A thin rod of mass 0.9 kg and length 1 m is suspended at rest from one end, so that it can freely os

Q: Why can't we just conserve linear momentum here?

Because the rod is hinged at the top! When the particle hits the rod, the hinge exerts a sudden, strong reaction force to keep the rod attached. This external force means the net force on the system is not zero, so linear momentum is not conserved.

Q: Why do we take the angular momentum specifically about the pivot point?

We choose the pivot point because the unknown reaction force from the hinge acts exactly at this point. The perpendicular distance from the pivot to this force is zero, meaning it creates zero torque. With zero external torque about the pivot, angular momentum is conserved.

Q: How did we get the moment of inertia of the rod as $Ml^2/3$?

The moment of inertia of a uniform rod of mass $M$ and length $l$ about an axis passing through its center of mass is $Ml^2/12$. Using the parallel axis theorem to shift the axis to one end (a distance of $l/2$), we get $I = Ml^2/12 + M(l/2)^2 = Ml^2/3$.

Q: Why is the initial angular momentum of the particle $mvl$?

For a point particle moving in a straight line, its angular momentum about a point is the product of its linear momentum ($mv$) and the perpendicular distance from the point to its line of motion. Here, the particle hits the bottom of the rod, so the perpendicular distance from the pivot is the rod's length, $l$.

Q: Is kinetic energy conserved in this collision?

No! The problem states that the particle 'sticks' to the rod, making it a perfectly inelastic collision. In such collisions, a significant amount of kinetic energy is lost as heat, sound, and deformation energy.

Question

A thin rod of mass $0.9	ext{ kg}$ and length $1	ext{ m}$ is suspended at rest from one end, so that it can freely oscillate in the vertical plane. A particle of mass $0.1	ext{ kg}$ moving in a straight line with velocity $80	ext{ m/s}$ hits the rod at its bottom-most point and sticks to it (see figure). The angular speed (in rad/s) of the rod immediately after the collision will be .........

Sigma Percentile · Accepted Answer

Imagine you are in a physics lab. You have a uniform thin rod hanging perfectly still from a pivot at its top end. Suddenly, a tiny particle, moving incredibly fast in a straight line, comes zooming in. It's heading straight for the very bottom tip of the rod. Let's break down the physics of this dramatic collision.

### Analyzing the Collision

Now, what happens the exact moment the particle crashes into the rod? It sticks! This is a classic perfectly inelastic collision. But here is the catch: can we conserve linear momentum?

**No!** The pivot at the top will exert a sudden, unknown reaction force to keep the rod attached. Because there is an external impulsive force acting on our system, linear momentum is not conserved.

However, if we take the torque about that very pivot, the torque of this reaction force is zero (since the distance from the pivot is zero). This means our **angular momentum about the pivot is perfectly conserved**.

### The Master Equation

Let's calculate the angular momentum just a millisecond before the crash. The rod is at rest, so it has zero angular momentum. The particle, however, is moving linearly. Its angular momentum about the pivot is simply its linear momentum ($mv$) multiplied by the perpendicular distance from the pivot, which is exactly the length of the rod ($l$).

$$L_i = m v l$$

Smash! The particle is now stuck to the rod. They become a single rotating system. To find their new angular momentum, we need the total moment of inertia of this new system about the pivot. That will be the moment of inertia of the rod about its end, plus the moment of inertia of the particle, which is now just a point mass at a distance $l$ from the pivot.

$$I_{	ext{system}} = I_{	ext{rod}} + I_{	ext{particle}} = \frac{M l^2}{3} + m l^2$$

So, the final angular momentum is:

$$L_f = I_{	ext{system}} \omega = \left( \frac{M l^2}{3} + m l^2 ight) \omega$$

### Final Calculation

We have our master equation. The initial angular momentum equals the final angular momentum. Let's carefully substitute the numbers given in the problem. The mass of the particle is $0.1	ext{ kg}$, its velocity is $80	ext{ m/s}$, and the length is $1	ext{ m}$. For the rod, its mass is $0.9	ext{ kg}$.

$$m v l = \left( \frac{M l^2}{3} + m l^2 ight) \omega$$

$$0.1 	imes 80 	imes 1 = \left( \frac{0.9 	imes 1^2}{3} + 0.1 	imes 1^2 ight) \omega$$

Let's crunch the numbers. Don't get intimidated by the fractions. $0.9 	imes 1^2 / 3$ is just $0.3$. Add the $0.1$ from the particle, and the total moment of inertia is $0.4$. On the left side, $0.1 	imes 80$ is simply $8$.

$$8 = (0.3 + 0.1) \omega$$

$$8 = 0.4 \omega$$

We are at the finish line. Dividing $8$ by $0.4$ gives us $20$.

$$\omega = \frac{8}{0.4} = 20	ext{ rad/s}$$

The rod and particle system will instantly start swinging with an angular speed of $20	ext{ rad/s}$. We just used a powerful conservation law to solve a complex collision!

Enter Numerical Value:

Visualized Solution

$S y s t e m S e t u p$

$C o n ser v a t i o n o f A n g u l a r M o m e n t u m$

$I ni t ia l A n g u l a r M o m e n t u m$

$F ina l A n g u l a r M o m e n t u m$

$E q u a t in g L_{i} an d L_{f}$

$S o l v in g f or ω$

$F ina l A n s w er$

$T h e W a y F or w a r d$

The Sigma Insight: Conservation of Angular Momentum

Analyzing the Collision

The Master Equation

Final Calculation

Similar Questions

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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

Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Angular Momentum

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

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

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

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

.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; } The Way Forward