A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertic

Q: Why don't we consider the reaction force at the pivot when calculating torque?

Torque is force multiplied by the perpendicular distance to the pivot. Since the reaction force acts exactly *at* the pivot, its perpendicular distance is zero. Therefore, it creates zero torque and doesn't affect the angular acceleration.

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

The formula $Ml^2/12$ is for a rod rotating about an axis passing through its center of mass. In this problem, the rod is pivoted at its *end*. Using the parallel axis theorem, $I = I_{cm} + Md^2 = Ml^2/12 + M(l/2)^2 = Ml^2/3$.

Q: Does the mass of the rod affect how fast it accelerates angularly?

No! Just like objects in free fall accelerate at $g$ regardless of their mass, the mass $M$ cancels out on both sides of the $\tau = I\alpha$ equation. A heavier rod has more torque, but also proportionally more rotational inertia.

Q: Is the angular acceleration constant as the rod falls?

No, it is not constant. The angular acceleration $\alpha$ depends on $\sin\theta$. As the rod falls, $\theta$ increases from $0^\circ$ to $90^\circ$, meaning $\sin\theta$ increases. Therefore, the angular acceleration increases as the rod falls.

Q: What is the angular acceleration the exact moment the rod is released from the vertical position?

At the exact moment of release, the rod is vertical, so $\theta = 0^\circ$. Since $\sin(0^\circ) = 0$, the torque is zero, and the angular acceleration is exactly zero. It only begins to accelerate once it is slightly perturbed from the vertical.

Question

A slender uniform rod of mass $M$ and length $l$ is pivoted at one end so that it can rotate in a vertical plane (see the figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle $	heta$ with the vertical, is

Options:

(A) $\frac{2g}{3l} \sin	heta$

(B) $\frac{3g}{2l} \cos	heta$

(C) $\frac{2g}{3l} \cos	heta$

(D) $\frac{3g}{2l} \sin	heta$

Sigma Percentile · Accepted Answer

### The Setup: A Rod in Motion

Imagine a slender, uniform rod of mass $M$ and length $l$. It is pivoted at one of its ends, completely free to rotate in a vertical plane without any friction holding it back. Initially, it is held perfectly vertical, pointing straight up. Then, it is released.

As it falls, gravity pulls it down, causing it to sweep through an angle $	heta$ with the vertical. Our goal is to find the exact angular acceleration $\alpha$ of the rod at this specific instant. This is a classic problem in rotational dynamics that beautifully connects forces, geometry, and motion.

### The Driving Force: Torque

Why does the rod rotate? The answer is **torque**. The primary force acting on the rod is its own weight, $Mg$. Because the rod is uniform, we can assume this entire weight acts exactly at its center of mass (CM), which is located at a distance of $\frac{l}{2}$ from the pivot.

However, force alone doesn't dictate rotation; the *leverage* of that force does. Torque $	au$ is defined as the force multiplied by the perpendicular distance from the pivot to the line of action of the force.

Looking at the geometry of the falling rod, the vertical line of action of the weight is offset from the pivot. Using basic trigonometry, this perpendicular distance is $\frac{l}{2} \sin	heta$. Therefore, the torque driving the rotation is:

$$	au = Mg \left(\frac{l}{2} \sin	hetaight)$$

### The Resistance: Moment of Inertia

Just as mass resists linear acceleration, the **moment of inertia** $I$ resists angular acceleration. The moment of inertia depends not just on the mass, but on how that mass is distributed relative to the axis of rotation.

For a uniform rod of mass $M$ and length $l$ rotating about an axis through its center, $I = \frac{Ml^2}{12}$. But our rod is pivoted at its *end*. Using the parallel axis theorem (or standard tables), the moment of inertia for a rod pivoted at its end is:

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

### The Master Equation

Now we bring in the rotational equivalent of Newton's Second Law. It states that the net torque on an object equals its moment of inertia multiplied by its angular acceleration:

$$	au = I\alpha$$

Substituting our expressions for torque and moment of inertia into this master equation, we get the raw setup:

$$Mg \left(\frac{l}{2}ight) \sin	heta = \left(\frac{Ml^2}{3}ight) \alpha$$

### The Final Result

This equation looks a bit dense, but it simplifies beautifully. Notice that the mass $M$ appears on both sides, meaning it cancels out entirely. The angular acceleration of the falling rod is completely independent of its mass! Furthermore, one power of the length $l$ also cancels out.

$$g \left(\frac{1}{2}ight) \sin	heta = \left(\frac{l}{3}ight) \alpha$$

Rearranging the terms to isolate the angular acceleration $\alpha$, we multiply both sides by 3 and divide by $l$:

$$\alpha = \frac{3g \sin	heta}{2l}$$

This elegant result tells us a lot about the physics of the falling rod. When the rod is perfectly vertical ($	heta = 0^\circ$), $\sin(0) = 0$, so the angular acceleration is zero. As it falls and approaches the horizontal position ($	heta = 90^\circ$), $\sin(90^\circ) = 1$, and the angular acceleration reaches its absolute maximum of $\frac{3g}{2l}$.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Identifying the Forces$

$Newton’s Second Law for Rotation$

$Calculating Torque (τ)$

$Moment of Inertia (I)$

$Equating Torque and I α$

$Simplifying the Equation$

$Final Answer$

$Physical Interpretation$

The Sigma Insight: Dynamics of Rigid Body Rotation

The Setup

The Driving Force

The Resistance

The Master Equation

The Final Result

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Select Answer:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Newton’s Second Law for Rotation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Torque (τ)

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Torque and Iα

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

.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; } Physical Interpretation