Problem 86 - Calculus Single Variable

Question

Suppose that the force \(R\) of air resistance on a \(0.2 \mathrm{~kg}\) object thrown straight up with initial velocity \(28 \mathrm{~m} / \mathrm{s}\) is given by the quadratic drag law, $$ R(v)=-k v^{2} $$ where \(k=4 \times 10^{-6} \mathrm{~kg} / \mathrm{m}\). Find an equation for the time \(\tau\) at which the object begins to fall.

Step-by-Step Solution

Answer

The time \( \tau \) at which the object starts falling is approximately given by \( \tau = \frac{0.2}{1.96} \cdot \ln\left(\frac{1.96}{1.96}\right), a detailed evaluation is needed.

Step 1: Identify Forces

Identify the forces acting on the object. These include the gravitational force, which is defined as \[ F_g = m \cdot g \]and the air resistance, given by\[ R(v) = -k \cdot v^2 \]where \( m = 0.2 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). The gravitational force is \( F_g = 0.2 \times 9.8 = 1.96 \text{ N} \).

Step 2: Set up Net Force Equation

Using Newton's second law, the net force on the object is given by\[ m \cdot a = F_g + R(v) \]where the air resistance \( R(v) = -4 \times 10^{-6} \cdot v^2 \). Thus the equation becomes \[ 0.2 \cdot a = 1.96 - 4 \times 10^{-6} \cdot v^2 \].

Step 3: Solve for Acceleration

Simplify and rewrite the equation to solve for acceleration \( a \): \[ a = \frac{1.96 - 4 \times 10^{-6} \cdot v^2}{0.2} \]. This specifies how the velocity changes over time.

Step 4: Recognize Zero Velocity Condition

At the peak of the motion, the velocity \( v = 0 \). Solve for the condition when the object begins to fall by setting \( a = 0 \) (as it's starting to accelerate downwards). At this point, \[ 0 = \frac{1.96 - 4 \times 10^{-6} \cdot 0^2}{0.2}\]. This confirms equilibrium at zero net force when velocity is zero.

Step 5: Integrate to Find Time

To find the time \( \tau \) when the object begins to fall, integrate \[ \frac{dv}{dt} = \frac{1.96 - 4 \times 10^{-6} \cdot v^2}{0.2} \]from the initial velocity \( v(0) = 28 \text{ m/s} \) to \( v(\tau) = 0 \). Solve this differential equation for time \( \tau \).

Step 6: Compute \( \tau \) Using Calculus

This involves a separable variable method:\[ \int_{28}^{0} \frac{0.2}{1.96 - 4 \times 10^{-6} \cdot v^2} \, dv = \int_{0}^{\tau} dt \]. Evaluating this integral yields \( \tau \). Conclude with the equation for \( \tau \):\[ \tau = \frac{0.2}{1.96} \cdot \ln\left(\frac{1.96}{1.96 - 4 \times 10^{-6} \cdot 28^2}\right) \].

Key Concepts

Quadratic Drag Law

The quadratic drag law describes the force of air resistance acting on an object moving through the air. It is particularly applicable in cases where an object is traveling at high speeds or through a dense fluid. The law is represented by the equation:

• \[ R(v) = -k v^2 \]

where:

• \( R(v) \) is the air resistance (force of drag),
• \( k \) is the drag coefficient, which depends on properties of the object and the fluid,
• \( v \) is the velocity of the object.

This formula indicates that the air resistance increases proportionally with the square of the velocity. In other words, as the object speeds up, the air resistance grows
quickly. For the problem at hand, the value of the drag coefficient, \( k \), is given as \( 4 \times 10^{-6} \text{ kg/m} \). Substituting this into the drag law helps us calculate the decelerating force acting against the object's motion while it is in the air.

Air Resistance

Air resistance, also known as drag, is the force opposing an object's motion through the air. It acts in the opposite direction to the velocity of the object.
In the given example, air resistance plays a crucial role in determining when and how an object moves. While the quadratic drag law portrays how this force scales with velocity, it's essential to understand its impact combined with other forces. Air resistance balances with gravitational force, leading to scenarios like terminal velocity where objects cease accelerating and
descend at a constant speed. Analyzing scenarios in air resistance begins with identifying the drag coefficient, which characterizes the influence of shape, surface texture, and air properties. Understanding air resistance aids in predicting when the object's motion changes, like reaching its peak or descending back due to gravity overpowering the air drag.

Gravitational Force

Gravitational force is the constant force pulling objects towards the center of the Earth. It acts on all objects with mass and is calculated using the formula:

• \[ F_g = m \cdot g \]

where:

• \( F_g \) is the gravitational force,
• \( m \) is the mass of the object,
• \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \).

In the exercise, the gravitational force is a major factor for the object thrown upwards, calculated as \( F_g = 0.2 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1.96 \, \text{N} \). Subtracting the counteracting air resistance gives us the net force affecting the motion,
dictating acceleration through Newton’s second law \( F = ma \). Understanding gravitational force is fundamental to know when the object will cease its upward travel and begin its descent.