Question 485: Centre of Mass and Rotational Motion

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
( $\mathrm{g}=$ acceleration due gravity)

Accepted Answer

Step 1: Concept
The rod swings like a physical pendulum about one end.
When its angular speed is
maximum
, it must be passing through the lowest vertical position (since at the lowest position, kinetic energy is max and potential energy min).
At the highest point, angular speed becomes zero (purely potential energy).
Thus, energy conservation applies:
$ 	ext{Maximum rotational KE} = 	ext{Gain in gravitational PE of CM}. $
Step 2: Energy expressions
For a rod of mass $M$, length $L$, pivot at one end:
Moment of inertia about one end is
$ I = \frac{1}{3} M L^2 . $
Rotational KE when angular speed = $\omega$:
$ K = \frac{1}{2} I \omega^2 = \frac{1}{2}\cdot \frac{1}{3}ML^2 \omega^2 = \frac{1}{6} M L^2 \omega^2 . $
Rise in potential energy of centre of mass:
CM is initially below pivot by distance $L/2$.
If it rises a vertical height $h$, gravitational potential energy increases by:
$ \Delta U = M g h . $
Step 3: Apply conservation of energy
$ \frac{1}{6} M L^2 \omega^2 = M g h . $
$ h = \frac{L^2 \omega^2}{6 g}. $
Step 4: Final Answer
$ \boxed{	frac{L^2 \omega^2}{6 g}} $
That corresponds to
Option D
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Answer

$\frac{L^2 \omega^2}{2 g}$

Answer

$\frac{\mathrm{L} \omega}{6 \mathrm{~g}}$

Answer

$\frac{\mathrm{L} \omega}{2 \mathrm{~g}}$

Answer

$\frac{\mathrm{L}^2 \omega^2}{6 \mathrm{~g}}$

Centre of Mass and Rotational Motion

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 ( \mathrm{g}= acceleration due gravity)

Choose the correct answer: