Introduction to Rotational Motion

Speaking of rotational motion, you may think of merry-go-round, a fan or even our rotating earth. Rotational motion is all about taking an object and spinning it. The object follows many rules while spinning just like the way they do when moving on a straight line. You may think it’s easy to get confused by those rules, but being analogous to physics of linear motion actually makes the rules of rotational motion very easy to memorize.

In this article, we are going to discuss some basic terms and a set of equations in rotational motion. AP® Physics 1 & 2 Crash Course Review will go through all you need to know about rotation and talk about rotation-related questions in AP® Physics and how those equations apply to solve real test questions.

Basic Elements in Rotation

Center of Mass

To tackle rotation-related questions, we should first expand the moving object from a point to a real object with shape and scale. In linear motion, we usually treat objects as if they are single particles, and we assume that all the forces are exerted only on a single point that represents the object. But how can we be sure? Why does that point summarize and represent the whole object?

If you have ever balanced a pen on your finger, you should be very familiar that you could always find a specific point to keep the pen from falling. The position of the point varies from pen to pen and are not always in the middle. That certain point is called center of mass. In other words, the center of mass is the point where all the mass of the object could be considered to be concentrated.

Pivot Point

A pivot point is a center of rotation. It could be the center of mass or any point that you select. Force exerted on this point will not contribute to the rotation.

Torque

Torque is the measure of a force’s effectiveness as making an object accelerate rotationally. Still don’t know what a torque is? To understand torque, first, let’s recall a time that you use a wrench to loosen a bolt.

Example 1

When exerted on the same spot, a greater force is usually more effective than a small force. It is easier to turn the wrench when we push it by a greater force.

Example 2

If exerted on a different spot, the same force may have different effects. You will find it more effective pushing the wrench at the end than in the middle. The closer the spot is to the rotation axis, the harder you are likely to turn the wrench.

Example 3

What about changing the angle of the force exerted on the wrench? As showed in the figure below, F1 is always perpendicular to the wrench while F2 has an angle and F3 is parallel to the wrench. F3 certainly will not contribute to turning the wrench and you may find F2 less efficient than F1 when doing the job.

To summarize what we discussed earlier, we can find that the effectiveness of turning the wrench is actually related to 3 factors: Force(in example 1), distance from the pivot point (in example 2) and the angle(example 3). So then people think of combining these factors together and…Boom, we get the torque equation:

\tau =rFsin { \theta }

Now let’s see what we are really calculating. Torque is technically a vector, but for AP® Physics 1&2, you only need to be able to calculate the magnitude of torque and the direction of rotation (clockwise or counterclockwise). The term Fsin { \theta } equals the magnitude of force perpendicular to the lever arm. We are therefore getting a variable that is the product of force and a distance, so the unit for torque is Newton∙meter (Nm). Using this equation, if θ is between 0 to 90 degree, τ is getting smaller and smaller as θ decreases. The change of radius and force also affect torque in the same way.

Angular Velocity and Angular Acceleration

If the object is already spinning, what are variables to characterize its movement? Just like velocity and acceleration in linear motion, angular velocity and angular acceleration depict the movement of an object or of a particular point on the object. Angular velocity depicts the rate an object is spinning. It equals to the angle turned divided by time, notated as \omega =\dfrac { \Delta \Phi }{ \Delta t }. The direction of angular velocity is perpendicular to the rotational plane, determined by the right-hand rule. Angular acceleration is the change rate of ω, denoted as α.

Moment of Inertia

Also known as rotational inertia, the moment of inertia is used to characterize the tendency of an object to continue rotating before a torque is exerted on it. To better understand inertia, let’s compare inertia to mass. Mass is the tendency of an object to resist changes in its motion. To make an object move, we add a force on it. To make an object rotate, we exert a torque. While F = ma, torque equals to moment of inertia times angular acceleration. Rotational inertia is denoted as I.

\tau =I\alpha

So what determine the moment of inertia? Let’s first look at an example in figure skating. If the athlete wants to speed up while spinning, they draw in their arms and legs. Why will this do? Because inertia is related to the distribution of mass. The further an object is from the pivot point or the axis, the harder it is to make it spin. So when the athletes fold their arms, they are reducing their rotational inertia, therefore reduce the torque required to make them spin and elevates the angular velocity.

About Rotational Equilibrium

After getting familiar with those terms in rotation, we can see how those terms apply in solving problems. First, we would like to determine when an object can reach the equilibrium. The translational equilibrium is reached when the sum of force acting on the object is zero. Similarly, the rotational equilibrium is reached when the sum of torque acting on the object is zero. Be careful, being in a rotational equilibrium state does not necessarily mean the object is not rotating, it could also be rotating around its center of mass at a constant speed. If the object is totally at rest, it is said to be in a static equilibrium.

\sum { \tau } =0

Tackling Rotation-related Questions

The first type of questions you might meet on test date is finding center of mass. The position of the center is related with distribution of mass in the object. In general, we use the equation below to compute the position of the center of mass.

{ x }_{ CM }=\dfrac { { x }_{ 1 }{ m }_{ 1 }+{ x }_{ 2 }{ m }_{ 2 }+{ x }_{ 3 }{ m }_{ 3 }+... ...+{ x }_{ n }{ m }_{ n } }{ { m }_{ 1 }+{ m }_{ 2 }+{ m }_{ 3 }+......+{ m }_{ n } }

For objects with uniform density, homogenous objects, the center of mass located at the geometric center. For example for a uniformed stick, the center of mass is the mid point of the stick. Always remember that center of mass can be located outside the body of object, such as a ring’s center of mass locate at its center. For other systems, simply use the equation above.

Example 4

Now we have a uniformed 2 kg bar with the length of L. M1 = 40kg, M2 = 60kg, M3 = 70kg, M2 is attached to a quarter point of the bar. If we attached it to the ceiling, where should the string attached in order to reach a equilibrium?

This is actually asking you to find the center of mass. First chose a point as the reference point. You can chose wherever you like as long as it’s easy to calculate. Then figure out the distance between the object and reference point. At last, make use of the formula. For example let’s chose the mid point to calculate.

{ x }_{ CM }=\dfrac { { x }_{ 1 }{ m }_{ 1 }+{ x }_{ 2 }{ m }_{ 2 }+{ x }_{ 3 }{ m }_{ 3 } }{ { m }_{ 1 }+{ m }_{ 2 }+{ m }_{ 3 } } = \dfrac { { L/2}{40kg}+{ L/4 }{ 60kg}-{ L/2}{70kg} }{ 40kg + 60kg+ 70kg } = 0

The result shows that the center of mass lies exactly on the mid point.

The second type of questions might be about how to reach equilibrium.

Example 5

A unified 2 kg bar is attached to the wall at the mid point. The thread has a 30 degree angle with the wall. M = 40kg. The system is at equilibrium. Find tension in the thread.

First identify the pivot point is the attached point at the left end of the bar. According to \sum { \tau } =0, the torque M exerted on the bar is MgL. Don’t forget to add the torque exerted by the bar’s weight, mg(L/2). To eliminate the effect of the sum of these 2 torques, the thread should exert another torque which equals to the magnitude of MgL but in opposite direction. Then we have

T(L/2)cos30 = MgL + mg(L/2)

T = 2Mg+mg/cos30 = 1620\dfrac { \sqrt { 3 } }{ 3 } N

You may wonder why normal force exerted by the wall is not contributing to rotation. It’s because normal force goes right through the pivot, so the torque it exerts is zero.

To sum up, rotational motion is very similar to linear motion by all means. This chart below will help you better memorize the relationship between linear motion and rotational motion.

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