Question

A thin uniform rod of length 'L' and mass 'M' is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is '$\omega$'. Its centre of mass rises to a maximum height of ______.

• A. $\frac{L^2 \omega^2}{2g}$
• B. $\frac{L \omega}{6g}$
• C. $\frac{L \omega}{2g}$
• D. $\frac{L^2 \omega^2}{6g}$

Hint:

For physical pendulums, potential energy $Mgh$ is ALWAYS calculated using the height change of the Center of Mass, not the physical end of the object. For a uniform rod, the COM is located at $L/2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We use the Law of Conservation of Energy. Rotational Kinetic Energy at the bottom is converted into Gravitational Potential Energy at the maximum height.

Step 2: Formula Application:
$K.E._{rot} = \frac{1}{2} I \omega^2$. For a rod rotating about one end, $I = \frac{1}{3} ML^2$. $P.E. = Mgh$, where $h$ is the vertical rise of the centre of mass.

Step 3: Explanation:
$\frac{1}{2} \left( \frac{1}{3} ML^2 \right) \omega^2 = Mgh$ $\frac{1}{6} ML^2 \omega^2 = Mgh$ $h = \frac{L^2 \omega^2}{6g}$.

Step 4: Final Answer:
The maximum height rise is $\frac{L^2 \omega^2}{6g}$.