Centripetal and Centrifugal Force

Difference Between Centripetal and Centrifugal Force

Centripetal force is the component of force acting on a rotating body which is directed towards the axis of rotation. Centrifugal force is a pseudo-force within the reference frame of the rotating body which acts outwards along the radius of rotation, pushing the body away from the axis of rotation.

The difference between the centripetal force and the centrifugal force is that while the centripetal force applies to any reference frame, the centrifugal force is only applicable in a rotating reference frame. The observer's reference frame is called an inertial reference frame, which means that the law of inertia, Newton's first law, holds true. A rotating reference frame is a non-inertial reference frame because the law of inertia does not hold true.

If we consider a ball placed on a rotating, frictionless platform, the ball will be pushed off of the platform. This makes sense for an observer watching the ball because the ball had initial velocity from the rotating platform. But from the ball's rotating reference frame, the ball begins to move radially outward from the platform without any forces acting on it. Thus in a non-inertial reference frame, Newton's laws of motion do not hold true. We can, however, use Newton's laws of motion in a non-inertial reference frame if we introduce "fictitious forces" such as centrifugal force. Fictitious forces make calculations in non-inertial reference frames easier since they allow us to use Newton's laws of motion.

Developing Formulas for Centripetal Force and Centrifugal Force

To find the formulas for centripetal force and centrifugal force, let's take a closer look at the equations that describe the centripetal acceleration. Consider the ball on a string at two different points,$P_1$and$P_2$, on a circle of radius$R$. At these points, the ball has corresponding velocities that we will call$v_1$and$v_2$. We will also call the distance that the ball traveled$∆x$and the angle$∆\theta$. The time it took for the ball to travel from$P_1$to$P_2$is$∆t$.

From the image shown above, we can draw two similar triangles for the change in velocity$∆v$and the change in distance$∆x$. Since the triangles are similar triangles, the ratio of similar sides of the triangles is equal and gives us:

$\frac{∆x}{R}=\frac{∆v}{v_1}$

$\frac{\left|\overrightarrow{∆v}\right|}{\overrightarrow{v_1}}=\frac{∆x}{R}\Rightarrow \left|\overrightarrow{∆v}\right|=\frac{∆x}{R}\overrightarrow{v_1}$

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:

$$\begin{gathered} a=\frac{∆v}{∆t} \\ =\frac{∆x}{R}\frac{v_1}{∆t} \\ =\frac{v_1}{R}\frac{∆x}{∆t} \end{gathered}$$

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$v_1$.

$\underset{∆t\to 0}{lim}\frac{∆x}{∆t}\Rightarrow v_1$

Now we can write the acceleration as:

$$\begin{gathered} a=\frac{v_1}{R}\underset{∆t\to 0}{lim}\frac{∆x}{∆t} \\ =\frac{v_1}{R}\left(v_1\right) \end{gathered}$$

Since we have taken the limit as the change in time approaches zero, we can drop the subscripts. We now arrive at the equation for the centripetal acceleration:

$a_c=\frac{v^2}{R}$

We find the equation for the magnitude of the centripetal force,$F_c$

, from Newton's second law as mentioned above:

$$\begin{gathered} F_c=m{a}_c \\ =m\frac{v^2}{R} \end{gathered}$$

The centripetal force points in the same direction as the centripetal acceleration, to the center of the circle.

The centrifugal force,$F_{cf}$, used only in a non-inertial reference frame, has the same magnitude as the centripetal force:

$F_{cf}=m\frac{v^2}{R}$

Though the centripetal force and the centrifugal force are equal in magnitude, they point in opposite directions. The centrifugal force is directed radially outwards.

Example of Centripetal Force and Centrifugal Force

Let's consider a ball on a string again, but this time we'll make it a pendulum, as shown below.

We will consider the pendulum in an inertial reference frame for now. 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!

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$T$is the hypotenuse and the force of gravity$F_g$is the adjacent side. Our last triangle divides the tension force into its x and y components. Using trigonometry, we obtain these equations:

$$\begin{gathered} \sin \theta =\frac{R}{L} \\ \cos \theta =\frac{F_g}{T} \\ \sin \theta =\frac{T_x}{T} \end{gathered}$$

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:

$\theta ={\sin }^{-1}\left(\frac{R}{L}\right)$

$$\begin{gathered} T=\frac{F_g}{\cos \theta } \\ =\frac{mg}{\cos \theta } \end{gathered}$$

$$\begin{gathered} T_x=T\sin \theta \\ =\left(\frac{mg}{\cos \theta }\right)\sin \theta \\ =mg\tan \theta \end{gathered}$$

Applications of Centripetal Force and Centrifugal Force

A common application that involves the centripetal force and centrifugal force is a car driving around a flat curve at a constant speed. In an inertial reference frame, the forces acting on the car are gravity, the normal force, and friction, which keeps the car from sliding. The centripetal force is supplied by the friction, keeping the car moving in a circle. A passenger in the car could argue that they feel an outward force pushing them into the door as the car goes around the curve, but in an inertial reference frame, there is no centrifugal force pushing outward. What the passenger is feeling comes from the law of inertia. The passenger has an initial velocity from the car in motion and wants to move in a straight line, but since the car is rounding a curve, the side of the car keeps the passenger from moving in a straight line. If we consider a non-inertial reference frame from the passenger's perspective, the passenger and car are at rest while everything else is moving. The outward force the passenger feels can be called the centrifugal force in their reference frame, but we need to remember that it is not a real force since there is no object applying it.