In what direction would a ball being swung in a circular motion on a string go if the string were suddenly cut? Since the string is no longer providing the tension force to keep the ball in a circular motion, the ball will fly off in the direction of its velocity, which is always tangent to the circle. In this article, we will discuss the force that keeps objects in a circular motion: the centripetal force. We will also talk about how centripetal force relates to the velocity and develop a formula for them.
Relationship between Centripetal Force and Velocity
In order to understand the relationship between the centripetal force, let's consider a ball on a string moving in a circle, as shown below.
Fig. 1 - A ball on a string stays in uniform circular motion because of the centripetal force from the tension in the string.
The tangential velocity \(\vec{v}\) of the ball is always in a direction tangent to the circle. If the magnitude of the velocity is constant, the ball is in uniform circular motion. Though the magnitude of the velocity may be constant, the direction of the velocity vector is constantly changing. This causes the ball to experience acceleration even if there is no change in the magnitude of the velocity. If the ball is in uniform circular motion, the acceleration vector \(\vec{a}\) is perpendicular to the velocity vector, and thus is always pointing to the center of the circle, as shown in the image. This radial component of the object’s acceleration is known as the centripetal acceleration, \(\vec{a}_c.\)
Centripetal acceleration: The radial component of the acceleration of an object moving in a circular motion.
If the ball is experiencing non-uniform circular motion in which the velocity of the ball is not constant, there will be acceleration vector components that do not point to the center of the circle. These acceleration components do not contribute to the centripetal acceleration.
For an object to remain in a circular motion, there must be a force acting on it. The net radial force acting on an object that keeps it in a circular motion can be referred to as the centripetal force. The centripetal force acting on the ball to keep it in a circular motion in our example comes from the tension force of the string. The centripetal force always points in the same direction as the acceleration vector and is perpendicular to the velocity vector.
Centripetal force: The total force acting radially on an object to keep it in a circular motion.
It is good to note that the centripetal force isn't an actual force, but rather we use the term centripetal force to describe the total force keeping the object in a circular motion. When drawing a free-body diagram, the centripetal force is not labeled as an actual force, but rather the real forces that provide the centripetal force are. In the example above, the centripetal force came from the tension force from the string. Gravity is another good example of a force that keeps an object such as a satellite in orbit around the earth.
Some people assume that an object in a circular motion experiences an outward "centrifugal force" that points in the direction opposite to the center of the circle. This is incorrect for an inertial reference frame. When we consider a rotational, non-inertial reference frame, we will discuss centrifugal force in greater detail. For the purposes of AP Physics C, we will only consider an inertial reference frame, so we recommend focusing on understanding centripetal force and acceleration and disregarding the term "centrifugal force".
Formula for Centripetal Force and Velocity
Let's take a closer look at the equations that describe the centripetal force and velocity. Consider the ball on a string at two different points on a circle of radius, \(r\). At these points, the ball has corresponding velocities that we will call, \(\vec{v}_1\) and \(\vec{v}_2\). We will also call the distance that the ball traveled \(\Delta x\) and the angle \(\Delta \theta\). The time it took for the ball to travel from one position to the other is \(\Delta t\).
Fig. 2 - A ball on a string traveling in uniform circular motion shown at two positions.
The angle between the initial and final velocity vectors is the same as the change in angle, \(\Delta \theta\), between the first and second position. This means that we can draw two similar triangles for the change in velocity, \(\Delta \vec{v},\) and the change in distance, \(\Delta x\). Since the triangles are similar triangles, the ratio of similar sides of the triangles is equal and gives us: \[\begin{align*}\frac{| \Delta \vec{v}|}{\vec{v}_1}&=\frac{\Delta x}{r}\\|\Delta \vec{v}|&=\frac{\Delta x}{r}\vec{v}_1.\end{align*}\]Now let's think about the average acceleration. Acceleration is defined as the change in velocity divided by the change in time. Thus, we can write: \[\begin{align*}\vec{a}&=\frac{\Delta \vec{v}}{\Delta t}\\&=\frac{\Delta x}{r} \frac{\vec{v}_1}{\Delta t}\\&=\frac{\vec{v}_1}{r}\frac{\Delta x}{\Delta t}.\end{align*}\]If we consider very small changes in distance and time, we can take the limit as the change in time approaches zero. As the change in time approaches zero, the change in distance over the change in time approaches \(\vec{v}_1.\) \[\lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} \to \vec{v}_1.\]Now we can write the acceleration as: \[\begin{align*}\vec{a}&=\frac{\vec{v}_1}{r}\lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t}\\&=\frac{\vec{v}_1}{r}\cdot \vec{v}_1.\end{align*}\]Since we have taken the limit as the change in time approaches zero, we can drop the subscripts and consider the magnitude of the velocityat any pointon the circle. We now arrive at the equation for the centripetal acceleration in terms of velocity: \[a_c=\frac{v^2}{r}.\]
Centripetal Force and Velocity Equation
The equation for the centripetal force in terms of the velocity is found by using Newton's second law: \[\begin{align*}F_c&=ma_c&=\frac{mv^2}{r}.\end{align*}\]We notice that the centripetal force is proportional to the centripetal acceleration and the square of the velocity.
It is important to remember that the equation above for centripetal acceleration is considering the acceleration of the center of mass of the system. The variables used above for the position, velocity and acceleration are all quantities based on the center of mass of the object.
Graph of Centripetal Force versus Angular Velocity
So far in this article, we have only discussed the linear velocity of an object in a circular motion. Now let’s talk a little about angular velocity. While the linear velocity is the change in position over the change in time, the angular velocity is the angular displacement over the change in time, \(\omega=\frac{\Delta \theta}{\Delta t}.\) Using angular velocity is sometimes more convenient than using linear velocity, like when an object is rotating about an axis. The magnitude of linear velocity is proportional to the angular velocity as \(v=r\omega.\) We can relate the centripetal force to the angular velocity by substituting this into our equation from before to get: \[\begin{align*}F_c&=m\frac{v^2}{r}\\&=m\frac{(r\omega)^2}{r}\\&=mr\omega^2.\end{align*}\]The graph below helps us to visualize the relationship between the centripetal force and the angular velocity, assuming the mass and the radius are constant. The shape of this graph is the same for the relationship between the centripetal force and the linear velocity.
Fig. 3 - Centripetal force with respect to angular velocity.
Centripetal Force and Velocity Examples
Using the equations we found above, we can find the centripetal force for an object in circular motion. Let’s do a couple of examples to get some practice!
A ball on a pendulum is attached to a \(10\,\mathrm{cm}\) string moving in a circle with a radius of \(5\,\mathrm{cm}\), as shown below. The mass of the ball is \(200\,\mathrm{g}\). Find the magnitude of the centripetal force acting on the ball.
Fig. 4 - The force of gravity and tension act on a pendulum.
The only forces acting on the ball are gravity and the tension force from the string. As discussed above, the centripetal acceleration points toward the center of the circle drawn in the image, and thus the centripetal force must also point in that direction. To calculate the centripetal force, we would need to find the x component of the tension force. Let's draw some triangles to help us!
Fig. 5 - Components of the tension force for a pendulum.
Our first triangle shows that the ball is at an angle \(\theta\) with respect to the normal. The hypotenuse of the triangle is given by the length of the string; we'll call it \(L\) for now. The radius of the circle, \(R\), also defines one side of the triangle. We can make our second triangle from the forces working on the ball. We have the same angle \(\theta\) with respect to the normal, the tension force \(\vec{T}\) is the hypotenuse and the force of gravity \(\vec{F}_g\) is the adjacent side. Our last triangle divides the tension force into its \(x\) and \(y\) components. Using trigonometry, we obtain these equations: \[\begin{align*}\mathrm{sin}\theta&=\frac{R}{L}\\\\\mathrm{cos}\theta&=\frac{F_g}{T}\\\\\mathrm{sin}\theta&=\frac{T_x}{T}.\end{align*}\]We have now dropped vector notation as we are considering the magnitude of the vectors. Since we're looking for the centripetal force, we need to solve for the x component of the tension force, as that is the component that is pointing to the center of the circle. Solving for \(T_x\) in terms of known variables gives us: \[\begin{align*}\theta&=\mathrm{sin}^{-1}\Big(\frac{R}{L}\Big)\\\\T&=\frac{F_g}{\mathrm{cos}\theta}\\&=\frac{mg}{\mathrm{cos}\theta}\\\\T_x&=T\mathrm{sin}\theta\\&=\Big(\frac{mg}{\mathrm{cos}\theta}\Big)\mathrm{sin}\theta\\&=mg\mathrm{tan}\theta.\end{align*}\]Substituting in the values given in the problem, we get: \[\begin{align*}T_x&=(0.2\,\mathrm{kg})\big(9.8\,\mathrm{\tfrac{m}{s^2}}\big)\mathrm{tan}\Big(\mathrm{sin}^{-1}\Big(\frac{0.05\,\mathrm{m}}{0.1\,\mathrm{m}}\Big)\Big)\\&=1.132\,\mathrm{N}.\end{align*}\]Thus, the magnitude of the centripetal force is \(1.132\,\mathrm{N}\).
A \(1000\,\mathrm{kg}\) car is going around a curve of radius \(20\,\mathrm{m}\) with a velocity of \(15\,\mathrm{\frac{m}{s}}\). What is the friction coefficient required to produce sufficient centripetal force?
We learned above that the magnitude of the centripetal force can be found by \(F_c=m\frac{v^2}{R}.\) The force supplying the centripetal in this problem is given by friction \(F_c=\mu F_n\), where \(\mu\) is the friction coefficient and \(F_n\) is the normal force. As stated above, the normal force is equal to the force of gravity, so that \(F_n=mg\). Substituting these into our equation for the centripetal force gives us:\[\begin{align*}F_c&=\mu F_n\\\frac{mv^2}{R}&=\mu mg\\ \mu&=\frac{v^2}{gR}\\&=\frac{\big(15\,\mathrm{\tfrac{m}{s}}\big)^2}{\big(9.8\,\mathrm{\tfrac{m}{s^2}}\big)(20\,\mathrm{m})}\\&=1.15.\end{align*}\]
Centripetal Force and Velocity - Key takeaways
- The centripetal force is the total force acting radially on an object to keep it in a circular motion.
- The centripetal acceleration is the radial component of the acceleration of an object moving in circular motion.
- The centripetal force and the centripetal acceleration are both proportional to the velocity squared.
- The centripetal force vector and the centripetal acceleration vector always point to the center of the circle.
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