Question

A falling object of mass $m$ can reach terminal velocity when the drag force is proportional to its velocity, with proportionality constant $k.$ Set up the differential equation and solve for the velocity given an initial velocity of 0 .

Short answer

The velocity of the object is $v=\frac{mg}{k}(1-e^{-\frac{kt}{m}})$.

Step-by-step solution

Understanding Forces

When an object falls, it experiences two main forces: the gravitational force pulling it downwards, and the drag force opposing its motion due to air resistance. The gravitational force is given by $F_g=mg$, where $m$ is the mass and $g$ is the acceleration due to gravity. The drag force is given by $F_d=kv$, where $k$ is the proportionality constant and $v$ is the velocity.

Setting Up the Differential Equation

The net force on the object can be calculated using Newton's second law: $F_{extnet}=mg-kv=m\frac{dv}{dt}$. The drag acts in the opposite direction of motion. Thus, the differential equation is $m\frac{dv}{dt}=mg-kv$.

Rearranging the Differential Equation

To solve the differential equation, rearrange it to $\frac{dv}{dt}=g-\frac{k}{m}v$. Here, the term $\frac{k}{m}$ is the drag coefficient per unit mass.

Separation of Variables

Separate variables to integrate: $\frac{dv}{g-\frac{k}{m}v}=dt$. We need to integrate both sides, with the left side in terms of $v$ and right side in terms of $t$.

Integrating Both Sides

Integrate the left side with respect to $v$ and the right side with respect to $t$. The left integral is $\int \frac{dv}{g-\frac{k}{m}v}=-\frac{m}{k}\ln |g-\frac{k}{m}v|$, and the right integral is $\int dt=t+C$, where $C$ is the integration constant.

Solving for the Constant

With initial conditions, when $t=0$ and $v=0$, substitute into the integrated solution: $-\frac{m}{k}\ln |g|=C$. This leads to $C=-\frac{m}{k}\ln |g|$.

Finding the Velocity Function

Substitute $C$ back and solve for $v$: $-\frac{m}{k}\ln |g-\frac{k}{m}v|=t-\frac{m}{k}\ln |g|$. Rearranging terms gives $\ln \left|\frac{g}{g-\frac{k}{m}v}\right|=\frac{-kt}{m}$. Finally, solving for $v$: $v=\frac{mg}{k}(1-e^{-\frac{kt}{m}})$.

Conclusion

The expression $v=\frac{mg}{k}(1-e^{-\frac{kt}{m}})$ describes how the velocity of the object approaches the terminal velocity $\frac{mg}{k}$ over time.

Key concepts

Terminal Velocity

Terminal velocity is an important concept when studying objects moving through a fluid, like air. It occurs when the force of gravity pulling the object downward is balanced by the drag force pushing upward.
This means the object stops accelerating and continues to fall at a constant speed. The point when this happens is known as the terminal velocity.

For terminal velocity, the equation $mg=kv$ is used, where:

• $mg$ is the gravitational force.
• $kv$ is the drag force, with $k$ as the drag coefficient and $v$ as the velocity.

Solving for $v$ gives terminal velocity $v_t=\frac{mg}{k}$. Understanding terminal velocity helps in predicting how fast objects fall under the influence of air resistance.

Drag Force

The drag force is the resistance force acting opposite to the relative motion of an object moving through a fluid. It's important in understanding how objects fall through the air.
This force is what slows down the object, alongside the gravitational pull. The formula for calculating drag force is $F_d=kv$, where:

• $k$ is a proportionality constant that depends on the characteristics of the fluid and the shape of the object.
• $v$ is the velocity of the object.

The drag force increases as the velocity increases, which is why objects don't continue to accelerate indefinitely. Instead, they reach a point where drag force equals gravitational force, leading to terminal velocity. Understanding drag force is essential for predicting and analyzing the motion of falling objects.

Newton's Second Law

Newton's second law of motion forms the foundation for setting up differential equations related to motion. It relates the net force acting on an object to its acceleration with the formula $F_{net}=ma$.
When discussing falling objects, this can be expanded to account for gravity and drag as follows:

• Net force $=mg-kv=m\frac{dv}{dt}$
• $m\frac{dv}{dt}$ represents the rate of change of velocity with respect to time, which is the acceleration.

This equation, derived from Newton's second law, is a differential equation that describes the motion of the object. Solving this equation gives us insight into how the velocity of the object changes over time.

Integration Technique

Solving differential equations often involves applying integration techniques, which are crucial in finding solutions to these equations. Integration helps in determining the exact relationship between variables over time.
In the context of our problem, solving the equation involves:

• Separating variables: rearrange terms to isolate those dependent on velocity and time.
• Integrating both sides: \( \int \frac{dv}{g - \frac{k}{m}v} = \int dt \)
• Resulting in solutions that describe how the velocity changes: \
  $$-\frac{m}{k}\ln |g-\frac{k}{m}v|=t+C$$

With initial conditions applied, we can solve for specific values and constants. Hence, integration techniques are indispensable for solving the differential equations that arise in physics problems, like those involving motion and resistant forces.