Introduction

Welcome to the FiveHive article for Unit 2.9 of AP Physics 1!

This article will cover the information needed to understand objects traveling in circular motion in almost every context, from orbits to roller coasters to racetracks, as well as the forces that cause these turns, loops, and revolutions. 

As usual, we will only cover the topics included in the CED for Unit 2.9.

Centripetal Acceleration

Acceleration is defined as an object’s rate of change of velocity. When moving linearly, this would only mean increasing or decreasing in magnitude. However, what happens when we introduce another dimension? 

When an object is moving in a circular path, such as a loop, its velocity can also change by changing direction. For instance, on a circular track, on one side of the track, you can be moving directly upward from a bird’s-eye view. As you get to the top of the track, your velocity is directed entirely in the leftward direction. This is another form of acceleration, and in circular motion, this is specifically centripetal acceleration, which will always be directed in a perpendicular direction from the object’s velocity, which is also towards the center point of the object’s path.

The object’s velocity, also referred to as the tangential speed, can be related to this centripetal acceleration using the following formula: $a_c=\frac{v^2}{r}$, where v is the velocity and r is the radius or the distance from the object to the center. In different contexts, this acceleration could be caused by varying types or quantities of forces. If an object is being spun around on a rope horizontally or is a pendulum, the rope would be the only force that pulls the object towards the center, so the tension of the rope is also the force of centripetal acceleration. If a satellite were orbiting around the Earth, gravity would be the force pulling it towards the center, so acceleration from gravity would also cause the centripetal acceleration. For an object rolling or a car driving and turning, the force acting towards the center would be static friction (it wouldn’t be kinetic friction, because then the tires would slip and the car would veer away from the circular road tangentially!). 

However, multiple forces can act on an object at the same time, in which case the centripetal acceleration is the result of all the forces going in the center-pointing direction combined. For instance, if a roller coaster does a loop, the main force seen acting upon the train would probably be the normal force from the track. However, when the train reaches the upper portion of the track, gravity would at least have a portion of its force going towards the center point. In the instant that the train is at the uppermost part of the track, the entirety of the acceleration of gravity plus the leftover normal force value from acceleration would equal the centripetal acceleration. This would mean that on the upper portions of the track, there is also less normal force acting on the train compared to the lower portions of the track. For objects undergoing circular motion, the centripetal acceleration value must be met. As such, there is a minimum speed for an object going on a vertical track to undergo circular motion, where gravity is equal to centripetal acceleration. With this, we can derive the minimum speed of an object to travel a vertical circular loop to be $v=\sqrt{gr}$. On the AP exam, you are expected to derive this equation from first principles as follows:

$F_c=F_N+F_g$

When a block is pressed on the track, the block will feel a downward normal force. When $F_N=0$, the block is no longer in contact with the loop. So the limiting speed for the block to pass the loop is:

$F_c=F_g$

$\frac{m{v}^2}{r_{\text{min}}}=mg$

Note that the mass cancels!

$v_{\text{min}}=\sqrt{gr}$

There are several other problem scenarios that could appear on the AP Physics 1 exam. For example, a ball spinning in a circle (i.e., what is the maximum speed until the string connecting the ball breaks?), a conical pendulum, and, rarely, banked curves. We will only focus on banked curves, as a mastery of such will result in a mastery of the other concepts.

Assuming a rough surface, when a block of mass $m$ travels along the banked curve without moving up or down the ramp in the diagram above, circular motion is present! In this example, components of the static friction (not kinetic, or slipping, occurs, which means no turning!) force and the normal force can contribute to the net force producing centripetal acceleration of an object travelling in a circle on a banked surface.

So how do the forces contribute to the centripetal force? We will draw a free-body diagram. 

This seems oddly similar to a block-on-a-ramp scenario encountered in Unit 2. However, things are a bit different here, as the frictional force is actually pointing downwards. To understand why, visualize that as a car is speeding up along a banked curve, it will want to go up the banked curve. Hence, friction points down the ramp to counteract such a tendency. The question we want to answer is, how fast does the mass need to go until it starts sliding up (away from the circle of “orbit”)?

We start by drawing the components of the forces on the block. For the purposes of the article, we will ignore friction in our diagram to simplify our analysis.

When there is no $+y$ movement, the block is in successful circular motion (no slipping!). For this to happen, the $y$ component of the normal force must balance out the force of gravity. This means the $x$ component will be the centripetal force. It is left to the reader as an exercise (hint: find the normal force in terms of $m$ and $\theta$, then equate the right equation to the centripetal force formula) to demonstrate that the speed of such a block in this scenario will be given by the equation:

$v=\sqrt{gr\tan \theta }$

This is as hard as it will get on the AP Physics 1 exam (this derivation would likely be an FRQ question, Mathematical Routines Q1). For those curious souls who are interested in "What if friction existed?" please take a look at this static simulator made by HyperLink Physics.

Link: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/carbank.html

Net Acceleration

Despite the introduction of centripetal force, there is still acceleration from the increase and decrease in the overall velocity of the object. In circular motion, this is referred to as tangential acceleration to differentiate it from centripetal acceleration, and as the name suggests, it will be directed in a tangential direction to the circle, just like velocity. As these are the two types of acceleration acting on an object moving circularly, net acceleration is the vector sum of centripetal acceleration and tangential acceleration (Unit 1). Below is an illustration of the two aforementioned forces in an oscillating pendulum.

Periods and Frequencies 

When dealing with objects that undergo repeated motions, period and frequency are terms that are often used, with period representing the time it takes to complete one set of that motion and frequency representing how often it happens, with units of s and $s^{-1}$ or Hertz (Hz), respectively. This would include objects undergoing circular motion at constant speeds. To convert between period and frequency, you would need to get the reciprocal of the other, so $T=\frac{1}{f}$, where $T$ is period and $f$ is frequency, and $f=\frac{1}{T}$. 

It is also possible to derive the period it takes for an object to undergo circular motion once, if given the radius and velocity of the object, by dividing the circumference of the track by the speed the object travels at, or $T=\frac{2\pi r}{v}$.

If an object of mass $m$ is in orbit around a celestial body of mass $M$ as the only force of centripetal acceleration, the period could be found if given the mass of the body and the radius of the orbit with the formula $T^2=\frac{4{\pi }^2}{GM}{R}^3$, where $T$ is the period, $G$ is the gravitational constant, and $R$ is the radius of the orbit. You must do the following derivation on the AP exam to obtain the orbital speed and/or orbital period:

$F_c=F_g$

$\frac{m{v}^2}{R}=G\frac{M_{J}m}{R^2}$

$v=+\sqrt{\frac{G{M}_J}{R}}$

$v=\frac{2\pi R}{T}$

$T=\frac{2\pi R}{v}=\frac{2\pi R}{\sqrt{\frac{G{M}_J}{R}}}$

$T=\frac{\sqrt{4{\pi }^2{R}^2}}{\sqrt{\frac{G{M}_J}{R}}}$

$T=\sqrt{\frac{4{\pi }^2{R}^3}{G{M}_J}}$

Practice