Question

When the play button is pressed, a CD accelerates uniformly from rest to 430 rev/min in 4.0 revolutions. If the CD has a radius of 7.0 cm and a mass of 17 g , what is the torque exerted on it?

Answer 1

The net torque exerted on CD is
`1.680 * 10^(-3)\,N\cdot m`.

Step-by-step explanation:

As CD is acceleration uniformly, the following equation of motion can be used to determine the angular acceleration:

`\dot n^(2) = \dot n_(o)^(2) + 2\cdot \ddot n \cdot \Delta n`

Where:

`\dot n_(o)` - Initial angular speed, measured in revolutions per minute.

`\dot n` - Final angular speed, measured in revolutions per minute.

`\ddot n` - Angular acceleration, measured in revolution per square minute.

`\Delta n` - Change in angular position, measured in revolutions.

The angular acceleration is cleared and calculated:

`\ddot n = (\dot n^(2)-\dot n_(o)^(2))/(2\cdot \Delta n)`

Given that
`\dot n_(o) = 0\,(rev)/(min)`,
`\dot n = 430\,(rev)/(min)` and
`\Delta n = 4\, rev`, the angular acceleration is:

`\ddot n = (\left(430\,(rev)/(min) \right)^(2)-\left(0\,(rev)/(min) \right)^(2))/(2\cdot (4\,rev))`

`\ddot n = 23112.5\,(rev)/(min^(2))`

The angular accelaration measured in radians per square second is:

`\alpha = \left(23112.5\,(rev)/(min^(2)) \right)\cdot \left(2\pi\,(rad)/(rev)\right)\cdot \left((1)/(3600)\,(min^(2))/(s^(2)) \right)`

`\alpha \approx 40.339\,(rad)/(s^(2))`

Net torque experimented by the CD during its accleration is equal to the product of its moment of inertia with respect to its axis of rotation and angular acceleration:

`\tau = I \cdot \alpha`

Where:

`I` - Moment of inertia, measured in
`kg \cdot m^(2)`.

`\alpha` - Angular acceleration, measured in radians per square second.

In addition, a CD has a form of a uniform disk, whose moment of inertia is:

`I = (1)/(2)\cdot m \cdot r^(2)`

Where:

`m` - Mass of the CD, measured in kilograms.

`r` - Radius of the CD, measured in meters.

If
`m = 0.017\,kg` and
`r = 0.07\,m`, then:

`I = (1)/(2)\cdot (0.017\,kg)\cdot (0.07\,m)^(2)`

`I = 4.165* 10^(-5)\,kg\cdot m^(2)`

Now, the net torque exerted on CD is:

`\tau = (4.165* 10^(-5)\,kg\cdot m^(2))\cdot \left(40.339\,(rad)/(s^(2)) \right)`

`\tau = 1.680* 10^(-3)\,N\cdot m`

The net torque exerted on CD is
`1.680 * 10^(-3)\,N\cdot m`.