Two uniform rods A and B of length 0.6 m each and of masses 0.01 kg and 0.02 kg, respectively are ri

Q: Why can't we conserve linear momentum during the collision?

Because the pivot point is fixed, it exerts an external impulsive reaction force on the rods during the sudden impact to keep them in place. This external force changes the total linear momentum of the system.

Q: Why is angular momentum conserved if there is an external force?

The external impulsive force acts exactly at the pivot point. Since torque is force times perpendicular distance, and the distance from the pivot to the line of action of this force is zero, the torque is zero. Hence, angular momentum about the pivot is conserved.

Q: How did we get the moment of inertia for rod B?

Rod B is not rotating about its own end or center. We must use the parallel axis theorem: $I = I_{\text{CM}} + md^2$. Its center of mass is at a distance of $l + l/2 = 1.5l$ from the pivot. So, $I_B = ml^2/12 + m(1.5l)^2$.

Q: Why do we track the center of mass for potential energy?

For an extended rigid body, the total gravitational potential energy is equivalent to the potential energy of a point mass (equal to the total mass) located exactly at the body's center of mass.

Q: What if the object bounced back elastically instead of sticking?

If the collision were perfectly elastic, kinetic energy would also be conserved during the impact. The object would rebound with some velocity, and the rods would swing up with a different angular velocity. You would need to set up two equations: one for angular momentum and one for kinetic energy.

Question

Two uniform rods $A$ and $B$ of length $0.6	ext{ m}$ each and of masses $0.01	ext{ kg}$ and $0.02	ext{ kg}$, respectively are rigidly joined end to end. The combination is pivoted at the lighter end, $P$ as shown in figure. Such that it can freely rotate about point $P$ in a vertical plane.

A small object of mass $0.05	ext{ kg}$, moving horizontally, hits the lower end of the combination and sticks to it. What should be the velocity of the object, so that the system could just be raised to the horizontal position?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a rigid pendulum made of two uniform rods, $A$ and $B$, hanging vertically from a pivot point $P$. Rod $A$ is attached to the pivot, and rod $B$ is attached to the bottom of rod $A$. Both rods have a length of $l = 0.6	ext{ m}$, but they have different masses: $m_A = 0.01	ext{ kg}$ and $m_B = 0.02	ext{ kg}$.

Suddenly, a small object of mass $m = 0.05	ext{ kg}$ comes flying in horizontally with a velocity $v$. It strikes the very bottom of rod $B$ and sticks to it. This is a classic perfectly inelastic collision. Our goal is to find the exact velocity $v$ required so that the entire combined system swings up and just reaches a perfectly horizontal position before momentarily coming to rest.

### The Master Equation: Angular Momentum

Here is a catch. During the collision, the pivot at $P$ will exert a sudden, unknown impulsive force on the rods to keep them attached. Because there is an external impulsive force acting on our system, we **cannot** conserve linear momentum.

However, since this impulsive force acts exactly at the pivot point $P$, its lever arm is zero. Therefore, the torque about $P$ is zero! This means we can safely apply the **Conservation of Angular Momentum** about the pivot point $P$.

Before we can equate the initial and final angular momenta, we need to find the moment of inertia $I$ of the new, combined system about $P$. The system consists of three parts: rod $A$, rod $B$, and the stuck mass $m$.

For rod $A$, which is pivoted at its end, the moment of inertia is standard:
$I_A = \frac{m_A l^2}{3}$

For rod $B$, we must use the parallel axis theorem. Its own center of mass is at a distance of $l + \frac{l}{2} = \frac{3l}{2}$ from the pivot $P$. 
$I_B = I_{	ext{CM}} + m_B d^2 = \frac{m_B l^2}{12} + m_B \left(\frac{3l}{2}ight)^2 = \frac{7 m_B l^2}{3}$

For the point mass $m$ stuck at the bottom (at a distance of $2l$ from $P$):
$I_m = m(2l)^2 = 4ml^2$

Adding them all up and substituting the given values ($l = 0.6	ext{ m}$, $m_A = 0.01	ext{ kg}$, $m_B = 0.02	ext{ kg}$, $m = 0.05	ext{ kg}$):
$I = \frac{0.01(0.6)^2}{3} + \frac{7(0.02)(0.6)^2}{3} + 0.05(1.2)^2$

$I = 0.0012 + 0.0168 + 0.072 = \mathbf{0.09 	ext{ kg-m}^2}$

Now, let's apply the conservation of angular momentum. Initially, only the mass $m$ has angular momentum about $P$:
$L_i = m v (2l)$

Right after the collision, the whole system rotates with an angular velocity $\omega$:
$L_f = I \omega$

Equating the two:
$m v (2l) = I \omega$

Substituting the values:
$0.05 	imes v 	imes 1.2 = 0.09 \omega$

$\omega = \frac{0.06}{0.09} v = \mathbf{\frac{2}{3} v}$

### Energy Conservation for the Swing

Now comes the main point. After the collision is over, the system swings upwards. During this swing, the only force doing work is gravity (a conservative force). Therefore, we can use the **Conservation of Mechanical Energy**. The rotational kinetic energy just after the collision will be completely converted into gravitational potential energy when the system reaches the horizontal position.

Let's calculate the total gain in potential energy ($\Delta U$). We must track the vertical shift of the center of mass for each component as they move from the vertical to the horizontal position.

Rod $A$'s center of mass rises by $\frac{l}{2}$.
Rod $B$'s center of mass rises by $l + \frac{l}{2} = \frac{3l}{2}$.
The mass $m$ rises by the full length $2l$.

$\Delta U = m_A g \left(\frac{l}{2}ight) + m_B g \left(\frac{3l}{2}ight) + m g (2l)$

Plugging in the numbers and using $g = 9.8	ext{ m/s}^2$:
$\Delta U = 0.01(9.8)(0.3) + 0.02(9.8)(0.9) + 0.05(9.8)(1.2)$

$\Delta U = 0.0294 + 0.1764 + 0.588 = \mathbf{0.7938 	ext{ J}}$

### Final Calculation

We equate the initial rotational kinetic energy to this potential energy gain:
$\frac{1}{2} I \omega^2 = \Delta U$

$\frac{1}{2} (0.09) \omega^2 = 0.7938$

$\omega^2 = \frac{1.5876}{0.09} = 17.64$

Taking the square root, we get a perfectly clean number:
$\omega = \mathbf{4.2 	ext{ rad/s}}$

Finally, we bring back our relationship between $v$ and $\omega$:
$v = \frac{3}{2} \omega$

$v = \frac{3}{2} (4.2) = \mathbf{6.3 	ext{ m/s}}$

The initial velocity of the object must be exactly $6.3	ext{ m/s}$ to make the system swing perfectly to the horizontal position!

Enter Numerical Value:

Visualized Solution

System Setup

Conservation of Angular Momentum

Moment of Inertia Setup

Moment of Inertia of Rods

Total Moment of Inertia

Applying Angular Momentum Conservation

Angular Velocity $ω$

Conservation of Mechanical Energy

Change in Potential Energy

Calculating $Δ U$

Solving for $ω$

Final Velocity $v$

The Way Forward

The Sigma Insight: Conservation of Angular Momentum

Analyzing the Setup

The Master Equation

Energy Conservation for the Swing

Final Calculation

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