Question

A CD has a mass of 16.0 g and a radius of 6.03 cm. When inserted into a player, the CD starts from rest and accelerates to an angular velocity of 19.5 rad/s in 0.685 s. Assuming the CD is a uniform solid disk, determine the net torque acting on it

Answer 1

Answer:

The net torque acting on the CD is $8.281\times 10^{-4}\,N\cdot m$.

Explanation:

According to the D'Alembert's Principle, net torque of a rigid body is equal to the product of the moment of inertia with respect to its center of gravity and angular acceleration. That is:

$\tau = I_{g}\cdot \alpha$

Where:

$\tau$ - Net torque, measured in Newton-meters.

$I_{g}$ - Moment of inertia, measured in kilogram-square meters.

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

The moment of inertia of a solid disk is:

$I_{g} = \frac{1}{2}\cdot m \cdot r^{2}$

Given that $m = 0.016\,kg$ and $0.0603\,m$, the moment of inertia of the CD is:

$I_{g} = \frac{1}{2}\cdot (0.016\,kg)\cdot (0.0603\,m)^{2}$

$I_{g} = 2.909\times 10^{-5}\,kg\cdot m^{2}$

Let suppose that CD accelerates constantly, so that motion equation is the following:

$\omega = \omega_{o} + \alpha \cdot t$

$\alpha = \frac{\omega - \omega_{o}}{t}$

$\alpha = \frac{19.5\,\frac{rad}{s}-0\,\frac{rad}{s} }{0.685\,s}$

$\alpha = 28.467\,\frac{rad}{s^{2}}$

Finally, the net torque acting on the CD is:

$\tau = (2.909\times 10^{-5}\,kg\cdot m^{2})\cdot \left(28.467\,\frac{rad}{s^{2}}\right)$

$\tau = 8.281\times 10^{-4}\,N\cdot m$

The net torque acting on the CD is $8.281\times 10^{-4}\,N\cdot m$.