The rate of rotation of a solid disk with a radius of #1 m# and mass of #5 kg# constantly changes from #16 Hz# to #27 Hz#. If the change in rotational frequency occurs over #12 s#, what torque was applied to the disk?

The torque is the rate of change of angular momentum

The moment of inertia of a solid disc is

The rate of change of angular velocity is

To calculate the torque applied to the disk, we can use the formula:

$\text{Torque} = I \times \alpha$

where $I$ is the moment of inertia of the disk and $\alpha$ is the angular acceleration.

The moment of inertia $I$ of a solid disk is given by the formula:

$I = \frac{1}{2} m r^2$

where $m$ is the mass of the disk and $r$ is the radius.

First, we calculate the initial and final angular velocities using the formula:

$\omega_1 = 2\pi f_1$ $\omega_2 = 2\pi f_2$

where $f_1 = 16 \, \text{Hz}$ and $f_2 = 27 \, \text{Hz}$.

Then, we calculate the initial and final angular accelerations using the formula:

$\alpha = \frac{\omega_2 - \omega_1}{t}$

where $t = 12 \, \text{s}$.

Now, we can plug in the values into the formulas to find the torque:

$\omega_1 = 2\pi \times 16 \, \text{Hz} = 32\pi \, \text{rad/s}$ $\omega_2 = 2\pi \times 27 \, \text{Hz} = 54\pi \, \text{rad/s}$

$\alpha = \frac{54\pi \, \text{rad/s} - 32\pi \, \text{rad/s}}{12 \, \text{s}} = \frac{22\pi \, \text{rad/s}}{12 \, \text{s}} = \frac{11}{6} \pi \, \text{rad/s}^2$

$I = \frac{1}{2} \times 5 \, \text{kg} \times (1 \, \text{m})^2 = \frac{5}{2} \, \text{kg} \cdot \text{m}^2$

$\text{Torque} = \left(\frac{5}{2} \, \text{kg} \cdot \text{m}^2\right) \times \left(\frac{11}{6} \pi \, \text{rad/s}^2\right) = \frac{55}{12} \pi \, \text{N} \cdot \text{m}$

Therefore, the torque applied to the disk is $\frac{55}{12} \pi \, \text{N} \cdot \text{m}$.