Answer:
The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].
Explanation:
As CD is acceleration uniformly, the following equation of motion can be used to determine the angular acceleration:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot \Delta n[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]\ddot n[/tex] - Angular acceleration, measured in revolution per square minute.
[tex]\Delta n[/tex] - Change in angular position, measured in revolutions.
The angular acceleration is cleared and calculated:
[tex]\ddot n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \Delta n}[/tex]
Given that [tex]\dot n_{o} = 0\,\frac{rev}{min}[/tex], [tex]\dot n = 430\,\frac{rev}{min}[/tex] and [tex]\Delta n = 4\, rev[/tex], the angular acceleration is:
[tex]\ddot n = \frac{\left(430\,\frac{rev}{min} \right)^{2}-\left(0\,\frac{rev}{min} \right)^{2}}{2\cdot (4\,rev)}[/tex]
[tex]\ddot n = 23112.5\,\frac{rev}{min^{2}}[/tex]
The angular accelaration measured in radians per square second is:
[tex]\alpha = \left(23112.5\,\frac{rev}{min^{2}} \right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}} \right)[/tex]
[tex]\alpha \approx 40.339\,\frac{rad}{s^{2}}[/tex]
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:
[tex]\tau = I \cdot \alpha[/tex]
Where:
[tex]I[/tex] - Moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
In addition, a CD has a form of a uniform disk, whose moment of inertia is:
[tex]I = \frac{1}{2}\cdot m \cdot r^{2}[/tex]
Where:
[tex]m[/tex] - Mass of the CD, measured in kilograms.
[tex]r[/tex] - Radius of the CD, measured in meters.
If [tex]m = 0.017\,kg[/tex] and [tex]r = 0.07\,m[/tex], then:
[tex]I = \frac{1}{2}\cdot (0.017\,kg)\cdot (0.07\,m)^{2}[/tex]
[tex]I = 4.165\times 10^{-5}\,kg\cdot m^{2}[/tex]
Now, the net torque exerted on CD is:
[tex]\tau = (4.165\times 10^{-5}\,kg\cdot m^{2})\cdot \left(40.339\,\frac{rad}{s^{2}} \right)[/tex]
[tex]\tau = 1.680\times 10^{-3}\,N\cdot m[/tex]
The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].
