Two heavy metallic plates are joined together at 90 degrees to each other. A laminar sheet of mass 3

Q: Why is the reaction at the hinge zero during the impact?

This happens because the obstacle is placed exactly at the 'center of percussion' of the swinging sheet. At this specific point, the impulsive force causes a rotation and a translation that perfectly cancel each other out at the hinge, resulting in zero reaction force.

Q: Why do we add the initial and final velocities in the impulse equation instead of subtracting them?

Impulse equals the change in momentum: $\Delta p = m(v_f - v_i)$. Because the sheet rebounds, its final velocity is in the opposite direction to its initial velocity. If we define the rebound direction as positive, then $v_i$ is negative. Thus, $v_f - (-v_i) = v_f + v_i$.

Q: Why did we reject the root $r = 0.4\text{ m}$?

When we substitute $r = 0.4\text{ m}$ back into our equations, we get a final angular velocity of $\omega = -0.5\text{ rad/s}$. A negative value means the sheet continued moving in its original direction, implying it passed right through the solid rubber obstacle. Since this is physically impossible, we reject this root.

Q: How do we know the sheet will never come to rest?

By solving the equations, we found that the rebound angular velocity is $1\text{ rad/s}$, which is exactly equal to the impact angular velocity. This means no kinetic energy is lost during the collision (it is perfectly elastic). Without energy loss, it will bounce between the obstacles infinitely.

Q: Can we just use conservation of energy to solve this?

No. We don't know a priori whether the collision is elastic or inelastic. We must use the impulse-momentum theorems to find the final state. Only after solving do we discover that energy happens to be conserved in this specific setup.

Question

Two heavy metallic plates are joined together at $90^\circ$ to each other. A laminar sheet of mass $30	ext{ kg}$ is hinged at the line $AB$ joining the two heavy metallic plates. The hinges are frictionless. The moment of inertia of the laminar sheet about an axis parallel to $AB$ and passing through its centre of mass is $1.2	ext{ kg-m}^2$. Two rubber obstacles $P$ and $Q$ are fixed, one on each metallic plate at a distance $0.5	ext{ m}$ from the line $AB$. This distance is chosen, so that the reaction due to the hinges on the laminar sheet is zero during the impact. Initially the laminar sheet hits one of the obstacles with an angular velocity $1	ext{ rad/s}$ and turns back. If the impulse on the sheet due to each obstacle is $6	ext{ N-s}$.
(a) Find the location of the centre of mass of the laminar sheet from $AB$.
(b) At what angular velocity does the laminar sheet come back after the first impact ?
(c) After how many impacts, does the laminar sheet come to rest ?

Sigma Percentile · Accepted Answer

### The Magic of the Center of Percussion

Imagine swinging a baseball bat and hitting the ball right on the "sweet spot." You feel absolutely no sting in your hands. This sweet spot is known in physics as the **center of percussion**. In this fascinating problem, a laminar sheet acts just like that baseball bat. It swings on a hinge and strikes a rubber obstacle. The problem explicitly states that the reaction force at the hinge is zero during the impact. This is a massive clue: the obstacle is positioned exactly at the center of percussion of the swinging sheet!

### Setting Up the Master Equations

To solve this, we need to analyze the collision using two fundamental principles: the linear impulse-momentum theorem and the angular impulse-momentum theorem.

First, let's look at the linear motion. When the sheet hits the obstacle, it receives a sharp impulse, $J$. This impulse causes a sudden change in the linear momentum of the sheet's center of mass. Because the sheet rebounds, its final velocity is in the opposite direction to its initial velocity. Therefore, the change in velocity is the sum of their magnitudes:

$$J = m(v_f + v_i)$$

Since linear velocity is related to angular velocity by $v = r\omega$, we can rewrite this as:

$$J = m(r\omega_f + r\omega_i)$$

Plugging in the given values ($J = 6	ext{ N-s}$, $m = 30	ext{ kg}$, $\omega_i = 1	ext{ rad/s}$), we get our first master equation:

$$6 = 30(r\omega + r(1)) \implies r(\omega + 1) = 0.2$$

Next, we apply the angular impulse-momentum theorem about the hinge. The impulse $J$ acts at a distance $d = 0.5	ext{ m}$ from the hinge, creating an angular impulse that changes the sheet's angular momentum:

$$J 	imes d = I_{AB}(\omega_f + \omega_i)$$

$$6 	imes 0.5 = I_{AB}(\omega + 1) \implies 3 = I_{AB}(\omega + 1)$$

### The Parallel Axis Theorem

To proceed, we need the moment of inertia about the hinge, $I_{AB}$. We are given the moment of inertia about the center of mass, $I_{CM} = 1.2	ext{ kg-m}^2$. Using the **parallel axis theorem**, we can express $I_{AB}$ in terms of the unknown distance $r$:

$$I_{AB} = I_{CM} + mr^2 = 1.2 + 30r^2$$

Substituting this back into our angular equation gives us our second master equation:

$$3 = (1.2 + 30r^2)(\omega + 1)$$

### Solving the Quadratic Mystery

We now have a system of two equations. By isolating $(\omega + 1)$ from the first equation and substituting it into the second, we eliminate the angular velocity entirely:

$$3 = (1.2 + 30r^2)\left(\frac{0.2}{r}ight)$$

Expanding and rearranging this yields a neat quadratic equation in terms of $r$:

$$6r^2 - 3r + 0.24 = 0$$

Solving this quadratic equation gives us two mathematically valid roots: $r = 0.4	ext{ m}$ and $r = 0.1	ext{ m}$.

### The Physical Constraint Check

Mathematics gives us possibilities, but physics demands reality. We must check which root makes physical sense by calculating the final angular velocity, $\omega$, for both cases.

If we use $r = 0.4	ext{ m}$, we find that $\omega = -0.5	ext{ rad/s}$. A negative angular velocity implies that the sheet did not rebound; instead, it continued moving forward, magically passing *through* the solid rubber obstacle! This is physically impossible.

However, if we use $r = 0.1	ext{ m}$, we find that $\omega = 1	ext{ rad/s}$. This positive value confirms a realistic rebound. Therefore, the center of mass is located $0.1	ext{ m}$ from the hinge.

### The Eternal Bounce

Finally, we look at the rebound speed. The sheet hits the obstacle at $1	ext{ rad/s}$ and rebounds at exactly $1	ext{ rad/s}$. Because the speeds are identical, no kinetic energy is lost during the collision. The impact is perfectly elastic. Consequently, the sheet will bounce back and forth between the two obstacles forever, never coming to rest. A beautiful, perpetual dance of physics!

Visualized Solution

Visualizing the Setup

Linear Impulse-Momentum Equation

Substituting Values into Linear Equation

Angular Impulse-Momentum Equation

Applying Parallel Axis Theorem

Solving the System of Equations

Finding the Roots for r

Checking Physical Constraints

Final Conclusions

The Sigma Insight: Dynamics of Rigid Body Rotation

The Magic of the Center of Percussion

Setting Up the Master Equations

The Parallel Axis Theorem

Solving the Quadratic Mystery

The Physical Constraint Check

The Eternal Bounce

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