Question

Mean Kinetic Energy of a Hanging Rod in SHM: A thin uniform rod of mass M and length L performs small oscillations about a horizontal axis passing through its upper end. Find the mean kinetic energy of the rod during its oscillations period if at t=0 it is deflected from vertical by an angle $\theta_0$ and imparted an angular velocity $\omega_0$.

Answer 1

$\frac{1}{12} ML^2 \omega_0^2 + \frac{1}{8} MLg \theta_0^2$

Answer 2

The rod is a uniform thin rod of mass M and length L, pivoted at its upper end. This system acts as a physical pendulum.

The distance of the center of mass (CM) from the pivot is $d = L/2$.

The moment of inertia of the rod about the pivot (its upper end) is $I = I_{CM} + Md^2 = \frac{1}{12}ML^2 + M(\frac{L}{2})^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2$.

For small oscillations, the motion is Simple Harmonic Motion (SHM). The angular frequency of small oscillations is given by $\omega = \sqrt{\frac{Mgd}{I}}$.

Substituting the values for the uniform rod:
$\omega = \sqrt{\frac{Mg(L/2)}{(1/3)ML^2}} = \sqrt{\frac{MgL/2}{ML^2/3}} = \sqrt{\frac{3g}{2L}}$.

The total energy of the rod during oscillation is conserved. At any time $t$, the total energy is the sum of kinetic energy $KE(t)$ and potential energy $PE(t)$.
$E = KE(t) + PE(t)$.

The kinetic energy of rotation is $KE(t) = \frac{1}{2} I \dot{\theta}(t)^2$, where $\dot{\theta}(t)$ is the angular velocity at time $t$.

The potential energy of the rod relative to its equilibrium position ($\theta=0$) is $PE(\theta) = Mgd(1 - \cos\theta)$. For small oscillations, $\cos\theta \approx 1 - \frac{\theta^2}{2}$, so $PE(\theta) \approx Mgd \frac{\theta^2}{2} = \frac{1}{2} (Mg \frac{L}{2}) \theta^2$.

The total energy is constant and can be calculated from the initial conditions at $t=0$, where $\theta(0) = \theta_0$ and $\dot{\theta}(0) = \omega_0$.
$E = KE(0) + PE(0) = \frac{1}{2} I \omega_0^2 + \frac{1}{2} Mg \frac{L}{2} \theta_0^2$.
Substitute $I = \frac{1}{3}ML^2$:
$E = \frac{1}{2} (\frac{1}{3}ML^2) \omega_0^2 + \frac{1}{4} MLg \theta_0^2 = \frac{1}{6} ML^2 \omega_0^2 + \frac{1}{4} MLg \theta_0^2$.

For any system undergoing SHM, the mean kinetic energy over one period is equal to the mean potential energy over one period, and each is equal to half of the total energy.
$\langle KE \rangle = \langle PE \rangle = \frac{1}{2} E$.

Therefore, the mean kinetic energy of the rod during its oscillation period is:
$\langle KE \rangle = \frac{1}{2} E = \frac{1}{2} (\frac{1}{6} ML^2 \omega_0^2 + \frac{1}{4} MLg \theta_0^2)$.
$\langle KE \rangle = \frac{1}{12} ML^2 \omega_0^2 + \frac{1}{8} MLg \theta_0^2$.