A uniform rod of length l is pivoted at one of its ends on a vertical shaft of negligible radius. Wh

Q: Why do we calculate torque about the center of mass instead of the pivot?

You can calculate torque about any point, but the problem specifically gives you the rate of change of angular momentum *about the center of mass*. To use this given information directly, we must calculate the net torque about the same point.

Q: Why is the horizontal force $F_H$ equal to the centripetal force?

The center of mass of the rod is moving in a uniform horizontal circle. According to Newton's second law, there must be a net horizontal force directed towards the center of this circle to provide the centripetal acceleration. $F_H$ is the only horizontal force available.

Q: How do we determine the direction of the torques?

Use the right-hand rule ($\vec{r} \times \vec{F}$). For $F_V$, the position vector from the CM points left, and the force points up. Left $\times$ Up = Out of the page. For $F_H$, the position vector points up, and the force points left. Up $\times$ Left = Into the page.

Q: What happens if the rod is massless?

If the rod were massless, there would be no weight ($mg=0$) and no angular momentum to change. The problem relies on the mass distribution of the uniform rod to create the dynamic balance between gravity and the rotational effects.

Q: Why does the rod stay vertical for small values of $\omega$?

The equation $\cos\theta = \frac{3g}{2l\omega^2}$ requires the right side to be $\le 1$. If $\omega$ is too small, the required $\cos\theta$ would be $> 1$, which is impossible. Physically, the centrifugal effect isn't strong enough to overcome gravity and lift the rod until $\omega$ reaches a critical threshold $\sqrt{3g/2l}$.

Question

A uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$, the rod makes an angle $	heta$ with it (see figure). To find $	heta$, equate the rate of change of angular momentum (direction going into the paper) $\frac{ml^2}{12}\omega^2\sin	heta\cos	heta$ about the centre of mass to the torque provided by the horizontal and vertical forces $F_H$ and $F_V$ about the centre of mass. The value of $	heta$ is then such that

Options:

(A) $\cos	heta = \frac{g}{l\omega^2}$

(B) $\cos	heta = \frac{g}{2l\omega^2}$

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

(D) $\cos	heta = \frac{3g}{2l\omega^2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a uniform rod of length $l$ and mass $m$, pivoted at its top end to a vertical shaft. As the shaft spins with an angular velocity $\omega$, the rod swings outward, making a steady angle $	heta$ with the vertical. This is a classic example of a conical pendulum, but with an extended rigid body instead of a simple point mass.

To understand this motion, we must first identify the forces acting on the rod. Gravity pulls straight down on the rod's center of mass (CM) with a force $mg$. At the pivot point, the hinge exerts a reaction force to keep the rod attached. We can resolve this reaction force into two components: a vertical force $F_V$ pointing upwards, and a horizontal force $F_H$ pointing inwards towards the axis of rotation.

### The Kinematics of the Center of Mass

Because the rod maintains a constant angle $	heta$, its center of mass travels in a perfect horizontal circle. The distance from the pivot to the CM is $l/2$. Therefore, the radius $r$ of this circular path is simply the horizontal projection of this distance:

$$r = \frac{l}{2} \sin	heta$$

Since the rod has no vertical acceleration, the vertical forces must perfectly balance. This gives us our first crucial relationship:

$$F_V = mg$$

In the horizontal direction, the force $F_H$ is the sole provider of the centripetal acceleration required to keep the CM moving in its circular path. Using Newton's second law for circular motion ($F_c = m\omega^2 r$), we find:

$$F_H = m\omega^2 \left( \frac{l}{2} \sin	heta ight)$$

### The Master Equation: Torque about the CM

The problem provides a massive shortcut: it gives us the rate of change of angular momentum about the center of mass, $\frac{dL}{dt} = \frac{ml^2}{12}\omega^2\sin	heta\cos	heta$, and specifies its direction is *into the paper*. According to the rotational analog of Newton's second law, the net external torque about the CM must equal this rate of change of angular momentum:

$$\vec{	au}_{net} = \frac{d\vec{L}}{dt}$$

Let's calculate the torques produced by $F_V$ and $F_H$ about the CM.

1.  **Torque due to $F_V$:** The vertical force $F_V$ acts at a horizontal distance of $\frac{l}{2}\sin	heta$ from the CM. Using the right-hand rule, this torque points *into the paper*. Its magnitude is $	au_V = F_V \left( \frac{l}{2} \sin	heta ight)$.
2.  **Torque due to $F_H$:** The horizontal force $F_H$ acts at a vertical distance of $\frac{l}{2}\cos	heta$ from the CM. Using the right-hand rule, this torque points *out of the paper*. Its magnitude is $	au_H = F_H \left( \frac{l}{2} \cos	heta ight)$.

Since the given $\frac{dL}{dt}$ is directed into the paper, we will take 'into the paper' as our positive direction. Thus, the net torque equation becomes:

$$	au_{net} = 	au_V - 	au_H = F_V \left( \frac{l}{2} \sin	heta ight) - F_H \left( \frac{l}{2} \cos	heta ight)$$

Equating this to the given rate of change of angular momentum yields our master equation:

$$F_V \left( \frac{l}{2} \sin	heta ight) - F_H \left( \frac{l}{2} \cos	heta ight) = \frac{ml^2}{12}\omega^2\sin	heta\cos	heta$$

### Final Calculation

Now, we substitute the expressions for $F_V$ and $F_H$ that we derived earlier into our master equation:

$$mg \left( \frac{l}{2} \sin	heta ight) - m\omega^2 \left( \frac{l}{2} \sin	heta ight) \left( \frac{l}{2} \cos	heta ight) = \frac{ml^2}{12}\omega^2\sin	heta\cos	heta$$

To solve for $	heta$, let's group all the terms containing $\omega^2$ on the right side:

$$mg \frac{l}{2} \sin	heta = \frac{ml^2}{12}\omega^2\sin	heta\cos	heta + m\omega^2 \frac{l^2}{4} \sin	heta\cos	heta$$

Factoring out the common terms on the right side:

$$mg \frac{l}{2} \sin	heta = m\omega^2\sin	heta\cos	heta \left( \frac{l^2}{12} + \frac{l^2}{4} ight)$$

Simplifying the fraction inside the parenthesis ($\frac{1}{12} + \frac{3}{12} = \frac{4}{12} = \frac{1}{3}$):

$$mg \frac{l}{2} \sin	heta = \frac{ml^2}{3}\omega^2\sin	heta\cos	heta$$

Assuming the rod is actually swinging outward ($	heta 
eq 0$), we can safely divide both sides by $m$, $l$, and $\sin	heta$:

$$\frac{g}{2} = \frac{l}{3}\omega^2\cos	heta$$

Finally, isolating $\cos	heta$, we arrive at the elegant result:

**$$\cos	heta = \frac{3g}{2l\omega^2}$$**

This equation beautifully captures the delicate balance between gravity trying to pull the rod down and the rotational inertia trying to fling it outward.

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Visualized Solution

$V i s u a l i z in g t h e S y s t e m$

$I d e n t i f y in g t h e F or ces$

$K in e ma t i cs o f t h e C e n t er o f M a ss$

$V er t i c a l E q u i l ib r i u m$

$H or i z o n t a l C e n t r i p e t a l F or ce$

$T or q u e ab o u t t h e C e n t er o f M a ss$

$T or q u e d u e t o F_{V}$

$T or q u e d u e t o F_{H}$

$N e t T or q u e E q u a t i o n$

$E q u a t in g T or q u e an d \frac{d L}{d t}$

$S u b s t i t u t in g F or ces$

$R e a r r an g in g t h e E q u a t i o n$

$S im pl i f y in g t h e R i g h t S i d e$

$F ina l C a l c u l a t i o n$

$P h y s i c a l I n t er p r e t a t i o n$

The Sigma Insight: Dynamics of Rigid Body Rotation

Analyzing the Setup

The Kinematics of the Center of Mass

The Master Equation

Final Calculation

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A uniform rod of length $l$ is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is $T (x)$ at a distance $x$ from the axis, then which of the following graphs depicts it most closely?

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Visualized Solution

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

.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; } Kinematics of the Center of Mass

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torque about the Center of Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torque due to FV​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torque due to FH​

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

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

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

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

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

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

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