### learning target

At the end of this section, you can:

- Describe the difference between rotational and kinetic energy
- Definition of the physical term for the moment of inertia in terms of the mass distribution of the axis of rotation
- Explain how the moment of inertia of a rigid body affects its rotational kinetic energy
- Using conservation of mechanical energy to analyze systems undergoing rotation and translation
- Calculate the angular velocity of a rotating system when energy is lost due to non-conservative forces

So far in this chapter we have discussed rotational kinematics: describing the motion of a rotating rigid body with a fixed axis of rotation. In this section, we define two new quantities that can be used to analyze the properties of rotating objects: moment of inertia and rotational kinetic energy. Now that these properties are defined, we have two important tools needed to analyze rotational dynamics.

### rotational kinetic energy

Every object in motion has kinetic energy. We know how to account for bodies undergoing translational motion, but what about rigid bodies undergoing rotation? This might seem complicated since every point on the rigidbody has a different velocity. However, we can express the kinetic energy of a rotating body in terms of angular velocity (which is the same for the entire rigid body).Figure 10.17An example of a high-energy rotating body is shown: a motorized grinding wheel driven by an electric motor. When the millstone is working, sparks fly, creating noise and vibration. The system has a lot of energy, some of which is in the form of heat, light, sound, and vibrations. However, most energy exists in the formrotational kinetic energy。

Lectra10.17 The rotational kinetic energy of the whetstone is converted into heat, light, sound and vibration. (Courtesy of Zachary David Bell, USN)

Energy in rotational motion is not a new form of energy; rather, it is energy associated with rotational motion, the same as kinetic energy in translational motion. However, since the kinetic energy is given by$K=\frac{1}{2}\mathrm{Medium}{v}^{2}$, and the velocity is different for every point on an object rotating around an axis, so it makes sense to find a way to represent kinetic energy in terms of variables$\mathrm{oh}$, which is the same for all points on the rigid body of revolution. For a single particle rotating about a fixed axis, this is easy to compute. Using this relationship, we can relate the angular velocity to the magnitude of the translational velocity${v}_{\text{time}}=\mathrm{oh}\mathrm{right}$, Where*right*is the distance of the particle from the axis of rotation i${v}_{\text{time}}$is the tangential velocity. Substituting kinetic energy into the equation, we find that

$$K=\frac{1}{2}\mathrm{Medium}{v}_{\text{time}}^{2}=\frac{1}{2}\mathrm{Medium}{\uff08\mathrm{oh}\mathrm{right}\uff09}^{2}=\frac{1}{2}\uff08\mathrm{Medium}{\mathrm{right}}^{2}\uff09{\mathrm{oh}}^{2}\u3002$$

In the case of rigid bodies of revolution, each body can be divided into a large number of smaller masses, each of which has a mass${\mathrm{Medium}}_{J}$and the distance from the axis of rotation${\mathrm{right}}_{J}$, making the total mass of the object equal to the sum of its individual masses:$\mathrm{Medium}={\displaystyle \underset{J}{S}{\mathrm{Medium}}_{J}}$. Each smaller mass has a tangential velocity${v}_{J}$, we delete the index*time*at this time. The total kinetic energy of the rigid body rotating body is

$$K={\displaystyle \underset{J}{S}\frac{1}{2}{\mathrm{Medium}}_{J}{v}_{J}^{2}=}{\displaystyle \underset{J}{S}\frac{1}{2}{\mathrm{Medium}}_{J}{\uff08{\mathrm{right}}_{J}{\mathrm{oh}}_{J}\uff09}^{2}}$$

since then${\mathrm{oh}}_{J}=\mathrm{oh}$For all the masses,

$$K=\frac{1}{2}\uff08{\displaystyle \underset{J}{S}{\mathrm{Medium}}_{J}{\mathrm{right}}_{J}^{2}}\uff09{\mathrm{oh}}^{2}\u3002$$

10.16

Unit isofficial 10.16is the Joule (J). This form of the equation is complete but tricky; we have to find a way to generalize it.

### moment of inertia

if we compareofficial 10.16the way we go into kinetic energywork and kinetic energy,$\uff08\frac{1}{2}\mathrm{Medium}{v}^{2}\uff09$, which indicates that we have a new rotation variable that can be added to the list of relationships between rotation and translation variables. quantity$\underset{J}{S}{\mathrm{Medium}}_{J}{\mathrm{right}}_{J}^{2}$is the mass counterpart in the rotational kinetic energy equation. This is an important new term for rotational motion. This quantity is calledmoment of inertia *and*,Unit is$\text{Kilogram}\xb7{\text{Medium}}^{2}$:

$$\mathrm{and}={\displaystyle \underset{J}{S}{\mathrm{Medium}}_{J}{\mathrm{right}}_{J}^{2}}\u3002$$

10.17

For now, we keep the expression as a sum representing the moment of inertia of a point particle system rotating about a fixed axis. Note that the moment of inertia of a single point particle about a fixed axis is simply$\mathrm{Medium}{\mathrm{right}}^{2}$, met*right*where is the distance of the point particle from the axis of rotation. In the next section we will study the integral form of this equation, which can be used to calculate the moment of inertia of some regularly shaped rigid bodies.

Moment of inertia is a quantitative measure of rotational inertia, and like translational motion, mass is a quantitative measure of linear inertia—that is, the more massive an object, the greater its inertia and resistance. to change the line speed. Similarly, the greater the moment of inertia of a rigid body or particle system, the greater its resistance to changes in angular velocity about a fixed axis of rotation. It is interesting to observe how the moment of inertia changes*,*Distance to the axis of rotation of the mass particleofficial 10.17. Rigid bodies and systems with particles of greater mass concentrated at greater distances from the axis of rotation have greater moments of inertia than rigid bodies and systems of particles of the same mass but concentrated near the axis of rotation. Thus, we can see that a hollow cylinder, rotating about an axis passing through its centre, has a greater moment of inertia than a solid cylinder of the same mass. alternativesofficial 10.17Come inofficial 10.16, which becomes the expression of the kinetic energy of the rotating rigid body

$$K=\frac{1}{2}\mathrm{and}{\mathrm{oh}}^{2}\u3002$$

10.18

From this equation we can see that the kinetic energy of a rotating rigid body is proportional to the moment of inertia and the square of the angular velocity. this is abusedflywheelEnergy storage devices designed to store large amounts of rotational kinetic energy. Many automakers are now testing flywheel energy storage devices, such as flywheels or kinetic energy recovery systems, in their vehicles, as shown hereFigure 10.18。

Lectra10.18 A KERS (Kinetic Energy Recovery System) flywheel used on a car. (Source: "cmonville"/Flickr)

The rotational and translational quantities of kinetic energy and inertia are summarized asTable 10.4. The ratio column is not included because there is no constant by which we can multiply the rotation to get the translation, as we do for the variables inTable 10.3。

turn around | translate eel |
---|---|

$\mathrm{and}={\displaystyle \underset{J}{S}{\mathrm{Medium}}_{J}{\mathrm{right}}_{J}^{2}}$ | $\mathrm{Medium}$ |

$K=\frac{1}{2}\mathrm{and}{\mathrm{oh}}^{2}$ | $K=\frac{1}{2}\mathrm{Medium}{v}^{2}$ |

table 10.4 Rotational and translational energy and inertia

### example 10.8

#### The moment of inertia of the particle system

On a rod of negligible mass and length 0.5 m, 6 small rings each having a mass of 20 g are placed at a distance of 10 cm, and the rod is rotated about an axis located at 25 cm as shownFigure 10.19. (a) What is the moment of inertia of the system? (b) If the two rings closest to the axis are removed, what are the moments of inertia of the remaining four rings? (c) If the six-ring system is rotating at 5 revolutions per minute, what is its rotational kinetic energy?

Lectra10.19 A rod of negligible mass has six rings spaced 10 cm apart and rotates about a vertical axis.

#### strategy

- We estimate this quantity using the definition of the moment of inertia of the particle system and summing. All masses are the same, so we can put that quantity in front of the addition sign.
- We also do similar calculations.
- We substitute the result of (a) into the expression for rotational kinetic energy.

#### solution

- $\mathrm{and}={\displaystyle \underset{J}{S}{\mathrm{Medium}}_{J}{\mathrm{right}}_{J}^{2}}=\uff08\mathrm{0,02}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\uff09\uff082\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{\uff08\mathrm{0,25}\phantom{\rule{0ex}{0ex}}\text{Medium}\uff09}^{2}+2\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{\uff08\mathrm{0,15}\phantom{\rule{0ex}{0ex}}\text{Medium}\uff09}^{2}+2\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{\uff08\mathrm{0,05}\phantom{\rule{0ex}{0ex}}\text{Medium}\uff09}^{2}\uff09=\mathrm{0,0035}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{Medium}}^{2}$。
- $\mathrm{and}={\displaystyle \underset{J}{S}{\mathrm{Medium}}_{J}{\mathrm{right}}_{J}^{2}}=\uff08\mathrm{0,02}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\uff09\uff082\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{\uff08\mathrm{0,25}\phantom{\rule{0ex}{0ex}}\text{Medium}\uff09}^{2}+2\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{\uff08\mathrm{0,15}\phantom{\rule{0ex}{0ex}}\text{Medium}\uff09}^{2}\uff09=\mathrm{0,0034}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{Medium}}^{2}$。
- $K=\frac{1}{2}\mathrm{and}{\mathrm{oh}}^{2}=\frac{1}{2}\uff08\mathrm{0,0035}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{Medium}}^{2}\uff09\uff08\mathrm{5,0}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}2\mathrm{PI}\phantom{\rule{0ex}{0ex}}\text{rad}\text{/}{\text{S}\uff09}^{2}=\mathrm{1,73}\phantom{\rule{0ex}{0ex}}\text{J}$。

#### significance

We can see the individual contribution to the moment of inertia. The contribution of masses close to the axis of rotation is very small. When we remove them, it has very little effect on the moment of inertia.

In the next section, we generalize the summation equation for point particles and develop a method for computing the moment of inertia of a rigid body. but now,Figure 10.20Returns the value of the moment of inertia of a common object shape about the specified axis.

Lectra10.20 Moments of inertia for common object shapes.

### Applied rotational kinetic energy

Let us now apply the ideas of rotational kinetic energy and moment of inertia tables to feel the energy associated with a pair of rotating bodies. The following examples will also help you become familiar with the use of these equations. First, let's look at a general strategy for solving rotational energy problems.

### problem solving strategies

#### rotational energy

- Determine if energy or work is involved in the rotation.
- Identify systems of interest. Often a sketch will help.
- Analyze the situation to determine what types of work and energy are involved.
- Mechanical energy is conserved if no energy is lost due to friction and other non-conservative forces, ie:${K}_{\text{and}}+{U}_{\text{and}}={K}_{\text{F}}+{U}_{\text{F}}$。
- When non-conservative forces are present, mechanical energy is not conserved and other forms of energy (such as heat and light) can enter or leave the system. Determine what they are and calculate them if necessary.
- Eliminate terminology wherever possible to simplify algebra.
- Evaluate the numerical solution to see if it makes sense for the physical situation posed in the problem statement.

### example 10.9

#### Calculating the Energy of a Helicopter

typical little rescuehelicopterThere are four blades: each blade is 4.00 m long and has a mass of 50.0 kg (Figure 10.21). A blade can be approximated as a thin rod that rotates at one end about an axis perpendicular to its length. The helicopter has a total load mass of 1000 kg. (a) Calculate the kinetic energy of rotation when the blade rotates at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter flying at 20.0 m/s and compare it to the rotational energy of the blades.

Lectra10.21 (a) Sketch of a quadrocopter. (b) Helicopter water rescue operations operated by Westpac Rescue Helicopter Services Auckland. (Image credit b: "111 Emergency"/Flickr job changes)

#### strategy

Rotational and translational energies can be calculated according to their definitions. The formulation of the problem provides all the necessary constants for evaluating the rotational and translational kinetic energy expressions.

#### solution

- The rotational kinetic energy is
$$K=\frac{1}{2}\mathrm{and}{\mathrm{oh}}^{2}\u3002$$

We need to convert the angular velocity to radians per second and calculate the moment of inertia to find it*K*. angular velocity$\mathrm{oh}$yes$$\mathrm{oh}=\frac{300\phantom{\rule{0ex}{0ex}}\text{Rotating speed}}{\mathrm{1,00}\phantom{\rule{0ex}{0ex}}\text{minute}}\phantom{\rule{0ex}{0ex}}\frac{2\mathrm{PI}\phantom{\rule{0ex}{0ex}}\text{rad}}{\text{1 district}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{1,00}\phantom{\rule{0ex}{0ex}}\text{minute}}{\mathrm{60,0}\phantom{\rule{0ex}{0ex}}\text{S}}=\phantom{\rule{0ex}{0ex}}31.4\phantom{\rule{0ex}{0ex}}\frac{\text{rad}}{\text{S}}\u3002$$

The moment of inertia of a blade is equal to the moment of rotation of a thin rod at its end as followsFigure 10.20. total*and*is four times the moment of inertia because there are four blades. so,$$\mathrm{and}=4\frac{\mathrm{Medium}{L}^{2}}{3}=4\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}\frac{\uff08\mathrm{50,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\uff09{\uff08\mathrm{4:00\; AM}\phantom{\rule{0ex}{0ex}}\text{Medium}\uff09}^{2}}{3}=\mathrm{1067,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{Medium}}^{2}\u3002$$

Nurse$\mathrm{oh}$U*and*In the expression for rotational kinetic energy$$K=\mathrm{0,5}\uff081067\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{Medium}}^{2}\uff09{\text{(31.4 rad/s)}}^{2}=5.26\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{5}\phantom{\rule{0ex}{0ex}}\text{J}\text{\u3002}$$

- Substituting the given values into the translational kinetic energy equation, we get
$$K=\frac{1}{2}\mathrm{Medium}{v}^{2}=\uff08\mathrm{0,5}\uff09\uff08\mathrm{1000,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\uff09{\uff0820.0\phantom{\rule{0ex}{0ex}}\text{Mrs}\uff09}^{2}=\mathrm{2,00}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{5}\phantom{\rule{0ex}{0ex}}\text{J}\u3002$$

To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is$$\frac{\mathrm{2,00}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{5}\phantom{\rule{0ex}{0ex}}\text{J}}{5.26\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{5}\phantom{\rule{0ex}{0ex}}\text{J}}=\mathrm{0,380}\u3002$$

#### significance

The ratio of translational kinetic energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter comes from the rotating blades.

### example 10.10

#### boomerang energy

roll oneboomerangIn air at a velocity of 30.0 m/s at an angle of$\mathrm{40,0}\text{\xb0}$relative to the horizontal plane (Figure 10.22). It has a mass of 1.0 kg and spins at 10.0 revolutions per minute. The moment of inertia of the boomerang is$\mathrm{and}=\frac{1}{12}\mathrm{Medium}{L}^{2}$Where$L=\mathrm{0,7}\phantom{\rule{0ex}{0ex}}\text{Medium}$. (a) What is the total energy of the boomerang when it leaves the hand? (b) Neglecting air resistance, how high does the boomerang rise from the height of the fist?

Lectra10.22 The boomerang is thrown into the air at an initial angle$40\text{\xb0}$。

#### strategy

We use the definitions of rotational kinetic energy and linear kinetic energy to calculate the total energy of the system. The problem is that we have to ignore air resistance so we don't have to worry about energy loss. In part (b), we exploit the conservation of mechanical energy to find the maximum height of the boomerang.

#### solution

- Moment of inertia:$\mathrm{and}=\frac{1}{12}\mathrm{Medium}{L}^{2}=\frac{1}{12}\uff08\mathrm{1,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\uff09\uff08\mathrm{0,7}{\text{Medium}\uff09}^{2}=\mathrm{0,041}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{Medium}}^{2}$。

Angular velocity:$\mathrm{oh}=\uff0810.0\phantom{\rule{0ex}{0ex}}\text{Rotating speed}\text{/}\text{S}\uff09\uff082\mathrm{PI}\uff09=\mathrm{62,83}\phantom{\rule{0ex}{0ex}}\text{rad}\text{/}\text{S}$。

Therefore the rotational kinetic energy is$${K}_{\text{right}}=\frac{1}{2}\uff08\mathrm{0,041}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{Medium}}^{2}\uff09{\uff08\mathrm{62,83}\phantom{\rule{0ex}{0ex}}\text{rad}\text{/}\text{S}\uff09}^{2}=\mathrm{80,93}\phantom{\rule{0ex}{0ex}}\text{J}\u3002$$

Translational kinetic energy is$${K}_{\text{time}}=\frac{1}{2}\mathrm{Medium}{v}^{2}=\frac{1}{2}\uff08\mathrm{1,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\uff09\uff0830.0\phantom{\rule{0ex}{0ex}}\text{Medium}\text{/}\text{S}{\uff09}^{2}=\mathrm{450,0}\phantom{\rule{0ex}{0ex}}\text{J}\u3002$$

Therefore, the total energy of the boomerang is:$${K}_{\text{total}}={K}_{\text{right}}+{K}_{\text{time}}=\mathrm{80,93}+\mathrm{450,0}=\mathrm{530,93}\phantom{\rule{0ex}{0ex}}\text{J}\u3002$$

- We use the law of conservation of mechanical energy. Since the boomerang is launched at an angle, we must write the total energy of the system in terms of its linear kinetic energy, using the velocity
*X*- in*j*- Travel instructions. The total energy of the boomerang when it leaves the hand is$${\mathrm{Second}}_{\text{for}}=\frac{1}{2}\mathrm{Medium}{v}_{X}^{2}+\frac{1}{2}\mathrm{Medium}{v}_{j}^{2}+\frac{1}{2}\mathrm{and}{\mathrm{oh}}^{2}\u3002$$

The total energy at the highest altitude is$${\mathrm{Second}}_{\text{Last}}=\frac{1}{2}\mathrm{Medium}{v}_{X}^{2}+\frac{1}{2}\mathrm{and}{\mathrm{oh}}^{2}+\mathrm{Medium}GH\u3002$$

By saving mechanical energy,${\mathrm{Second}}_{\text{for}}={\mathrm{Second}}_{\text{Last}}$So, after removing similar conditions, we$$\frac{1}{2}\mathrm{Medium}{v}_{j}^{2}=\mathrm{Medium}GH\u3002$$

from${v}_{j}=30.0\phantom{\rule{0ex}{0ex}}\text{Medium}\text{/}\text{S}\uff08\text{Slam}\phantom{\rule{0ex}{0ex}}40\text{\xb0}\uff09=19.28\phantom{\rule{0ex}{0ex}}\text{Medium}\text{/}\text{S}$, we discover$$H=\frac{{\uff0819.28\phantom{\rule{0ex}{0ex}}\text{Medium}\text{/}\text{S}\uff09}^{2}}{2\uff089.8\phantom{\rule{0ex}{0ex}}\text{Medium}\text{/}{\text{S}}^{2}\uff09}=18.97\phantom{\rule{0ex}{0ex}}\text{Medium}\u3002$$

#### significance

In part (b), the solution shows how energy conservation can be an alternative to solving problems that are usually solved using kinematics. In the absence of air resistance, rotational kinetic energy plays no role in the maximum altitude solution.

### check your understanding 10.4

The moment of inertia of nuclear submarine propeller is$\mathrm{800,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{Medium}}^{2}$. If the speed of the propeller underwater is 4.0 rpm when the engine is off, what is the speed of the propeller after 5.0 seconds when water resistance removes 50,000 J from the system?