Circular Motion and Free-Body Diagrams

Drawing a Free-Body Diagram

To begin, let's talk about what a free-body diagram is and why they are important. A free-body diagram, also known as a force diagram, is a diagram that shows all of the vector forces acting on an object. It is helpful to draw the object separate from the system and identify each force acting on the object. Do not include any forces that the object exerts on other objects, only those acting on the object. If there are multiple objects in a system, you will need to make a free-body diagram for each object.

We represent the forces on a free-body diagram by using vectors. We show each force acting on the object by drawing a vector whose length and direction represent the magnitude and direction of the force, respectively. We draw free-body diagrams n two different ways. The first way is to draw the force vectors acting on the object coming from a dot representing the object's center of mass. The second way is to draw each force vector on the object at the point where the force acts. The image below is an example of a free-body diagram for a box on a ramp. The vectors for the force of gravity, the normal force and the friction force come from the center of mass of the box and point in the direction of the force.

Free-body diagram for a box on a ramp, StudySmarter Originals

Free-Body Diagram for Circular Motion

Now that we understand how to draw a free-body diagram, we can discuss free-body diagrams for objects in uniform circular motion. Let's consider a satellite in orbit around the Earth. In the image below, the satellite is shown moving in a uniform circle around the Earth. What force is acting on the satellite to keep it in orbit? Gravity! Without gravity, the satellite's high speed would cause it to fly off in a straight line. The satellite is constantly falling toward the Earth because of gravity. The satellite's high velocity and gravity keep the satellite in a circular orbit.

The free-body diagram for the satellite is pretty simple since the only force acting on it is gravity. We represent the force of gravity with a vector. The direction of the vector points toward the center of the Earth.

Objects in uniform circular motion must have a centripetal force that points toward the center of the circle; this force is what keeps the object moving in a circle. The centripetal force is the sum of all of the forces acting radially on the object.

When making a free-body diagram for an object in uniform circular motion, it is important to identify which forces contribute to the centripetal force. The centripetal force is proportional to the centripetal acceleration. This will help us later when we write our equation.

Let's take a closer look at two free-body diagrams: a ball on a string being swung in a vertical circular motion and a horizontal circular motion.

Free-Body Diagram for Vertical Circular Motion

Consider a ball being swung in a vertical circular motion, as shown in the image below. We will draw three free-body diagrams for the ball at positions 1, 2, and 3.

Free-body diagram for a ball being swung in a vertical motion at three positions, StudySmarter Originals

The ball constantly has two forces acting on it: the force of gravity and the tension force. The force of gravity at all three positions is in the downward direction, but the tension force points down at position 1, to the right at position 2, and up at position 3. We also must note the magnitude of the vectors. The magnitude of the gravity force depends on the mass of the ball, which remains constant. Thus the magnitude of the gravity force is the same at all three positions.

The magnitude of the tension force will not be the same at all three positions. In order for the ball to remain in a circular motion, the ball's velocity must increase along the lower part of the circle. At the top of the circle, the ball slows down due to gravity, and thus the string isn't pulling it as hard. It's important to note that since the velocity of the ball is changing, the ball is not in uniform circular motion in this case. If the velocity of the ball becomes too small, the string will go limp and the tension force will be zero. Then the ball will fall out of its circular motion. We will say for this problem that the ball maintains sufficient velocity to keep the ball from falling out of circular motion. The magnitude of the tension force will be the smallest at position 1 and increase to the largest magnitude at position 3.

We can use Newton's second law,$F_{net}=ma$, to describe the motion of the ball at each position. Even though the ball isn't in uniform circular motion, we can still use the equation$a_c=\frac{v^2}{r}$for centripetal acceleration at each point, where$r$is the radius of the circle.

Let's start with position 1. We begin by setting the centripetal force equal to the mass multiplied by the centripetal acceleration. The net force in this equation is the centripetal force, so we need to include all of the forces contributing to the centripetal force. For position 1, we have:

$F_{net}=m{a}_c$

$T_1+mg=m\frac{{\left(v_1\right)}^2}{r}$

Notice that when we define our coordinate system at this position, we choose the downward direction to be the positive direction since this is the direction that the centripetal acceleration points.

Now for position 2, we will define the direction to the right of the ball to be the positive direction. Since the force of gravity is perpendicular to the tension force, it has no components that contribute to the centripetal force. Our force equations for this position are:

$T_2=m\frac{{\left(v_2\right)}^2}{r}$

Lastly, for position 3 we define the upward direction as positive. Our force equations become:

$T_3-mg=m\frac{{\left(v_3\right)}^2}{r}$

We can use these equations to find the magnitude of the tension force or the velocity at each position.

Free-Body Diagram for Horizontal Circular Motion

The case for a ball swinging in horizontal circular motion is a bit simpler than the vertical circular motion. Since the velocity of the ball is constant, the ball experiences uniform circular motion. The weight of the ball makes it so we could never swing the ball completely horizontally, but we will approximate that the ball is light and treat the situation as if it were completely horizontal. We only need to make one free-body diagram in this case because the force of gravity and the centripetal for are always perpendicular to each other, as shown in the image below. This means that the force of gravity will not contribute to the centripetal force.

Since the only centripetal force comes from the tension in the string, our force equation is:

$\begin{array}{rcl}{F}_{net}& =& m{a}_c\\ T& =& m\frac{v^2}{r}\end{array}$

Solving a Problem with Horizontal Circular Motion

Now let's do a practice problem for horizontal circular motion. Consider a car driving on a banked curve traveling in circular motion . At a certain angle, no friction is required to keep the car in circular motion because the horizontal component of the normal force will be the force causing the centripetal acceleration. What is this angle for a$1000\mathrm{kg}$car going around a curve with a radius of$65\mathrm{m}$and a velocity of$15\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$?

We can draw the free-body diagram of the car as shown below. The components of the normal force are shown with dotted lines so that we can see how the components contribute to the system.

The car is not experiencing motion in the vertical direction, so we know that the vertical acceleration is zero. Our force equation in the vertical direction gives us:

$\begin{array}{rcl}{F}_{net}& =& ma\\ F_n\cos \theta -F_g& =& 0\\ F_n\cos \theta & =& mg\\ F_n& =& \frac{mg}{\cos \theta }\end{array}$

The only force contributing to the centripetal acceleration is the horizontal component of the normal force:

$\begin{array}{rcl}{F}_{net}& =& m{a}_c\\ F_n\sin \theta & =& m\frac{v^2}{r}\\ F_n& =& \frac{m{v}^2}{r\sin \theta }\end{array}$

Setting these equal to each other gives us:

$\begin{array}{rcl}\frac{mg}{\cos \theta }& =& \frac{m{v}^2}{r\sin \theta }\\ \tan \theta & =& \frac{v^2}{rg}\end{array}$

We can then use this equation to find the necessary angle.

$\begin{array}{rcl}\theta & =& {\tan }^{-1}\left(\frac{v^2}{rg}\right)\\ & =& {\tan }^{-1}\left(\frac{{(15\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.)}^2}{(65\mathrm{m})(9.8\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{s}}^2$}\right.)}\right)\\ & =& 0.34\mathrm{rad}\\ & =& 19.45°\end{array}$

Circular Motion and Gravitation Free-Body Diagram Examples

Here are a couple more examples just for practice!