The Mathematics of a Ball's Descent on a Staircase
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Chapter 1: Understanding the Problem
Recently, I engaged with an interesting applied mathematics problem and was pleasantly surprised by the elegance of the solution. This scenario involves a particle launched from the top of a staircase, bouncing downwards and striking each step. It is a familiar sight, making it a delightful challenge to model mathematically. We will utilize classical mechanics while disregarding complications such as air resistance and friction, leading to a somewhat idealized yet fascinating outcome.
The Problem Statement
Consider a staircase comprising N smooth horizontal steps, each at height h from the one below. A particle descends from the top stair with an initial speed U, moving straight downwards, and bounces on each step. The coefficient of restitution between the particle and each stair is denoted as e, with e < 1. Our goal is to derive an expression for the horizontal distance covered between the n-th and (n+1)-th bounces. Additionally, for a very large N and a step length L, we aim to find an expression for U.
Illustrative Diagram
To effectively tackle this classical mechanics issue, I sketched a preliminary diagram. A clear diagram is essential for visualizing the problem.
My illustration highlights the distinction between the horizontal and vertical velocities of the particle as it descends the staircase. Given that the only forces in play are gravity and restitution (the bounce), both acting vertically, we can assert that the particle's horizontal velocity remains constant at U. The vertical velocity at the moment of impact on each step is represented as v_n.
Thus, the crux of the problem lies in two main steps:
- Calculate the time t_n required for the particle to transition from the n-th bounce to the (n+1)-th bounce, focusing on its vertical motion.
- Use this time to determine the horizontal distance d_n between the n-th and (n+1)-th bounces, expressed as U × t_n.
The First Bounce
Initially, our particle has zero vertical velocity but accelerates downwards at a rate of g (approximately 9.8 m/s²). Consequently, the vertical velocity v_1 at the first bounce is:
In this video, we observe a ball bouncing down a staircase, illustrating the concept of motion and the coefficient of restitution.
Integrating this, we can establish the vertical distance traveled, noting that the distance to the first bounce is h. Thus, substituting back into our equation provides an expression for the vertical velocity at the first bounce.
Alternatively, we can apply the motion equation v² = u² + 2as, where a is acceleration and s is the distance. Since the initial velocity u = 0, and the vertical acceleration is g while the distance is h, we arrive at the above expression for v_1.
The Second Bounce
Upon impact, a particle typically loses some kinetic energy, which affects its velocity. This energy loss is encapsulated by the coefficient of restitution. When the particle strikes the surface with vertical velocity v_1, it rebounds with velocity -ev_1, where the negative sign indicates the direction change.
Following this, the particle's upward vertical velocity will be slowed by gravity until it reaches a peak height before descending again. At the height from which it bounced, it regains the speed of ev_1 and continues to accelerate downwards at g until it strikes the next step.
Using the motion equation v² = u² + 2as again, we can express the situation mathematically.
The n-th Bounce
By extending this reasoning, we can formulate a general expression for the vertical velocity at the n-th bounce:
The vertical velocities form a geometric series with n terms, starting from 2gh and having a common ratio of e². We can apply the formula for the sum of the first n terms of a geometric series, leading us to a simplified expression.
Finding the Horizontal Distance
Next, we can determine the time t_n between the n-th and (n+1)-th bounces. The initial vertical velocity after the n-th bounce is -ev_n, and the final velocity at the next impact is v_{n+1}, with g as the acceleration. Using v = u + at, we can derive our equations.
Finally, recognizing that the particle maintains a constant horizontal velocity U, we find:
d_n = U × t_n, yielding our expression for the distance between the n-th and (n+1)-th bounces.
What if There are Infinitely Many Steps?
If there are an infinite number of steps, ensuring the particle only bounces once on each step necessitates that d_n = L. Thus, the distance between bounces must equal the step length.
As n approaches infinity, the sums of the geometric series converge due to the condition that the common ratio e² < 1, allowing us to derive a limiting case.
This leads us to a specific value for U, necessary for the scenario to hold.
Calculating Initial Speeds
Let’s utilize realistic values for L, h, and e to compute the necessary initial speed for the particle to bounce indefinitely down the stairs without missing a step.
# function to calculate initial speed U to 3 decimal places
calculate_U <- function (L, h, e, g = 9.8) {
U <- sqrt((L^2*g)/(2*h))*sqrt((1-e)/(1+e))
paste0("Initial speed U = ", round(U, 3), "m/s" )
}
# stair length 0.5m, stair height 0.5m, coefficient of restitution 0.66
calculate_U(L = 0.5, h = 0.5, e = 0.66)
[1] "Initial speed U = 0.708m/s"
# stair length 0.5m, stair height 0.3m, coefficient of restitution 0.5
calculate_U(L = 0.5, h = 0.3, e = 0.5)
[1] "Initial speed U = 1.167m/s"
I hope you found this exploration of the problem and its solution engaging. Feel free to share any insights or critiques regarding my methodology.
Chapter 2: Additional Insights
As we dive deeper into the dynamics of a bouncing tennis ball, we can observe its behavior in various scenarios.
This video showcases a tennis ball bouncing down a flight of stairs, striking each step sequentially, and rebounding to the height it fell from, providing a visual representation of the principles discussed.