A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through

Q: Why do we use $I = \frac{ml^2}{3}$ and not $\frac{ml^2}{12}$?

The moment of inertia depends heavily on the axis of rotation. $\frac{ml^2}{12}$ is the moment of inertia for an axis passing through the center of mass. Since the rod in this problem is pivoted at one end, we must use the parallel axis theorem to find the moment of inertia about the end, which evaluates to $\frac{ml^2}{3}$.

Q: Why does the mass $m$ cancel out in the final equation?

Both the rotational kinetic energy and the gravitational potential energy are directly proportional to the mass of the rod. Therefore, when you equate them, the mass appears as a linear multiplier on both sides and cancels out. The maximum height depends only on the geometry (length) and the initial kinematics (angular speed).

Q: Is the velocity of the center of mass zero at the maximum height?

Yes! At the maximum height, the rod momentarily stops before swinging back down. This means its angular velocity $\omega$ becomes zero, and consequently, the linear velocity of the center of mass also becomes zero.

Q: What if the rod was replaced by a point mass on a massless string of length $l$?

That setup describes a simple pendulum. The moment of inertia for a point mass at distance $l$ is $ml^2$. The energy equation would then be $\frac{1}{2}(ml^2)\omega^2 = mgh$, resulting in a different height: $h = \frac{l^2\omega^2}{2g}$.

Q: Why do we consider the height of the center of mass and not the tip of the rod?

The gravitational potential energy of an extended rigid body is calculated by assuming all its mass is concentrated exactly at its center of mass. Therefore, the total change in potential energy of the system is simply $mg\Delta h_{cm}$.

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

Options:

(A) $\frac{1}{3} \frac{l^2 \omega^2}{g}$

(B) $\frac{1}{6} \frac{l \omega}{g}$

(C) $\frac{1}{2} \frac{l^2 \omega^2}{g}$

(D) $\frac{1}{6} \frac{l^2 \omega^2}{g}$

Sigma Percentile · Accepted Answer

### The Swinging Rod: A Dance of Energy

Imagine you are at a playground, looking at a rigid swing. If you give it a strong push, it arcs upwards, fighting gravity until it momentarily stops and falls back down. This classic physics problem captures that exact scenario, but instead of a swing, we have a uniform rigid rod pivoted at one end.

Our goal is to find out exactly how high the center of mass of this rod will rise when given an initial angular velocity $\omega$.

### The Master Principle: Conservation of Energy

In the absence of non-conservative forces like friction at the pivot or air resistance, the mechanical energy of the system remains perfectly conserved. This means the kinetic energy the rod possesses at its lowest point will be entirely converted into gravitational potential energy at its highest point.

At the lowest point, the rod is purely rotating about its pivot. Therefore, its initial kinetic energy is entirely rotational:

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

At the maximum height, the rod momentarily comes to rest. Its kinetic energy becomes zero, and all that energy is now stored as gravitational potential energy. If the center of mass rises by a height $h$, the potential energy is:

$$PE_f = mgh$$

Equating the two, we get our master equation:

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

### The Crucial Detail: Moment of Inertia

Here is where many students make a classic mistake. What is the moment of inertia $I$ of the rod?

If the rod were rotating about its center of mass, $I$ would be $\frac{ml^2}{12}$. However, our rod is pivoted at its **end**. Using the parallel axis theorem, the moment of inertia of a uniform rod of mass $m$ and length $l$ about an axis passing through its end is:

$$I = \frac{ml^2}{3}$$

### The Final Calculation

Now, we substitute this correct moment of inertia back into our energy conservation equation:

$$\frac{1}{2} \left( \frac{ml^2}{3} \right) \omega^2 = mgh$$

Notice something beautiful? The mass $m$ appears on both sides of the equation. This means we can cancel it out!

$$\frac{l^2 \omega^2}{6} = gh$$

Finally, isolating $h$, we arrive at our answer:

$$h = \frac{l^2 \omega^2}{6g}$$

### The Takeaway

The fact that the mass $m$ cancels out is a profound physical insight. It tells us that whether the rod is made of lightweight plastic or heavy solid steel, as long as its length $l$ and initial angular speed $\omega$ are the same, its center of mass will rise to the exact same height. Physics is elegant like that!

Animated Solution for Physics - 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 $ω$ . Its centre of mass rises to a maximum height of

Select Answer:

Visualized Solution

$Visualizing the Swing$

$Conservation of Mechanical Energy$

$Energy Equation$

$Moment of Inertia$

$Solving for h$

$Conclusion$

The Sigma Insight: Work and Energy in Rotational Motion

The Swinging Rod

The Master Principle

The Crucial Detail

The Final Calculation

The Takeaway

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