One end of a straight uniform 1 m long bar is pivoted on horizontal table. It is released from rest

Q: Why do we use the center of mass to calculate potential energy instead of the whole length?

Gravity acts on every particle of the rod, but mathematically, the total work done by gravity on an extended rigid body is exactly equivalent to the work done as if all its mass were concentrated at its center of mass.

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

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

Q: Can we solve this using torque and angular acceleration?

You can, but it's much harder! The torque ($\tau = mg \frac{L}{2} \cos\theta$) changes continuously as the angle $\theta$ changes. You would need to integrate the angular acceleration over the angle to find the final angular velocity. Energy conservation is much faster.

Q: Does the mass of the rod affect the final angular speed?

No! As you can see in the equations, the mass $m$ cancels out from both sides. A heavier rod has more potential energy, but it also has proportionally more rotational inertia.

Q: What if the table wasn't frictionless?

The pivot is a single point, so even if there is friction at the pivot, it does no work because the displacement of the pivot point is zero. However, if there was air resistance, some energy would be lost, and the final $\omega$ would be smaller.

Question

One end of a straight uniform 1 m long bar is pivoted on horizontal table. It is released from rest when it makes an angle 30º from the horizontal (see figure). Its angular speed when its hits the table is given as $\sqrt{n}	ext{ s}^{-1}$, where $n$ is an integer. The value of $n$ is ......

Sigma Percentile · Accepted Answer

The Falling Rod: A Dance of Energy

Have you ever watched a tall tree being chopped down? It starts falling slowly, but as it gets closer to the ground, it whips through the air with terrifying speed. This everyday phenomenon is a perfect demonstration of the conservation of mechanical energy in rotational dynamics.

In this problem, we are looking at a miniature version of that falling tree: a uniform $1	ext{ m}$ long bar pivoted at one end on a horizontal table. It is released from rest at an angle of $30^\circ$ to the horizontal. Our goal is to find its angular speed right before it smacks into the table.

### Analyzing the Setup

Let's break down the physical situation. We have a rigid body (the rod) that is constrained to rotate about a fixed axis (the pivot). Because the pivot is fixed, the rod cannot translate; it can only rotate.

When the rod is released, the only force doing work on it is gravity. The pivot force does no work because the pivot point doesn't move. Since gravity is a conservative force, the total mechanical energy of the rod is conserved throughout its fall.

### The Master Equation: Conservation of Energy

The principle of conservation of mechanical energy states that the loss in gravitational potential energy must equal the gain in kinetic energy.

Initially, the rod is at rest, so its kinetic energy is zero. As it falls, its center of mass drops, losing potential energy. This lost potential energy is entirely converted into rotational kinetic energy.

Mathematically, we write this as:

$$\Delta PE = \Delta KE_{rot}$$

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

Here, $m$ is the mass of the rod, $h$ is the vertical distance the center of mass falls, $I$ is the moment of inertia of the rod about the pivot, and $\omega$ is the final angular speed.

### Pinpointing the Center of Mass

To calculate the change in potential energy, we must track the center of mass (CM) of the rod. For a uniform rod of length $L$, the CM is located exactly at its midpoint, a distance of $L/2$ from the pivot.

Initially, the rod is at an angle of $30^\circ$ to the horizontal. Using basic trigonometry, the initial height $h$ of the CM above the table is:

$$h = \frac{L}{2} \sin 30^\circ = \frac{L}{2} 	imes \frac{1}{2} = \frac{L}{4}$$

When the rod hits the table, its CM is at height zero. So, the total vertical drop of the CM is exactly $L/4$.

### The Rotational Inertia

Next, we need the moment of inertia $I$. The rod is rotating about its end, not its center. The moment of inertia of a uniform rod of mass $m$ and length $L$ about an axis through its end is a standard result derived using the parallel axis theorem:

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

### Equating and Solving

Now we have all the pieces of the puzzle. Let's substitute $h$ and $I$ into our energy conservation equation:

$$mg\left(\frac{L}{4}ight) = \frac{1}{2}\left(\frac{mL^2}{3}ight)\omega^2$$

Notice something beautiful here? The mass $m$ appears on both sides of the equation. This means the mass cancels out entirely! A heavier rod would have more potential energy, but it would also be harder to rotate (more inertia), and these two effects perfectly balance each other out.

Let's simplify the equation after cancelling $m$ and one $L$:

$$\frac{g}{4} = \frac{L}{6}\omega^2$$

Rearranging to solve for $\omega^2$:

$$\omega^2 = \frac{6g}{4L} = \frac{3g}{2L}$$

### Final Calculation

We are given that $L = 1	ext{ m}$ and we can take $g = 10	ext{ m/s}^2$. Plugging these numbers in:

$$\omega^2 = \frac{3 	imes 10}{2 	imes 1} = \frac{30}{2} = 15$$

Taking the square root gives us the final angular speed:

$$\omega = \sqrt{15}	ext{ s}^{-1}$$

The problem states that the angular speed is $\sqrt{n}	ext{ s}^{-1}$. Comparing our result, we can clearly see that:

$$n = 15$$

And there we have it! By simply tracking the energy of the center of mass, we bypassed the complex calculus of changing torques and arrived elegantly at the final answer.

Enter Numerical Value:

Visualized Solution

$Visualizing the Fall$

$Conservation of Energy$

$Identifying Variables$

$Equating Energies$

$Simplifying the Equation$

$Calculating ω$

$Food for Thought$

The Sigma Insight: Work and Energy in Rotational Motion

Analyzing the Setup

The Master Equation

Pinpointing the Center of Mass

The Rotational Inertia

Equating and Solving

Final Calculation

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