A thin smooth rod of length L and mass M is rotating freely with angular speed w_0 about an axis per

Q: Why is the initial moment of inertia of the beads zero?

The moment of inertia of a point mass is given by $mr^2$, where $r$ is the perpendicular distance from the axis of rotation. Since the beads are initially exactly at the center (the axis itself), $r=0$, making their moment of inertia zero.

Q: Why is angular momentum conserved in this specific case?

Angular momentum is conserved because there is no external torque acting on the system in the plane of rotation. The rod is smooth, and no outside twisting force is applied as the beads slide.

Q: Why does the angular speed decrease?

As the beads move outwards, the mass of the system is distributed further away from the axis. This increases the total moment of inertia ($I$). Since angular momentum ($L = I\omega$) must remain constant, an increase in $I$ forces a proportional decrease in $\omega$.

Q: Does the mass of the rod matter?

Absolutely! The rod has its own moment of inertia $\frac{ML^2}{12}$ which remains constant throughout the process. It acts as the 'base' inertia of the system and heavily influences the final angular speed.

Q: What if the rod was not smooth and had friction?

If there was friction between the beads and the rod, the beads would still eventually reach the ends (due to centrifugal effects). Friction is an internal force here; it cannot exert a net external torque on the system as a whole. Therefore, the total angular momentum would still be conserved, and the final angular speed would be exactly the same!

Question

A thin smooth rod of length $L$ and mass $M$ is rotating freely with angular speed $\omega_0$ about an axis perpendicular to the rod and passing through its centre. Two beads of mass $m$ and negligible size are at the centre of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be

Options:

(A) $\frac{M\omega_0}{M+3m}$

(B) $\frac{M\omega_0}{M+m}$

(C) $\frac{M\omega_0}{M+2m}$

(D) $\frac{M\omega_0}{M+6m}$

Sigma Percentile · Accepted Answer

Imagine you are watching a figure skater spinning on the ice. When they pull their arms in, they spin incredibly fast. But the moment they throw their arms outward, their spin dramatically slows down. This beautiful phenomenon is governed by one of the most profound laws in physics: **The Conservation of Angular Momentum**.

In this problem, we have a mechanical version of our ice skater. A smooth rod of mass $M$ and length $L$ is spinning freely with an initial angular velocity $\omega_0$. Right at the center of this rod, sitting quietly on the axis of rotation, are two tiny beads, each of mass $m$.

### Analyzing the Setup

Because the rod is smooth and there are no external twisting forces (torques) acting on the system, the net external torque is zero ($	au_{	ext{ext}} = 0$). This is our golden ticket! It tells us that the total angular momentum of the system must remain perfectly constant throughout the motion.

Mathematically, this is written as:
$$L_i = L_f$$
$$I_i \omega_i = I_f \omega_f$$

### The Initial State

Let's look at the system before the beads start sliding. The moment of inertia of a uniform rod rotating about its center is a standard result:
$$I_{	ext{rod}} = \frac{ML^2}{12}$$

What about the beads? Since they are located exactly at the center, their distance from the axis of rotation is $r = 0$. Therefore, their contribution to the moment of inertia ($mr^2$) is zero. The initial moment of inertia is simply:
$$I_i = \frac{ML^2}{12}$$

### The Final State

Now, the beads slide outward until they hit the ends of the rod. They are now at a distance of $r = \frac{L}{2}$ from the axis. The new moment of inertia of the system must include the rod AND the two beads at their new positions.

$$I_f = I_{	ext{rod}} + I_{	ext{beads}}$$
$$I_f = \frac{ML^2}{12} + 2m\left(\frac{L}{2}ight)^2$$

Let's simplify this expression. Squaring the distance gives us $\frac{L^2}{4}$. Multiplying by the two beads yields:
$$I_f = \frac{ML^2}{12} + \frac{2mL^2}{4} = \frac{ML^2}{12} + \frac{mL^2}{2}$$

### The Master Equation

Now we bring it all together using our conservation law. We equate the initial angular momentum to the final angular momentum:

$$\left(\frac{ML^2}{12}ight)\omega_0 = \left(\frac{ML^2}{12} + \frac{mL^2}{2}ight)\omega$$

Notice how $L^2$ appears in every single term? We can divide the entire equation by $L^2$ to cancel it out. To make the algebra even cleaner, let's multiply the entire equation by $12$ to eliminate the fractions:

$$M\omega_0 = \left(M + 6might)\omega$$

### Final Calculation

All that's left is to isolate our final angular velocity, $\omega$:

$$\omega = \frac{M\omega_0}{M + 6m}$$

And there we have it! Just like the ice skater throwing their arms out, as the mass of the beads moved further from the axis, the system's moment of inertia increased. To compensate and keep the angular momentum conserved, the angular velocity had to decrease. The physics works out beautifully!

Select Answer:

Visualized Solution

$Visualizing the System$

$τ_{ext} = 0 \Rightarrow L_{i} = L_{f}$

$I_{i} = \frac{M L ^{2}}{12}$

$I_{f} = \frac{M L ^{2}}{12} + 2 m (\frac{L}{2})^{2}$

$I_{f} = \frac{M L ^{2}}{12} + \frac{m L ^{2}}{2}$

$(\frac{M L ^{2}}{12}) ω_{0} = (\frac{M L ^{2}}{12} + \frac{m L ^{2}}{2}) ω$

$M ω_{0} = (M + 6 m) ω$

$ω = \frac{M ω _{0}}{M + 6 m}$

$Food for Thought$

The Sigma Insight: Conservation of Angular Momentum

Analyzing the Setup

The Initial State

The Final State

The Master Equation

Final Calculation

Similar Questions

Consider a uniform rod of mass $M = 4 m$ and length $l$ pivoted about its centre. A mass $m$ moving with velocity $v$ making angle $θ = \frac{π}{4}$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is

A rod of length $L$ and mass $M$ is hinged at point $O$ . A small bullet of mass $m$ hits the rod as shown in the figure. The bullet gets embedded in the rod. Find angular velocity of the system just after impact.

A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with an angular speed $ω$ . Two particles having mass $m$ each are now attached at diametrically opposite points. The angular speed of the ring will become

A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $ω$ . Two objects, each of mass $m$ , are attached gently to the opposite ends of a diameter of the ring. The wheel now rotates with an angular velocity

A thin circular ring of mass $m$ and radius $R$ is rotating about its axis with a constant angular velocity $ω$ . Two objects each of mass $M$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $ω^{'}$ =

Initial angular velocity of a circular disc of mass $M$ is $ω_{1}$ . Then, two small spheres of mass $m$ are attached gently to two diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?

A hoop of radius $r$ and mass $m$ rotating with an angular velocity $ω_{0}$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

Select Answer:

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; } τext​=0⇒Li​=Lf​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ii​=12ML2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If​=12ML2​+2m(2L​)2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If​=12ML2​+2mL2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (12ML2​)ω0​=(12ML2​+2mL2​)ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mω0​=(M+6m)ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω=M+6mMω0​​

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