### Unit 6: Rotational Statics and Dynamics

What do desks, bridges, buildings, trees, and mountains have in common – at least in the eyes of a physicist? The answer is that they are ordinarily motionless relative to the Earth. Consequently, their acceleration, with respect to the Earth as a frame of reference, is zero. Newton's second law states that net F = ma, so the net external force is zero on all stationary objects and for all objects moving at constant velocity. There are forces acting, but they are balanced. That is, the forces are in equilibrium.

**Completing this unit should take you approximately 1 hour.**

Upon successful completion of this unit, you will be able to:

- define the conditions necessary for a rigid body to be in equilibrium;
- define the vector quantity torque;
- define rotational inertia;
- compare and contrast the dynamics of linear and rotational motion;
- Solve statics problems with rotational motion; and
- solve dynamics problems involving rotational motion.

### 6.1: Conditions for Equilibrium

When an object is in equilibrium, the forces acting upon the object are balanced. That is, the net force on the object is zero. For this to occur, the object must either not be moving, or it must be moving at a constant velocity.

There are two types of equilibrium: static equilibrium and dynamic equilibrium.

1. Static equilibrium describes a system that is balanced and does not rotate. An example is a seesaw where two children sitting at either end are exactly the same weight. Since the seesaw will not move, it is in static equilibrium.

2. Dynamic equilibrium describes a system that is balanced, but also moving (without any angular acceleration). An example is a planet in perfect circular orbit around its parent star. There is no torque acting on the planet making it orbit faster or slower, but it will keep orbiting for a very long time. The system is in equilibrium, and because it is moving, it is dynamic.

As you read, pay attention to the illustration of static equilibrium in Figure 9.3 and the illustration of dynamic equilibrium in Figure 9.4. An object in static equilibrium is completely motionless. An object in dynamic equilibrium is moving at constant velocity.

The study of statics is the study of objects that are in equilibrium. Two important conditions must be met for an object to be in equilibrium. First, the net force on the object must be zero. Secondly, a rotating object does not experience rotational acceleration. That is, a rotating object can be in equilibrium if its rotational velocity does not change.

Watch this video, which accompanies what you just read.

### 6.2: Torque

A system is not in equilibrium when a rotational force is acting on it to make it accelerate in its rotation. We call this rotational force torque. We define torque as the force to turn or twist an object, thus changing its rotational velocity. The unit for torque is the Newton-meter (Nm).

We can write the definition of torque as , where (the Greek letter tau) is torque, is how far the force is applied from the axis of rotation, is the force magnitude, and is the angle between the force and radial vector from the axis of rotation and where the force is being applied.

As you read, pay attention to the diagram of an object rotating on a frictionless table in Figure 10.11. We can see the radius from the center of the table (the pivot point) and the mass at the end of the radius.

When solving dynamics problems, we first need to identify the system and draw a free-body diagram of all the forces acting upon the system. Once the forces acting upon the system are defined, we can use the torque equation and angular acceleration equations to solve the problem: , where is the moment of inertia, is torque, and is the rotational acceleration due to the torque.

Example 10.7 shows how to use these equations to determine the angular acceleration of a person pushing a merry-go-round. Here, the first step is to calculate torque. The next step is to calculate the moment of inertia. Finally, torque and moment of inertia are used to calculate the angular acceleration on the merry-go-round.

Watch this video, which accompanies what you just read. Note that Greg Clements discusses the moment of inertia or rotational inertia, and Figure 9.6 which is in our next reading.

We define the moment of inertia, or rotational inertia, as for a point mass, where is the mass of the object being rotated and is the radius from the pivot point to the end of the mass.

However, the moment of inertia for a distribution of mass that makes a shape, such as a rotating cylinder or sphere, applies different equations. For example, for a solid sphere rotating about a central axis going through the core of the sphere, the moment of inertia is where is the mass of the entire sphere and is the sphere's radius.

In Section 6.1, we said that an object in equilibrium must have no rotational acceleration. We can restate this by saying that an object in equilibrium must have a torque of zero. When no torque is acting on a system, no rotational acceleration is given to the system, and it remains in equilibrium.

As you read, pay attention to Figure 9.6, which shows the torque on a rotating plank of wood secured at a pivot point at one end. This diagram shows how the direction of the force impacts the rotation of the plank of wood.

When the force is perpendicular to the length of the plank of wood, the plank experiences torque, and it rotates. When the force is parallel to the length of the plank of wood, it does not experience a net force and therefore does not rotate or experience torque. When the force is at an angle other than 90° from the length of the plank, the plank experiences less torque than if the force was at 90° from the plank's length.

- This video discusses the concept of moment of inertia.
- This video offers demonstrations of rotational inertia, the property of an object that deals with the resistance to a change in the state of rotational motion. This depends on the mass of the object and the way that mass is distributed from the axis of rotation.

### 6.3: Applications of Statics

When performing calculations, the first step is to determine if the system is, in fact, in equilibrium. Recall from the previous section that two conditions must be met for a system to be in equilibrium: the system must not be accelerating and the torque must be zero. The second step is to draw a free-body diagram of the system. It is important to determine all of the forces acting upon the system. The third step is to solve the problem by applying the relevant conditions of equilibrium: force is zero, and torque is zero.

As you read, pay attention to Example 9.1, which shows how to do a statics problem. Here, children are balanced on a seesaw. We are given information about the masses of both children, and how far from the pivot point one child is sitting. We are asked to determine where the second child is sitting to balance.

In Figure 9.8 we see that the children are balanced and therefore are at equilibrium. The free-body diagram shows that there is no net force, and no net rotational acceleration. To determine the distance of the second child from the pivot point, we use the torque equation, and set torque equal to zero. To determine the upward balancing force from the pivot point, we use the fact that net force equals zero to solve for the individual force at the pivot point.

As you read, notice that Example 9.2 shows a similar worked example of a statics problem. Here, a pole vaulter holds a pole at one end and we are asked to calculate the forces from each of the pole vaulter's hands. We take the same approach as in Example 9.1.

### Unit 6 Assessment

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- This assessment
- Receive a grade
Take this assessment to see how well you understood this unit.

- This assessment
**does not count towards your grade**. It is just for practice! - You will see the correct answers when you submit your answers. Use this to help you study for the final exam!
- You can take this assessment as many times as you want, whenever you want.

- This assessment