A long horizontal rod has a bead which can slide along its length and is initially placed at a dista

Q: Wait, why is the normal reaction $N$ equal to $m \alpha L$ and not $mg$?

The problem explicitly states that gravity is neglected, so we don't consider vertical forces. In the horizontal plane, the rod pushes the bead tangentially to give it an angular acceleration $\alpha$. This tangential pushing force exerted by the rod is the normal reaction $N = m a_t = m \alpha L$.

Q: Why does the friction act towards the center?

The bead tends to slide outwards (away from the pivot) due to inertia as the rod spins faster. To prevent this outward sliding and maintain the circular path, static friction must act inwards, towards the center, providing the necessary centripetal force.

Q: Why do we use $\omega = \alpha t$?

The rod is set into motion with a *constant* angular acceleration $\alpha$. Just like linear kinematics where $v = u + at$, in rotational kinematics, $\omega = \omega_0 + \alpha t$. Since it starts from rest, $\omega_0 = 0$, giving $\omega = \alpha t$.

Q: What happens if the time $t$ is less than $\sqrt{\frac{\mu}{\alpha}}$?

If $t$ is less than this threshold, the required centripetal force $m \omega^2 L$ is strictly less than the maximum available static friction $\mu N$. The static friction will simply self-adjust to equal $m \omega^2 L$, and the bead will not slip.

Q: Does the mass of the bead or its distance $L$ affect the time it starts slipping?

Surprisingly, no! If you look at the final equation $m \alpha^2 t^2 L = \mu m \alpha L$, both the mass $m$ and the distance $L$ cancel out completely. The time of slipping depends only on the coefficient of friction $\mu$ and the angular acceleration $\alpha$.

Question

A long horizontal rod has a bead which can slide along its length and is initially placed at a distance $L$ from one end $A$ of the rod. The rod is set in angular motion about $A$ with a constant angular acceleration $\alpha$. If the coefficient of friction between the rod and bead is $\mu$, and gravity is neglected, then the time after which the bead starts slipping is

Options:

(A) $\sqrt{\frac{\mu}{\alpha}}$

(B) $\frac{\mu}{\sqrt{\alpha}}$

(C) $\frac{1}{\sqrt{\mu\alpha}}$

(D) infinitesimal

Sigma Percentile · Accepted Answer

### The Thrill of the Spinning Rod

Imagine you are standing on a giant, horizontal merry-go-round that is spinning faster and faster. You would feel a mysterious force trying to fling you outwards, right? In this fascinating problem, we explore a similar scenario: a small bead of mass $m$ resting on a long horizontal rod.

The rod is pivoted at one end, point $A$, and the bead is initially placed at a distance $L$ from this pivot. Suddenly, the rod is set into motion with a constant angular acceleration $\alpha$. Our mission is to find the exact moment when the bead loses its grip and starts slipping along the rod. Since gravity is neglected, our entire focus shifts to the horizontal plane.

### Analyzing the Setup: The Geometry of Motion

Before we dive into the forces, we must understand the kinematics of the bead. The rod starts from rest and accelerates angularly at a constant rate $\alpha$.

Using the rotational equivalent of the first equation of motion, we can determine the angular velocity $\omega$ at any given time $t$. Since the initial angular velocity is zero, the equation simplifies beautifully:

$$ \omega = \alpha t $$

This tells us that the rod is spinning faster with every passing second. Consequently, the bead, which is forced to move in a circle of radius $L$, experiences a tangential acceleration $a_t$. This tangential acceleration is simply the product of the angular acceleration and the radius:

$$ a_t = \alpha L $$

### The Master Equation: Unmasking the Forces

Now, let's talk about dynamics. Why does the bead accelerate tangentially? It doesn't have an engine! The answer lies in the rod itself. As the rod sweeps through the horizontal plane, it physically pushes against the side of the bead.

This pushing force is the **normal reaction** $N$. In most textbook problems, the normal force points vertically upwards to counter gravity. But here, it acts horizontally, perpendicular to the length of the rod. According to Newton's Second Law, this force is responsible for the tangential acceleration:

$$ N = m a_t = m \alpha L $$

But there is another crucial force at play. For the bead to travel in a circle, it requires a **centripetal force** $F_c$ directed towards the pivot $A$. Without this force, the bead would simply fly off in a straight line due to inertia. The required centripetal force is given by:

$$ F_c = m \omega^2 L $$

Substituting our earlier expression for $\omega$, we get:

$$ F_c = m (\alpha t)^2 L = m \alpha^2 t^2 L $$

Notice how this required force grows quadratically with time. As the rod spins faster, the demand for centripetal force skyrockets!

### The Threshold of Slipping: When Friction Gives Up

So, what provides this ever-increasing centripetal force? It is the **static friction** $f_r$ between the bead and the rod. Friction acts like an invisible tether, pulling the bead towards the center to keep it on its circular path.

However, static friction is not infinitely strong. It has a strict upper limit, determined by the coefficient of friction $\mu$ and the normal reaction $N$. The maximum available friction is:

$$ f_{	ext{max}} = \mu N $$

Substituting our expression for the normal reaction $N$, we find the absolute maximum force friction can muster:

$$ f_{	ext{max}} = \mu m \alpha L $$

The bead will remain stationary relative to the rod as long as the required centripetal force is less than or equal to this maximum friction. The exact moment of slipping occurs at the threshold where the required force perfectly matches the maximum available friction:

$$ F_c = f_{	ext{max}} $$

### Final Calculation: The Elegance of Cancellation

Now, we simply equate the two expressions we derived:

$$ m \alpha^2 t^2 L = \mu m \alpha L $$

This is where the magic of physics reveals itself. Notice how the mass $m$ and the distance $L$ appear on both sides of the equation? They completely cancel out!

$$ \alpha^2 t^2 = \mu \alpha $$

We can also divide both sides by $\alpha$, leaving us with a remarkably simple relationship:

$$ \alpha t^2 = \mu $$

Finally, isolating $t^2$ and taking the square root, we arrive at our answer:

$$ t = \sqrt{\frac{\mu}{\alpha}} $$

This elegant result tells us that the time until slipping depends solely on the coefficient of friction and the angular acceleration. Whether the bead is heavy or light, or whether it is placed near the pivot or far away, it will always start slipping at this exact moment. Therefore, the correct option is **(a)**.

Select Answer:

Visualized Solution

Visualizing the Setup

Angular Velocity

Tangential Force and Normal Reaction

Centripetal Force Requirement

Centripetal Force as a Function of Time

Limiting Friction

Condition for Slipping

Simplifying the Equation

Solving for Time

Final Conclusion

The Sigma Insight: Dynamics of Circular Motion

The Thrill of the Spinning Rod

Analyzing the Setup

The Master Equation

The Threshold of Slipping

Final Calculation

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