The 2-lb Box Slides On The Smooth Circular Ramp.

The graceful descent of a 2-lb box down a smooth circular ramp presents a captivating problem in classical mechanics, uniting concepts of gravitational potential energy, kinetic energy, and constrained motion. Analyzing this seemingly simple scenario allows us to explore the interplay of these fundamental principles and provides insights into the dynamics of curvilinear motion.

Understanding the Setup

Imagine a small box, weighing 2 pounds, resting at the top of a circular ramp. The ramp is perfectly smooth, meaning we can neglect friction. As the box begins to slide, gravity acts upon it, pulling it downwards. However, the ramp constrains the box's motion, forcing it to follow a circular path. This combination of gravity and constraint results in a fascinating dynamic behavior. To fully understand the box's movement, we need to consider several key factors:

• Gravitational Potential Energy (GPE): The box initially possesses potential energy due to its height above a reference point (typically the bottom of the ramp). As it descends, this potential energy is converted into kinetic energy.
• Kinetic Energy (KE): As the box gains speed, it accumulates kinetic energy, which is the energy of motion. This energy is directly related to the box's velocity.
• Conservation of Energy: Since we're dealing with a smooth ramp (no friction), the total mechanical energy (GPE + KE) of the box remains constant throughout its motion. This principle of conservation of energy is crucial for solving the problem.
• Centripetal Force: As the box moves along the circular path, it experiences a centripetal force directed towards the center of the circle. This force is what keeps the box moving in a circle rather than a straight line. The ramp provides this force.
• Normal Force: The ramp exerts a normal force on the box, perpendicular to the surface of the ramp. This force is a reaction to the component of the box's weight that is perpendicular to the ramp's surface.

Defining the Variables

Before diving into the calculations, let's define the variables we'll be using:

• m: Mass of the box (in slugs, since we're using pounds as the unit of weight)
• g: Acceleration due to gravity (approximately 32.2 ft/s²)
• R: Radius of the circular ramp
• θ: Angle between the vertical and the radial line connecting the center of the circle to the box's current position (measured in radians)
• v: Velocity of the box at a given angle θ
• N: Normal force exerted by the ramp on the box

Applying Conservation of Energy

The principle of conservation of energy is the cornerstone of our analysis. We can express it mathematically as:

Initial Energy = Final Energy

At the start, the box is at rest at the top of the ramp. Its initial kinetic energy is zero, and its initial potential energy is mgR (mass * gravity * height). As the box slides down to a position at angle θ, its potential energy decreases to mgRcos(θ), and its kinetic energy increases to (1/2)*mv²:

mgR = mgRcos(θ) + (1/2)*mv²

We can simplify this equation by dividing both sides by m and rearranging to solve for v²:

v² = 2gR(1 - cos(θ))

This equation gives us the velocity of the box at any angle θ as it slides down the ramp.

Analyzing the Forces

To determine the normal force exerted by the ramp on the box, we need to consider the forces acting on the box in the radial direction (the direction pointing from the box to the center of the circle). These forces are:

• Component of Weight: The component of the box's weight acting along the radial direction is mgcos(θ). This component acts inwards, towards the center of the circle.
• Normal Force: The normal force, N, exerted by the ramp acts outwards, away from the center of the circle.

The net force in the radial direction must equal the centripetal force required to keep the box moving in a circle:

N - mgcos(θ) = mv²/R

Now, we can substitute the expression for v² that we derived earlier:

N - mgcos(θ) = m(2gR(1 - cos(θ)))/R

Simplifying this equation, we get:

N - mgcos(θ) = 2mg(1 - cos(θ))

Finally, we can solve for the normal force N:

N = mg(3cos(θ) - 2)

This equation tells us how the normal force changes as the box slides down the ramp.

Determining the Point of Departure

A critical question arises: at what angle θ does the box lose contact with the ramp? This happens when the normal force N becomes zero. Intuitively, this makes sense – the ramp can only exert a pushing force (normal force), not a pulling force. When the required normal force becomes zero, the box is no longer constrained by the ramp and will detach.

Setting N = 0 in the equation for the normal force:

0 = mg(3cos(θ) - 2)

Dividing by mg:

0 = 3cos(θ) - 2

Solving for cos(θ):

cos(θ) = 2/3

Therefore, the angle at which the box leaves the ramp is:

θ = arccos(2/3) ≈ 48.19 degrees

This means the box will detach from the ramp when it has descended approximately 48.19 degrees from the vertical.

Post-Departure Trajectory

Once the box leaves the ramp, it becomes a projectile. Its subsequent motion is governed by the laws of projectile motion, influenced only by gravity. To determine its trajectory, we need to know its velocity at the point of departure. We already have the equation for velocity as a function of angle:

v² = 2gR(1 - cos(θ))

Substituting θ = arccos(2/3):

v² = 2gR(1 - 2/3) = (2/3)*gR

So, the velocity at the point of departure is:

v = sqrt((2/3)*gR)

At the point of departure, the box has both a horizontal and a vertical component of velocity. Let's call these v_x and v_y:

• v_x = vsin(θ) = sqrt((2/3)*gR) * sin(arccos(2/3))
• v_y = vcos(θ) = sqrt((2/3)*gR) * cos(arccos(2/3)) = sqrt((2/3)*gR) * (2/3)

To find the horizontal distance the box travels after leaving the ramp, we need to determine the time it spends in the air. This time depends on its initial vertical velocity and the acceleration due to gravity. We can use the following kinematic equation:

Δy = v_yt + (1/2)at²

Where:

• Δy is the vertical displacement (the height from the point of departure to the ground, which is R*cos(arccos(2/3)) = (2/3)*R)
• v_y is the initial vertical velocity (sqrt((2/3)*gR) * (2/3))
• a is the acceleration due to gravity (-g, negative since it acts downwards)
• t is the time in the air

Substituting these values, we get a quadratic equation in terms of t:

(2/3)*R = sqrt((2/3)*gR) * (2/3) * t - (1/2)gt²

Solving this quadratic equation for t (using the quadratic formula) will give us the time the box spends in the air.

Once we have the time, we can calculate the horizontal distance traveled by the box:

Δx = v_xt = sqrt((2/3)*gR) * sin(arccos(2/3)) * t

This will tell us how far the box lands from the point where it left the ramp.

Numerical Example

Let's consider a concrete example to illustrate these concepts. Assume the radius of the ramp, R, is 2 feet. We'll continue using g = 32.2 ft/s² and a box weight of 2 lbs. Remember to convert the weight to mass by dividing by g: m = 2/32.2 slugs.

First, let's find the velocity of the box at θ = 30 degrees (π/6 radians):

v² = 2 * 32.2 * 2 * (1 - cos(π/6)) ≈ 2 * 32.2 * 2 * (1 - 0.866) ≈ 17.25

v ≈ sqrt(17.25) ≈ 4.15 ft/s

Next, let's calculate the normal force at the same angle:

N = (2/32.2) * 32.2 * (3 * cos(π/6) - 2) ≈ 2 * (3 * 0.866 - 2) ≈ 1.196 lbs

As expected, the normal force is positive, indicating that the ramp is pushing on the box.

We already calculated the angle at which the box leaves the ramp: θ = arccos(2/3) ≈ 48.19 degrees. At this point, the velocity is:

v² = (2/3) * 32.2 * 2 ≈ 42.93

v ≈ sqrt(42.93) ≈ 6.55 ft/s

The horizontal and vertical components of velocity at the point of departure are:

• v_x = 6.55 * sin(arccos(2/3)) ≈ 6.55 * 0.745 ≈ 4.88 ft/s
• v_y = 6.55 * (2/3) ≈ 4.37 ft/s

Now, we solve for the time of flight t using the quadratic equation:

(2/3)*2 = 4.37 * t - (1/2) * 32.2 * t²

1. 33 = 4.37 * t - 16.1 * t²

2. 1 * t² - 4.37 * t + 1.33 = 0

Using the quadratic formula, we get two possible values for t:

t ≈ (4.37 ± sqrt((-4.37)² - 4 * 16.1 * 1.33)) / (2 * 16.1) ≈ (4.37 ± sqrt(19.09 - 85.65)) / 32.2

Since the discriminant is negative, there's an error in the calculation. We need to re-examine the problem setup. The initial y-position of the box when it leaves the ramp is R*cos(θ) and it falls to a y-position of zero. Therefore, Δy = -R*cos(θ) = -(2/3)*R. The equation becomes:

-(2/3)*R = v_y*t - (1/2)*g*t^2

-(2/3)*2 = 4.37*t - (1/2)*32.2*t^2

-1.33 = 4.37*t - 16.1*t^2

16.1*t^2 - 4.37*t - 1.33 = 0

t = (4.37 +/- sqrt(4.37^2 - 4*16.1*(-1.33)))/(2*16.1)

t = (4.37 +/- sqrt(19.0969 + 85.652))/32.2

t = (4.37 +/- sqrt(104.7489))/32.2

t = (4.37 +/- 10.2347)/32.2

We take the positive root for t:

t = (4.37 + 10.2347)/32.2 = 14.6047/32.2 ≈ 0.454 seconds

Finally, the horizontal distance traveled is:

Δx = 4.88 * 0.454 ≈ 2.22 feet

Therefore, the box lands approximately 2.22 feet horizontally from the point where it left the ramp.

Factors Affecting the Trajectory

Several factors influence the box's trajectory:

• Radius of the Ramp (R): A larger radius will result in a higher velocity at the point of departure and a greater horizontal distance traveled.
• Acceleration due to Gravity (g): This constant directly affects both the velocity of the box and the time it spends in the air.
• Initial Conditions: Starting the box from a different point on the ramp (not the very top) would change the initial potential energy and consequently affect the entire trajectory.
• Friction (in a Non-Ideal Scenario): Introducing friction would complicate the analysis significantly. Friction would dissipate energy, reducing the box's velocity and altering its trajectory. The conservation of energy principle would no longer be directly applicable, requiring a more complex treatment using work-energy theorem.

Conclusion

The seemingly simple problem of a 2-lb box sliding down a smooth circular ramp reveals a rich interplay of fundamental physics principles. By applying conservation of energy and analyzing the forces acting on the box, we can accurately predict its velocity, the normal force exerted by the ramp, and the point at which the box loses contact with the ramp. Furthermore, by understanding projectile motion, we can determine the box's trajectory after it leaves the ramp. This analysis provides valuable insights into the dynamics of constrained motion and highlights the power of classical mechanics in describing and predicting the behavior of physical systems. This understanding can be extended to various real-world applications, such as designing safe and efficient roller coasters, analyzing the motion of objects in curved paths, and understanding the dynamics of rotating machinery.