A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to ro

Q: Why did we take the fall in height as $l$ and not $2l$?

Gravitational potential energy depends on the position of the **center of mass**. The center of mass starts at a height of $l/2$ above the pivot and ends at $l/2$ below the pivot. The total vertical displacement of the center of mass is $l/2 - (-l/2) = l$.

Q: Why didn't we use the mass $m = 2 \text{ kg}$ given in the question?

In the energy conservation equation, mass $m$ appears on both sides (in potential energy $mgl$ and in the moment of inertia $ml^2/3$). It cancels out completely, meaning the final speed is independent of the rod's mass!

Q: Why is the moment of inertia $ml^2/3$ and not $ml^2/12$?

The rod is rotating about a pivot at its **end**, not its center. By the parallel axis theorem, $I_{end} = I_{cm} + m(l/2)^2 = ml^2/12 + ml^2/4 = ml^2/3$.

Q: Can we use equations of motion like $v^2 = u^2 + 2as$ here?

No, because the angular acceleration $\alpha$ is not constant. As the rod falls, the torque due to gravity changes with the angle, making the acceleration variable. Energy conservation is the best approach.

Question

A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is ........... $	ext{ms}^{-1}$. (Take, $g = 10 	ext{ ms}^{-2}$)

Sigma Percentile · Accepted Answer

### The Setup: A Precarious Balance

Imagine standing a pencil perfectly upright on a table. It is a precarious balance, a state of unstable equilibrium. The slightest nudge, and gravity takes over, pulling it down in a sweeping arc.

This is exactly what happens to our $2 	ext{ kg}$ steel rod, but on a much grander scale! The rod is clamped at the bottom, meaning it can only rotate around that pivot point like a hinge. As it falls from its highest vertical position to its lowest vertical position, it traces out a perfect semicircle.

### The Physics: Energy is the Currency

In the frictionless world of this problem, energy is the ultimate currency. As the rod falls, it loses gravitational potential energy. But this energy doesn't just vanish; it transforms entirely into rotational kinetic energy.

To calculate the loss in potential energy, we must track the **center of mass**. The center of mass starts at a height of $l/2$ above the pivot and ends at a depth of $l/2$ below the pivot. The total vertical displacement is exactly $l$. Therefore, the loss in potential energy is simply $mgl$.

### The Math: Unlocking the Speed

We set up our master equation by equating the lost potential energy to the gained rotational kinetic energy:

$$mgl = \frac{1}{2} I \omega^2$$

We know the moment of inertia of a rod rotating about its end is $I = \frac{ml^2}{3}$. Substituting this into our equation gives:

$$mgl = \frac{1}{2} \left(\frac{ml^2}{3}ight) \omega^2$$

Notice how the mass $m$ appears on both sides? It cancels out completely! This is a beautiful quirk of physics—the final speed of the rod doesn't depend on how heavy it is. Solving for the angular velocity $\omega$, we get $\omega = \sqrt{\frac{6g}{l}}$.

### The Grand Finale

The question asks for the linear speed of the **free end** of the rod. We use the bridge equation $v = \omega r$. For the free end, the radius $r$ is the full length of the rod, $l$.

$$v = \sqrt{\frac{6g}{l}} \cdot l = \sqrt{6gl}$$

Now, we just plug in the given numbers: $g = 10 	ext{ m/s}^2$ and $l = 0.6 	ext{ m}$.

$$v = \sqrt{6 	imes 10 	imes 0.6} = \sqrt{36}$$

The square root of $36$ is a perfect $6$. The free end of the rod whips past the lowest point at a thrilling speed of $6 	ext{ m/s}$!

Enter Numerical Value:

Visualized Solution

$Initial State$

$The Fall$

$Conservation of Energy$

$Energy Equation$

$Moment of Inertia$

$Angular Velocity$

$Linear Speed of Free End$

$Numerical Substitution$

$Final Answer$

$The Way Forward$

The Sigma Insight: Work and Energy in Rotational Motion

The Setup

The Physics

The Math

The Grand Finale

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Fall

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moment of Inertia

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Linear Speed of Free End

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Numerical Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward