A \(24-\mathrm{cm}\) -long pen is tossed up in the air, reaching a maximum height of \(1.2 \mathrm{~m}\) above its release point. On the way up, the pen makes 1.8 revolutions. Treating the pen as a thin uniform rod, calculate the ratio between the rotational kinetic energy and the translational kinetic energy at the instant the pen is released. Assume that the rotational speed does not change during the toss.

To find the velocity of the pen when it is released, we can use the following kinematic equation: \(v^2 = u^2 + 2as\) Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration due to gravity, and \(s\) is the maximum height reached. The final velocity at the maximum height will be 0, and the acceleration due to gravity is -9.81 \(m/s^2\). So, the equation can be written as: \(0 = u^2 - 2(9.81)(1.2)\) Now, we can solve for the initial velocity \(u\): \(u = \sqrt{2(9.81)(1.2)}\approx 4.84 \;\mathrm{m/s}\)

We know that the pen makes 1.8 revolutions on its way up. To find the angular velocity of the pen, we will first convert it to radians. \(1.8\;\mathrm{revolutions} = 1.8\times 2\pi\;\mathrm{radians} = 3.6\pi\;\mathrm{radians}\) Now, we will use the equation for angular velocity: \(\omega = \dfrac{\theta}{t}\) To find the time \(t\), we can use the equation of motion for vertical displacement: \(h = ut - \frac{1}{2}gt^2\) Here, \(h\) is the maximum height reached, \(u\) is the initial-vertical velocity and \(g\) is the acceleration due to gravity. We've already calculated the initial-vertical velocity, \(u\approx 4.84 \;\mathrm{m/s}\)and \(h = 1.2\;\mathrm{m}\), and \(g= 9.81\;\mathrm{m/s^2}\): \(1.2 = 4.84 t - \frac{1}{2}(9.81)t^2\) Solving the quadratic equation for \(t\), we get two possible values, one of which is the time taken to reach the maximum height \(t\approx 0.494\;\mathrm{s}\). Now, we can calculate the angular velocity: \(\omega = \dfrac{3.6\pi}{0.494}\approx 22.9\;\mathrm{rad/s}\)

Now that we have the initial velocity \(u\) and angular velocity \(\omega\), we can calculate the rotational and translational kinetic energies. The translational kinetic energy is given by: \(K_t = \frac{1}{2}mv^2 = \frac{1}{2}m(4.84)^2\) The rotational kinetic energy of a rotating rod is given by: \(K_r = \frac{1}{2}I\omega^2\) Here, \(I\) denotes the moment of inertia of the rod about an axis perpendicular to the rod and passing through its center of mass. For a uniform rod of length \(L\) and mass \(m\), the moment of inertia is given by: \(I = \frac{1}{12}mL^2 = \frac{1}{12}m(0.24)^2\) Now, substituting the value of \(I\) in the equation for rotational kinetic energy: \(K_r = \frac{1}{2}\left(\frac{1}{12}m(0.24)^2\right)(22.9)^2\)

We are now ready to calculate the ratio between the rotational and translational kinetic energies. Notice that the mass \(m\) will cancel out in the ratio: \(\dfrac{K_r}{K_t} = \dfrac{\frac{1}{2}\left(\frac{1}{12}(0.24)^2\right)(22.9)^2}{\frac{1}{2}(4.84)^2} \approx 0.383\) Hence, the ratio between the rotational kinetic energy and the translational kinetic energy at the instant the pen is released is approximately 0.383.