Question

An object of mass \(m\) falls from rest under gravity subject to an air resistance proportional to its velocity. Taking the \(y\) axis as positive down, show that the differential equation of motion is \(m(d v / d t)=m g-k v\), where \(k\) is a positive constant. Find \(v\) as a function of \(t\), and find the limiting value of \(v\) as \(t\) tends to infinity; this limit is called the terminal relocity. Can you find the terminal velocity directly from the differential equation without solving it? Hint: What is \(d v / d t\) after \(v\) has reached an essentially constant value? Consider the following specific examples of this problem. (a) A person drops from an airplane with a parachute. Find a reasonable value of \(k\). (b) In the Millikan oil drop experiment to measure the charge of an electron, tiny electrically charged drops of oil fall through air under gravity or rise under the combination of gravity and an electric field. Measurements can be made only after they have reached terminal velocity. Find a formula for the time required for a drop starting at rest to reach \(99 \%\) of its terminal velocity.

Short answer

The differential equation is \( m \frac{dv}{dt} = mg - kv \). The solution is \( v(t) = \frac{mg}{k}(1 - e^{-\frac{k}{m}t}) \). Terminal velocity is \( \frac{mg}{k} \).

Step-by-step solution

Set Up the Differential Equation of Motion

The forces acting on the object are gravity, which is downward, and air resistance, which is proportional to the velocity and acts in the opposite direction of motion. Using Newton's second law, we write the net force as: downward force due to gravity: \[ F_{gravity} = mg \] upward force due to air resistance: \[ F_{air} = -kv \] The net force acting on the object is therefore: \[ F_{net} = mg - kv \] Using Newton's second law, \[ F_{net} = ma \], where \[ a = \frac{dv}{dt} \]. Substituting in, we get: \[ m \frac{dv}{dt} = mg - kv \] This is the differential equation of motion.

Solving for Velocity as a Function of Time

To solve the differential equation, we first rearrange it: \[ m \frac{dv}{dt} + kv = mg \] Dividing through by m gives: \[ \frac{dv}{dt} + \frac{k}{m}v = g \] This is a first-order linear differential equation. We can solve it using the integrating factor method.

Finding the Integrating Factor

The integrating factor \( \mu(t) \) is given by: \[ \mu(t) = e^{\int \frac{k}{m} dt} = e^{\frac{k}{m}t} \] We multiply through the differential equation by the integrating factor: \[ e^{\frac{k}{m}t} \frac{dv}{dt} + \frac{k}{m}e^{\frac{k}{m}t}v = ge^{\frac{k}{m}t} \] The left-hand side is the derivative of \( e^{\frac{k}{m}t}v \): \[ \frac{d}{dt} \left( e^{\frac{k}{m}t} v \right) = ge^{\frac{k}{m}t} \] We integrate both sides with respect to \( t \): \[ e^{\frac{k}{m}t} v = \int ge^{\frac{k}{m}t} dt \]

Integrating and Solving for v(t)

The right-hand side integral is: \[ \int ge^{\frac{k}{m}t} dt = \frac{mg}{k}e^{\frac{k}{m}t} + C \] Thus, \[ e^{\frac{k}{m}t} v = \frac{mg}{k} e^{\frac{k}{m}t} + C \] Dividing by \( e^{\frac{k}{m}t} \): \[ v = \frac{mg}{k} + Ce^{-\frac{k}{m}t} \] Using the initial condition that the object starts from rest, \( v(0) = 0 \): \[ 0 = \frac{mg}{k} + C \Rightarrow C = -\frac{mg}{k} \] Therefore, the solution is: \[ v(t) = \frac{mg}{k} \left( 1 - e^{-\frac{k}{m}t} \right) \]

Finding the Terminal Velocity

The limiting value of \( v \) as \( t \to \infty \) is: \[ v_{terminal} = \lim_{t \to \infty} \frac{mg}{k} \left( 1 - e^{-\frac{k}{m}t} \right) = \frac{mg}{k} \] This is the terminal velocity.

Finding Terminal Velocity Directly

To find the terminal velocity directly, observe that at terminal velocity, the object is no longer accelerating, so \( \frac{dv}{dt} = 0 \). Substituting into the differential equation: \[ m \frac{dv}{dt} = mg - kv \Rightarrow 0 = mg - kv_{terminal} \Rightarrow v_{terminal} = \frac{mg}{k} \]

Calculating a Reasonable Value of k (Example A)

For a person dropping with a parachute, let's assume a mass of \( 80 \ kg \). The terminal velocity for a parachuting person is roughly \( 5.5 \ m/s \). Using the terminal velocity formula: \[ \frac{mg}{k} = 5.5 \Rightarrow k = \frac{80 \times 9.8}{5.5} = 142.18 \ kg/s \]

Time to Reach 99% of Terminal Velocity (Example B)

We know that \( v(t) = v_{terminal} \left( 1 - e^{-\frac{k}{m}t} \right) \). To find the time required to reach \( 99% \) of terminal velocity: \[ 0.99 \left( \frac{mg}{k} \right) = \frac{mg}{k} \left( 1 - e^{-\frac{k}{m}t} \right) \Rightarrow 0.99 = 1 - e^{-\frac{k}{m}t} \Rightarrow e^{-\frac{k}{m}t} = 0.01 \] Taking the natural logarithm of both sides: \[ -\frac{k}{m}t = \ln(0.01) \Rightarrow t = -\frac{m}{k} \ln(0.01) \] Simplifying, we get: \[ t = \frac{4.605}{k/m} = \frac{4.605 m}{k} \]

Key concepts

Velocity Under Gravity

When an object falls under the influence of gravity, it accelerates downward due to the gravitational force. This force can be calculated using Newton's second law of motion, which states that the force acting on an object is equal to its mass times its acceleration. In mathematical terms:

• Gravity force, \(\text{F}_{\text{gravity}} = m \times g\)
• Where \(m\) is the mass of the object and \(g\) is the acceleration due to gravity (\(\text{approx.} 9.8 \text{ m/s}^2\)).

As the object accelerates, it encounters air resistance which acts in the opposite direction. This resistance increases with velocity, leading to a net force described by:

• Net force, \(\text{F}_{\text{net}} = m \times g - k \times v\)
• Where \(k\) is a proportional constant for air resistance, and \(v\) is the object's velocity.

Thus, the differential equation of motion considering both forces is given by:

• \(m \frac{dv}{dt} = mg - kv\)

Terminal Velocity

As an object falls, the air resistance against it gradually increases. At a certain speed, the force due to air resistance equals the gravitational force. When these forces are balanced, the object no longer accelerates and falls at a constant speed. This constant speed is known as terminal velocity.
To find terminal velocity mathematically, set the acceleration to zero (since the velocity is constant) in the differential equation:

• \(0 = mg - kv_{\text{terminal}}\)

Solving this gives:

• \(v_{\text{terminal}} = \frac{mg}{k}\)

Terminal velocity depends on both the mass of the object and the resistance constant. In example A from the exercise, for a parachutist with a mass of 80 kg, a reasonable value for terminal velocity might be 5.5 m/s. Using this, we can calculate \(k\) as:

• \(k = \frac{80 \times 9.8}{5.5} = 142.18 \text{ kg/s}\)

Integration Factor Method

The integration factor method is a technique used to solve linear first-order differential equations of the form \[ \frac{dy}{dx} + P(x) y = Q(x) \].
In this approach, we multiply through the differential equation by an integrating factor, \( \: \mu(x) = e^{\int P(x) dx}\).
For the given equation \(m \frac{dv}{dt} + kv = mg\), dividing by \(m\):

• \( \frac{dv}{dt} + \frac{k}{m} v = g \)
• Here, \( P(t) = \frac{k}{m}\)

The integrating factor is:

• \( \mu(t) = e^{\int \frac{k}{m} dt} = e^{\frac{k}{m} t} \)

Multiplying through the equation by \(e^{\frac{k}{m}t}\) gives us a form where the left-hand side becomes the derivative of \( e^{\frac{k}{m}t}v\). Integrating both sides, we can then solve for \(\text{v(t)}\):

• \(v(t) = \frac{mg}{k} \big( 1 - e^{- \frac{k}{m} t} \big)\)

Newton's Second Law

Newton's second law of motion forms the foundation for understanding the differential equation of motion in this problem. It states that the force on an object is equal to the mass of the object multiplied by its acceleration: \( \text{F} = m \times \text{a} \).
In the context of the exercise:

• Downward force (gravity): \( \text{F}_{\text{gravity}} = mg \)
• Upward force (air resistance): \( \text{F}_{\text{air}} = -kv \)
• Net force: \( \text{F}_{\text{net}} = m \frac{dv}{dt} = mg - kv \)

Using Newton's second law, we can derive the differential equation \( m \frac{dv}{dt} = mg - kv\). This equation describes the motion of an object falling under gravity with air resistance considered. It is essential to grasp Newton's second law to understand how forces result in the acceleration or velocity of objects in motion with varying forces acting upon them.