7.3: Two-Dimensional Kinematics with Rectangular Coordinates

Two-dimensional motion (also called planar motion) is any motion in which the objects being analyzed stay in a single plane. When analyzing such motion, we must first decide the type of coordinate system we wish to use. The most common options in engineering are rectangular coordinate systems, normal-tangential coordinate systems, and polar coordinate systems. Any planar motion can potentially be described with any of the three systems, though each choice has potential advantages and disadvantages.

The rectangular coordinate system (also sometimes called the Cartesian coordinate system) is the most intuitive approach to describing motion. In rectangular coordinate systems we have an \(x\)-axis and a \(y\)-axis. These axes remain fixed to some origin point in the environment and they do not change over time. Instead, the bodies we are analyzing usually move relative to these fixed axes. An example of a body with a rectangular coordinate systems is shown in the figure below.

Figure \(\PageIndex{1}\): In the rectangular coordinate system we have a fixed origin point at \(o\), the particle at point \(p\), and the \(x\) and \(y\) directions, which must be perpendicular to one another. The vector \(\vec{r}\) is the vector going from \(o\) to \(p\). The component of this vector in the \(x\) direction is \(x\) and the component of this vector in the \(y\) direction is \(y\). We usually describe position in terms of \(x\) and \(y\) at any given point in time. The vectors \(\hat{i}\) and \(\hat{j}\) represent unit vectors (vectors with a length of one) in the \(x\) and \(y\) directions respectively.

Rectangular coordinates work best for systems where all forces maintain a constant direction. The most common example of this is projectile motion, where gravity (the only force in these systems) maintains a constant downward direction. An example of a system where the forces change direction over time would be something like a car going around a curve in the road. In this case, the friction force at the tires is going to be rotating with the car. The car problem will therefore be better suited to the use of normal-tangential or polar coordinate systems.

When describing the position of a point in rectangular coordinate systems, we are going to start by describing both \(x\) and \(y\) coordinates in a vector form. For this, the values \(x\) and \(y\) represent distances and the unit vectors \(\hat{i}\) and \(\hat{j}\) are used to indicate which distance corresponds with which direction. This may seem redundant, but remember when solving actual problems, \(x\) and \(y\) will just be numbers.

\[ \text{Position:} \quad r_{p/o} (t) = x(t) \ \hat{i} + y(t) \ \hat{j} \]

Just like with one-dimensional problems, if we take the derivative of the position equation, we will find the velocity equation. If we take the derivative of the velocity equation we will wind up with the acceleration equation. Also like one-dimensional problems, we can use integration to move in the other direction, moving from an acceleration equation to a velocity equation to a position equation.

The unit vectors add a new element in two dimensions, but since the unit vectors don't change over time (they are constants), we treat them like we would any other constant for derivatives and integrals. The resulting velocity and acceleration equations are as follows.

\begin{align} \text{Velocity:} \quad &\, v(t) = \dot{x}(t) \ \hat{i} + \dot{y}(t) \ \hat{j} \\[5pt] \text{Acceleration:} \quad &\, a(t) = \ddot{x}(t) \ \hat{i} + \ddot{y}(t) \ \hat{j} \end{align}

The above equations are vector equations with velocities and accelerations broken down into \(x\) and \(y\) components. Since the \(x\) and \(y\) directions are perpendicular, they are also independent (movement in the \(x\) direction doesn't impact movement in the \(y\) direction, and vice versa). This essentially means we can split our vector equation into a set of two scalar equations. To do this we just put everything in front of the \(\hat{i}\) unit vectors in the \(x\) equations and everything in front of the \(\hat{j}\) unit vectors in the \(y\) equations.

\begin{align} \text{Position:} \quad &\, x(t) = \dots \quad\quad y(t) = \dots \\[5pt] \text{Velocity:} \quad &\, \dot{x}(t) = \dots \quad\quad \dot{y}(t) = \dots \\[5pt] \text{Acceleration:} \quad &\, \ddot{x}(t) = \dots \quad\quad \ddot{y}(t) = \dots \end{align}

Once we have everything split into \(x\)- and \(y\)-direction equations, we can just use the same processes we used for one dimensional motion to move from \(x\) to \(\dot{x}\) to \(\ddot{x}\), and from \(y\) to \(\dot{y}\) to \(\ddot{y}\). The variable linking the two equations is time \((t)\).

Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source: https://youtu.be/qKqUNuSPWT4.

Example \(\PageIndex{1}\)

A motorcycle drives off a one-meter-tall ramp at an angle of 30 degrees as shown below. Determine the equations for the acceleration, velocity, and position over time. How far does the motorcycle travel in the \(x\) direction before hitting the ground?

Figure \(\PageIndex{2}\): problem diagram for Example \(\PageIndex{1}\). A motorcycle drives off the edge of a ramp with an initial velocity of 22 m/s at 30° above the \(x\)-axis.

Solution

Video \(\PageIndex{2}\): Worked solution to example problem \(\PageIndex{1}\), provided by Dr. Majid Chatsaz. YouTube source: https://youtu.be/ujM4LIQlG1Q.

Example \(\PageIndex{2}\)

A basketball is thrown towards a hoop that is three feet higher in the \(y\) direction and 25 feet away in the \(x\) direction. If the ball is thrown at an initial angle of 50 degrees, what must the initial velocity be for the ball to make it into the hoop?

Figure \(\PageIndex{3}\): problem diagram for Example \(\PageIndex{2}\). A ball is thrown at some initial velocity, at an angle of 50° above the horizontal.

Solution

Video \(\PageIndex{3}\): Worked solution to example problem \(\PageIndex{2}\), provided by Dr. Majid Chatsaz. YouTube source: https://youtu.be/IGBB1YXRNr0.