Consider an object traveling along a circle of radius $r$ on the $(x,y)$ plane with constant angular velocity $\omega$ (see figure 1). The circle is centered at the origin $O$ of the plane.

In a previous post, I presented a description of this problem from a pure kinematical perspective. In other words, the causes of the motion were ignored. Here, I will present this same problem from a pure mechanical perspective, meaning the main focus will be on the forces which cause the object to move along a circular path.

On the kinematical description, I showed that when an object travels along a circle of radius $r$ with constant angular velocity $\omega$, the acceleration points always to the center of the circle. The magnitude $a$ of this acceleration is given by: $$ a = \omega^2 r = \displaystyle\frac{v^2}{r} $$ where $v$ is the magnitude of the velocity. In vector form, letting ${\bf x}(t) := (x(t),y(t))$ be the position vector of the object at time $t$ (see figure 1), I showed that the acceleration ${\bf a}(t)$ of the object satisfies: $$ {\bf a}(t) = -\omega^2 {\bf x}(t) $$

Newton's second law of motion states that the acceleration ${\bf a}(t)$ of a body and the net force ${\bf F}(t)$ acting on that body (the net force on a body is the sum of all forces acting on it) at time $t$ are related through: $$ {\bf F}(t) = m{\bf a}(t) $$ where $m$ is the mass of the body. Therefore, for uniform circular motion, the net force ${\bf F}(t)$ must also point to the center of the circle since it is parallel to ${\bf a}(t)$. This net force is called the centripetal force. It causes the object to always travel along the circle. For completeness, the magnitude of ${\bf F}(t)$ is such that: $$ \boxed{ F(t) := \|{\bf F}(t)\| = m\omega^2 r = \displaystyle\frac{mv^2}{r} } $$

Let's then consider an interesting problem. Imagine you are washing your clothes on your washing machine. The interior of the washing machine is a cylindrical body of radius $r$ (see figure 2) which spins very fast (hopefully not so fast as to annoy your neighbor with the vibration it transfers to the ground...). The clothes seem to be "compressed against" the walls of the washing machine, right?

One could think "well, there is clearly some kind of centrifugal force pushing the clothes against the walls of the washing machine". However, as we have seen above, this is wrong! In fact, the force acting on each set of clothes must point to the center of the machine. But then who is providing this force? Well... the walls of the washing machine. They provide a contact force, called the normal force, which is always perpendicular to the contact surface (these forces are shown in figure 2 for each set of clothes). It is this contact force which acts as the centripetal force for each set of clothes (gravity and friction with the walls are also present, but washing machines typically rotate so fast that these forces become negligible). To be precise, if $m_g$, $m_r$ and $m_b$ are the masses of the green, red and blue set of clothes respectively: $$ N_g := \|{\bf N}_g\| = \displaystyle\frac{m_gv^2}{r} \quad N_r := \|{\bf N}_r\| = \displaystyle\frac{m_rv^2}{r} \quad N_b := \|{\bf N}_b\| = \displaystyle\frac{m_bv^2}{r} $$ where $v = \omega r$. This type of contact force is what prevents you from falling through your chair or from sinking into the ground towards the center of the Earth due the gravitational force which constantly pulls you down. Microscopically, this force is the result of the repulsive electromagnetic forces which prevent atoms from overlapping. In our example, the normal forces prevent the clothes from going through the machine walls.

Another interesting example is a car making a turn. Consider for instance a car traveling along an arc of radius $r$ with its speed $v$ kept constant (see figure 3). The net force acting on the car must point to the center of the circle which contains the arc the car is traveling on. But where is this force coming from? From the friction with the ground! The friction forces between the tires and the ground point to the left of the car (towards the center of the circle as shown in figure 3) and, when summed, yield the centripetal force ${\bf F}_{\textrm{cp}}$ ("cp" here stands for "centripetal") which keeps the car on the arc. This is the reason why icy roads are very dangerous: the tires do not properly adhere to the ground and therefore the friction forces are not strong enough to act as the centripetal force the cars need to stay on the arc.

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