There is a rectangular plate of mass M kg of dimensions (a x b). The plate is hinged and held in hor

Q: Wait, why is the change in momentum $2mv$ and not just $mv$?

The collision is perfectly elastic. The ball hits the plate with velocity $v$ upwards and rebounds with velocity $v$ downwards. The change in velocity is $v - (-v) = 2v$. Therefore, the change in momentum is $2mv$.

Q: Why does the force from the balls act at $3b/4$ from the hinge?

The balls strike uniformly over the shaded region, which extends from $x = b/2$ to $x = b$. For a uniform force distribution, the effective point of application is the geometric center of that region. The midpoint between $b/2$ and $b$ is $\frac{b/2 + b}{2} = \frac{3b}{4}$.

Q: Why don't we consider the torque about the center of mass of the plate?

We can calculate torque about any point for a body in equilibrium. However, calculating it about the hinge is the smartest choice because the unknown reaction force at the hinge passes right through it, making its torque zero. This eliminates an unknown variable from our equation!

Q: Does the length $a$ of the plate affect the final velocity?

Yes, it does! The total number of balls striking the plate depends on the area of the shaded region, which is $a \times \frac{b}{2}$. A longer plate means more area, more balls, and thus a greater upward force for the same velocity.

Q: What would happen if the balls were striking the entire plate instead of just the shaded half?

If the balls struck the entire plate, the area would be $ab$, and the effective force would act at the center of the plate, $b/2$. The torque equation would become $Mg \frac{b}{2} = (nab \cdot 2mv) \frac{b}{2}$, leading to a different required velocity.

Question

There is a rectangular plate of mass $M$ kg of dimensions $(a 	imes b)$. The plate is hinged and held in horizontal position by striking $n$ small balls uniformly each of mass $m$ per unit area per unit time. These are striking in the shaded half region of the plate. The balls are colliding elastically with velocity $v$. What is $v$ ?

It is given $n = 100$, $M = 3 	ext{ kg}$, $m = 0.01 	ext{ kg}$; $b = 2 	ext{ m}$; $a = 1 	ext{ m}$; $g = 10 	ext{ m/s}^2$.

Sigma Percentile · Accepted Answer

### The Setup: A Battle of Torques

Imagine a heavy rectangular plate of mass $M$ and dimensions $a 	imes b$, hinged at one of its edges. Left to its own devices, gravity would immediately pull it down, causing it to swing like a trapdoor. However, this plate is being held perfectly horizontal by a rather unconventional method: a continuous, relentless barrage of tiny balls striking its outer half from below.

For the plate to remain perfectly horizontal, it must be in a state of **rotational equilibrium**. This means that the clockwise torque exerted by gravity must be exactly counterbalanced by the counter-clockwise torque generated by the impact of the balls.

### Analyzing the Gravity

The first step is to understand the force trying to pull the plate down. The entire weight of the uniform plate, $Mg$, acts precisely at its center of mass. Since the total width of the plate is $b$, the center of mass is located at a distance of $\frac{b}{2}$ from the hinge.

Therefore, the torque due to gravity about the hinge is:

$$	au_{	ext{gravity}} = Mg 	imes \frac{b}{2}$$

### The Barrage of Balls: Momentum in Action

Now, let's analyze the upward force keeping the plate afloat. The balls strike the plate and undergo perfectly elastic collisions. This means a ball hitting the plate with an upward velocity $v$ will rebound with a downward velocity $v$. The change in momentum for a single ball is:

$$\Delta p = p_{	ext{final}} - p_{	ext{initial}} = (-mv) - (mv) = -2mv$$

The magnitude of the momentum transferred to the plate per ball is $2mv$.

To find the total force, we need to know how many balls strike the plate every second. We are given that $n$ balls strike per unit area per unit time. The balls only strike the shaded outer half of the plate. The area of this shaded region is $a 	imes \frac{b}{2}$.

Thus, the total number of balls striking per second is $n 	imes \left(a 	imes \frac{b}{2}ight)$. The total upward force $F$ is the total momentum transferred per second:

$$F = \left(n \frac{ab}{2}ight) 	imes (2mv) = nabmv$$

### The Point of Application: A Geometric Catch

Here is where many students make a critical error. Where does this total upward force act? Because the balls strike uniformly over the shaded half, the effective force acts at the **geometric center** of this specific region.

The shaded region spans from $x = \frac{b}{2}$ to $x = b$. The midpoint of this region is:

$$x_{	ext{center}} = \frac{\frac{b}{2} + b}{2} = \frac{3b}{4}$$

So, the lever arm for the upward force is $\frac{3b}{4}$. The counter-clockwise torque due to the balls is:

$$	au_{	ext{balls}} = F 	imes \frac{3b}{4} = (nabmv) 	imes \frac{3b}{4}$$

### The Master Equation: Balancing the Torques

Equating the downward torque to the upward torque gives us our master equation:

$$Mg \frac{b}{2} = nabmv \frac{3b}{4}$$

We can cancel one $b$ from both sides and rearrange the terms to isolate the required velocity $v$:

$$Mg = nabmv \frac{3}{2}$$

$$v = \frac{2Mg}{3nabm}$$

### The Final Execution

All that remains is to substitute the given numerical values into our elegantly derived formula:
- $M = 3 	ext{ kg}$
- $g = 10 	ext{ m/s}^2$
- $n = 100 	ext{ balls/m}^2	ext{s}$
- $a = 1 	ext{ m}$
- $b = 2 	ext{ m}$
- $m = 0.01 	ext{ kg}$

$$v = \frac{2 	imes 3 	imes 10}{3 	imes 100 	imes 1 	imes 2 	imes 0.01}$$

The denominator simplifies beautifully: $3 	imes 100 	imes 2 	imes 0.01 = 6$. The numerator is $60$.

$$v = \frac{60}{6} = 10 	ext{ m/s}$$

The balls must strike with a velocity of **$10 	ext{ m/s}$** to keep the plate perfectly horizontal.

Enter Numerical Value:

Visualized Solution

The Physical Setup

Rotational Equilibrium

Torque due to Gravity

Force Exerted by Balls

Total Upward Force

Point of Application of Force

Torque due to Balls

Equating the Torques

Solving for Velocity $v$

Substituting the Values

Final Calculation

The Sigma Insight: Torque and Angular Momentum

The Setup

Analyzing the Gravity

The Barrage of Balls

The Point of Application

The Master Equation

The Final Execution

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