1.8 - Equations of Motion

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Introduction

Newton's Second Law, $\mathbf{F}_\text{net} = m\mathbf{a}$, provides the fundamental relationship between forces and motion. However, to fully understand how an object moves, we need to solve this equation to find the position, velocity, or other quantities as functions of time. When we express Newton's Second Law in terms of derivatives, we obtain a differential equation known as an equation of motion. Later in this course, will discover other ways to generate an equation of motion for a given physical problem. But for now, we will focus on using Newton's Second Law.

Equation of Motion:

An equation of motion is a differential equation that describes how a physical system evolves over time. To generate an equation of motion from Newton's Second Law, we write the acceleration as a second time derivative of the position vector $\mathbf{a} = \ddot{\mathbf{r}}$.

For a particle of mass $m$ subject to a net force $\mathbf{F}_\text{net}$, the most general form of the equation of motion is:

$$m\ddot{r} = \mathbf{F}_\text{net}(\mathbf{r}, \mathbf{v}, t)$$

where $\mathbf{r}$ is the position vector, the double-dot notation signifies a second time derivative, and the force may depend on position, velocity, time, or a combination of these.

In one dimension, the equation of motion becomes a single, second-order differential equation:

$$m\ddot{x} = F_\text{net}(x, v, t)$$
where the net force $F_\text{net}$ can, in general, be a function of the position $x$, the velocity $v$, and the time $t$. The details of the physics problem will determine the specific form of the net force. From there, one often solves the differential equation (either analytically or numerically) to find the position $x(t)$ of the object as a function of time. Sometimes, we may want to solve for $v(t)$ or $v(x)$ instead.

In two dimensions, the equation of motion becomes a system of two second-order differential equations:

$$m\ddot{x} = F_\text{net,x}(x,y, v_x, v_y, t)$$ $$m\ddot{y} = F_\text{net,y}(x,y, v_x, v_y, t)$$
where the net force $F_\text{net,x}$ and $F_\text{net,y}$ can, in general, be a function of the vector position $(x,y)$, the vector velocity $(v_x, v_y)$, and the time $t$. If the $F_\text{net,x}$ only depends on $x$ and $v_x$, and the $F_\text{net,y}$ only depends on $y$ and $v_y$, then we can solve the two differential equations independently, greatly simplifying the problem.

In three dimensions, the equation of motion becomes a system of three second-order differential equations that have a similar form to the two-dimensional case:

$$m\ddot{x} = F_\text{net,x}(x,y, z, v_x, v_y, v_z, t)$$ $$m\ddot{y} = F_\text{net,y}(x,y, z, v_x, v_y, v_z, t)$$ $$m\ddot{z} = F_\text{net,z}(x,y, z, v_x, v_y, v_z, t).$$

Solving the Equation of Motion

Often, when we say we want to solve the equation of motion, we mean solving for the position $x(t)$ of the object as a function of time. However, we may also want to solve for the velocity $v(t)$ or the velocity $v(x)$ instead. In two and higher dimensions, we may want to solve for tragectory of the object, such as $y(x)$.

The approach to solving the equation of motion depends on the functional form of the net force. In the following cases, we will assume we want to find $x(t)$ for one-dimensional motion. Each case describes the general proceedure to follow. When solving your own problems, you should follow all the steps of the proceedure. Don't try to simply plug in your force into the final equation. Doing so will be difficult to follow and will likely lead to mistakes.

Case 1: Net Force depends only on time $F_\text{net} = F_\text{net}(t)$

If the net force is a function of time, our equation of motion becomes:

$$m\ddot{x} = F_\text{net}(t).$$
To solve it, we write the acceleration as a velocity derivative, divide by $m$, and integrate to solve for $v(t)$:
$$\frac{dv}{dt}= \frac{1}{m}F_\text{net}(t).$$
Use the chain rule (or simply multiply both sides by $dt$) and integrate:
$$\int_{v_0}^v dv'= \frac{1}{m}\int_{t_0}^t F_\text{net}(t') dt'.$$
This gives:
$$v(t) = v_0 + \int_0^t \frac{F_\text{net}(t')}{m} dt'$$.
If this integral is easy to evaluate, we can solve for $v(t)$ and then integrate again to find $x(t)$. If not, we may need to use a numerical method to solve for $v(t)$. Once we have $v(t)$, we can integrate again to find $x(t)$. We start with the definition of velocity:
$$v = \frac{dx}{dt}.$$
Multiply both sides by $dt$ and integrate both sides:
$$\int_{x_0}^x dx'= \int_{t_0}^t v(t') dt'.$$
After integrating the left side, we get:
$$x(t) = x_0 + \int_{t_0}^t v(t') dt'.$$
This gives us the desired result: $x(t)$ as a function of $t$.

Case 2: Net Force depends only on position $F_\text{net} = F_\text{net}(x)$

If the net force is a function of position $F_\text{net} = F_\text{net}(x)$ and the force is conservative, then energy will be conserved and we can use conservation of energy to write down the equation of motion. We will see examples of this later in the course.

However, if we wish to proceed with Newton's Second Law, our equation of motion is

$$m\ddot{x} = F_\text{net}(x).$$
The problem is that we have a second-order differential equation in $x$ and $v$, which is difficult to solve. If we replace the acceleration with a velocity derivative, we get a first-order differential equation that contains three variables: $t$, $x$ and $v$:
$$m\frac{dv}{dt} = F_\text{net}(x).$$
To make progress, we must eliminate one of the variables (time) using the following trick. We write the acceleration as a velocity derivative and then use the chain rule to write the velocity as a position derivative:
$$\begin{aligned} a &= \frac{dv}{dt} \\ &= \frac{dx}{dt} \frac{dv}{dx} \\ &= v \frac{dv}{dx} \end{aligned} $$
We now substitute this expression for the acceleration into Newton's Second Law:
$$m v \frac{dv}{dx} = F_\text{net}(x).$$
This gives us a first-order differential equation in $v$ and $x$. We can solve this equation by separating variables and integrating:
$$m \int_{v_0}^v v' dv'= \int_{x_0}^x F_\text{net}(x') dx'.$$
This gives:
$$\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = \int_{x_0}^x F_\text{net}(x') dx'.$$
Notice the left side of the equation is the change in kinetic energy of the object and the right side is the work done by the net force. Thus, we have just derived the work-kinetic energy theorem! We can use this expression to solve for $v(x)$ in terms of $F_\text{net}(x)$.
$$ v(x) = \pm \sqrt{v_0^2 +\frac{ 2}{m}\int_{x_0}^x F_\text{net}(x') dx'}.$$
Once we have $v(x)$, we use the definition of velocity to write:
$$ \frac{dx}{dt} = v(x) $$
Multiply both sides by $dt$, use separation of variables, and integrate:
$$\int_{x_0}^x \frac{dx'}{v(x')}= \int_0^t dt'.$$
This lets us solve for $t(x)$ instead of $x(t)$:
$$t(x) = \int_{x_0}^x \frac{dx'}{v(x')}.$$
Whether we can solve this integral depends on the functional form of $v(x)$. Even if we can solve it, we are not guaranteed that we can invert the solution to get $x(t)$. But at least we have a proceedure to follow if the equations play nice.

Example: Find $v(x)$ for a Spring

Find $v(x)$ for a spring-mass system with force $F(x) = -kx$. Assume the mass starts from rest with initial displacement $x_0$.

Solution:

Use the trick to write the acceleration as $a = v(dv/dx)$: This produces the equation of motion:

$$mv \, dv = -kx \, dx$$

Integrating both sides from initial conditions ($v = 0$ at $x = x_0$) to general values:

$$\int_0^v mv' \, dv' = \int_{x_0}^x -kx' \, dx'$$
$$\frac{1}{2}mv^2 = \frac{1}{2}k(x_0^2 - x^2)$$

Solving for $v(x)$:

$$v(x) = \pm\sqrt{\frac{k}{m}(x_0^2 - x^2)}$$

The $\pm$ sign indicates the velocity can be positive or negative depending on the direction of motion.

Alternate Solution: Use Conservation of Energy

Since we know the spring force is conservative, we can use conservation of energy to solve for $v(x)$. The potential energy is $U = \frac{1}{2}kx^2$, the kinetic energy is $K = \frac{1}{2}mv^2$ and the total energy is $E=K+U$. The total energy is conserved, so $E_\text{final}=E_\text{init}$

$$U + K = U_0 + K_0$$

Substituting expressions for the kinetic and potentials energies with $v_0=0$ gives:

$$ \frac{1}{2}kx^2 + \frac{1}{2}mv^2 = \frac{1}{2}kx_0^2 $$

Solving for the velocity gives the same result as above:

$$v(x) = \pm\sqrt{\frac{k}{m}(x_0^2 - x^2)}$$
Note: If you want to solve for $x(t)$, you can continue following the proceedure for Case 2.

Case 3: Net Force depends only on velocity $F_\text{net} = F_\text{net}(v)$

If the net force is a function of velocity, our equation of motion becomes:

$$m\ddot{x} = F_\text{net}(v).$$
To solve it, we write the acceleration as a velocity derivative and use separation of variables to integrate:
$$m\frac{dv}{dt}= F_\text{net}(v)$$ $$m\int_{v_0}^v\frac{dv'}{F_\text{net}(v')}= \int_0^t dt'.$$
This gives $t(v)$:
$$t(v) =m \int_{v_0}^v \frac{dv'}{F_\text{net}(v')}.$$
If you can invert the solution to get $v(t)$, you can then integrate again to find $x(t)$.
$$\frac{dx}{dt}= v(t)$$
Integrating both sides gives $x(t)$:
$$x(t) = x_0 + \int_0^t v(t') dt'.$$

Example: Find $x(t)$ for a velocity-dependent drag force.

Because Chapter 2 extensively covers this problem, we will not solve it here. Please refer to Chapter 2.

Factors in Choosing a Solution Approach:
  • What the problem asks for: If asked for position vs. time, solve for $x(t)$; if asked for speed at a location, solve for $v(x)$.
  • Form of the force:
    • If $F = F(t)$ only (time-dependent), solve for $v(t)$ or $x(t)$ directly
    • If $F = F(v)$ only (velocity-dependent), solve for $v(t)$ first, then integrate for $x(t)$
    • If $F = F(x)$ only (position-dependent), consider solving for $v(x)$ using the energy method
    • If $F = F(x,v)$ (coupled), may need numerical methods or special techniques
  • Complexity: Sometimes solving for $v(x)$ first and then finding $t(x)$ is easier than solving for $x(t)$ directly.
  • Physical insight: Energy methods ($v(x)$) provide physical insight about turning points and motion ranges.

General Solution Strategy

Here's a general flowchart for approaching equations of motion:

General Workflow:
  1. Write down Newton's Second Law: $m\frac{d^2\mathbf{r}}{dt^2} = \mathbf{F}_\text{net}$
  2. Identify the form of the force: Does it depend on $t$, $x$, $v$, or combinations?
  3. Determine what you need to find: $x(t)$, $v(t)$, $v(x)$, etc.
  4. Choose the appropriate method:
    • Direct integration if simple
    • Separation of variables if the equation can be separated
    • Energy method if the force is conservative
    • Numerical methods if analytical solution is difficult
  5. Apply initial conditions: Use given initial position and velocity
  6. Solve and verify: Check that your solution satisfies the original equation and initial conditions

Summary

Newton's Second Law, $\mathbf{F}_\text{net} = m\mathbf{a}$, provides the foundation for generating equations of motion. By expressing acceleration as $\frac{d^2\mathbf{r}}{dt^2}$ or $\frac{d\mathbf{v}}{dt}$, we obtain differential equations that describe the system's evolution. The choice of solving for $x(t)$, $v(t)$, $v(x)$, or $t(x)$ depends on the problem requirements, the form of the force, and which approach is most mathematically tractable. Understanding these different perspectives allows us to tackle a wide variety of mechanical problems effectively.