To find the ratio of the professor's final kinetic energy to his initial kinetic energy, you can use the conservation of angular momentum. Angular momentum is conserved when no external torques are acting on a system. In this case, the professor is standing on a freely rotating platform, so we can assume there are no external torques.
The angular momentum (L) of an object can be calculated as:
\(L = I \cdot \omega\)
Where: \(L\) is the angular momentum. \(I\) is the moment of inertia. \(\omega\) is the angular velocity.
The moment of inertia for a system of two dumbbells can be calculated as:
\(I_{\text{initial}} = 2 \cdot (m \cdot r_{\text{initial}}^2)\)
\(I_{\text{final}} = 2 \cdot (m \cdot r_{\text{final}}^2)\)
Where: \(m\) is the mass of each dumbbell. \(r_{\text{initial}}\) is the initial distance from the axis of rotation (when arms are outstretched). \(r_{\text{final}}\) is the final distance from the axis of rotation (when arms are pulled in).
The angular momentum is conserved, so we can set up the following equation:
\(I_{\text{initial}} \cdot \omega_{\text{initial}} = I_{\text{final}} \cdot \omega_{\text{final}}\)
Substitute the given values:
\(2 \cdot (m \cdot r_{\text{initial}}^2) \cdot 1.5 \, \text{rad/sec} = 2 \cdot (m \cdot r_{\text{final}}^2) \cdot 5.0 \, \text{rad/sec}\)
Now, we can cancel out some common factors:
\(r_{\text{initial}}^2 \cdot 1.5 = r_{\text{final}}^2 \cdot 5.0\)
Now, solve for the ratio of \(r_{\text{final}}^2\) to \(r_{\text{initial}}^2\):
\(\frac{r_{\text{final}}^2}{r_{\text{initial}}^2} = \frac{1.5}{5.0} = 0.3\)
Now, we have the ratio of the squares of the final and initial distances from the axis of rotation. To find the ratio of kinetic energies, we can use the fact that kinetic energy (\(K\)) is proportional to the square of angular velocity:
\(K = 0.5 \cdot I \cdot \omega^2\)
So, the ratio of final kinetic energy (\(K_{\text{final}}\)) to initial kinetic energy (\(K_{\text{initial}}\)) is:
\(\frac{K_{\text{final}}}{K_{\text{initial}}} = \frac{0.5 \cdot I_{\text{final}} \cdot \omega_{\text{final}}^2}{0.5 \cdot I_{\text{initial}} \cdot \omega_{\text{initial}}^2}\)
Now, plug in the values:
\(\frac{K_{\text{final}}}{K_{\text{initial}}} = \frac{r_{\text{final}}^2 \cdot 25.0}{r_{\text{initial}}^2 \cdot 2.25}\)
The \(0.5\), \(2\), and \(m\) cancel out:
\(\frac{K_{\text{final}}}{K_{\text{initial}}} = \frac{7.5}{2.25}\)
\(\frac{K_{\text{final}}}{K_{\text{initial}}} = 3.33\) (approximately)
So, the ratio of the professor's final kinetic energy to his initial kinetic energy is approximately 3.33.