic Energy of CD

`W=\Delta K`

`W=(1)/(2)I\omega_f^2-(1)/(2)I\omega_i^2`

where
`\omega =`angular velocity

`v_t=\omega * r`

`W=(1)/(2)* 6.572* 10^(-5)\left ( \left ( (1.15)/(0.0209)\right )^2-0\right )`

`W=0.0994\ J`