### learning target

By the end of this section, you will be able to:

- Calculating the Moment of Inertia of a Rigid Body of Uniform Shape
- Apply the parallel axis theorem to find the moment of inertia about any axis parallel to the known axis
- Calculating the Moment of Inertia of Complex Objects

In the previous section, we defined the moment of inertia, but did not explain how it is calculated. In this section we show how to calculate the moments of inertia for several standard object types, and how to use known moments of inertia to calculate the moments of inertia for translational axes or complex objects. This section is useful for learning how to apply general equations to complex objects (an essential skill for more advanced physics and engineering courses).

### Moment of inertia

We define the moment of inertia*and*object to be$\mathrm{and}={\displaystyle \sum _{\mathrm{and}}{\mathrm{rice}}_{\mathrm{and}}{R}_{\mathrm{and}}^{2}}$For all the particles that make up the object. because*R*is the distance from each mass of the object to the axis of rotation. The moment of inertia of each object depends on the selected axis. To see this, let's take the simple example of two masses (with negligibly small masses) at the end of a massless rod (Figure 10.23) and calculate the moments of inertia of the two different axes. Summing by mass is easy in this case, since the two masses at the end of the weights can be approximated as point masses, so the summation is only two terms.

In the case of an axle, at the center of weight, each of the two masses*rice*is the distance*R*axis, the moment of inertia given by

$${\mathrm{and}}_{1}=\mathrm{rice}{R}^{2}+\mathrm{rice}{R}^{2}=2\mathrm{rice}{R}^{2}.$$

With the axis at the end of the weight passing through one of the masses, the moment of inertia is

$${\mathrm{and}}_{2}=\mathrm{rice}{(0)}^{2}+\mathrm{rice}{(2R)}^{2}=4\mathrm{rice}{R}^{2}.$$

From this result we can conclude that the rotation around the end weighs twice as much as the rotation around the middle.

I like10.23 (a) a weight with an axis of rotation in the middle; (b) a weight with an axis of rotation at one end.

In this example we have two mass points and the sum is easy to calculate. However, to deal with non-point objects, we need to carefully consider each term in the equation. This equation requires us to add each "mass" at a given distance from the axis of rotation. But what exactly does each "mass" mean? Remember that as we derive this equation, each piece of mass has the same velocity, which means that the entire mass must have a single distance*R*on the axis of rotation. However, this is not possible unless we take a very small mass*division of rice*,as the picture showsFigure 10.24.

I like10.24 Use an infinitesimal mass to calculate the contribution to the total moment of inertia.

Requires the use of extremely small masses*division of rice*Suggests that instead of creating discrete sums over finite masses, we can write the moment of inertia by computing the integral of an infinitely small mass:

$$\mathrm{and}={\displaystyle \sum _{\mathrm{and}}{\mathrm{rice}}_{\mathrm{and}}{R}_{\mathrm{and}}{}^{2}}\phantom{\rule{0ex}{0ex}}\text{become}\phantom{\rule{0ex}{0ex}}\mathrm{and}={\displaystyle \int {R}^{2}\mathrm{Man}\mathrm{rice}.}$$

October 19th

This is actually the form we need to generalize complex shape equations. The best way to understand how to calculate the moment of inertia of a particular shape is to study a specific example in detail. This is the focus of most of the rest of this section.

#### A uniform thin rod with a shaft running through the middle

Consider a thin mass of uniformity (density and shape)*rice*and length*large*as the picture showsFigure 10.25.We want a thin rod, so we can assume that the cross-sectional area of the rod is small, and the rod can be thought of as a series of masses along a 1D line. In this example, for simplicity, the axis of rotation is perpendicular to the rod and passes through the center point. Our task is to calculate the moment of inertia about this axis. We align the axes like this*z*-Axis is the axis of rotation and*X*- The length of the shaft through the rod as shown. This is an appropriate choice because then we can integrate*X*-operating system.

I like10.25 Calculation of moment of inertia*and*For a uniform thin rod about an axis passing through the center of the rod.

we define*division of rice*are the small mass units that make up the rod. The integral of the moment of inertia is the integral over the mass distribution. However, we know how to integrate through space, not mass. So we need to find a way to relate mass to spatial variables. we will dolinear mass density $\mathrm{Lift}$The mass of an object, that is, the mass per unit length. Since the mass density of this object is uniform, we can write

$$\mathrm{Lift}=\frac{\mathrm{rice}}{\mathrm{Lift}}\phantom{\rule{0ex}{0ex}}\text{or}\phantom{\rule{0ex}{0ex}}\mathrm{rice}=\mathrm{Lift}\mathrm{Lift}.$$

If we differentiate on each side of this equation, we find

$$\mathrm{Man}\mathrm{rice}=\mathrm{Man}(\mathrm{Lift}\mathrm{Lift})=\mathrm{Lift}(\mathrm{Man}\mathrm{Lift})$$

outside of$\mathrm{Lift}$is a constant. we chose to walk along the bar*X*-Axis for practical reasons - this option is very useful here. watch out for that stick*dividing line*completely wrong*X*-os and has length*dx*; Strictly speaking,$\mathrm{Man}\mathrm{Lift}=\mathrm{Man}X$In this case. That's why we can write$\mathrm{Man}\mathrm{rice}=\mathrm{Lift}(\mathrm{Man}X)$, which gives us an integration variable to work with. The distance of each mass block*division of rice*The slave axis is given by the variable*X*,as the picture shows. Putting everything together, we get

$$\mathrm{and}={\displaystyle \int {R}^{2}\mathrm{Man}\mathrm{rice}={\displaystyle \int {X}^{2}\mathrm{Man}\mathrm{rice}={\displaystyle \int {X}^{2}\mathrm{Lift}\mathrm{Man}X}}}.$$

The final step is to take care of our integration boundaries. bar extends from$X=\text{\u2212}\mathrm{large}\text{/}2$again$X=\mathrm{large}\text{/}2$, since the axis is in the middle of the bar$X=0$.which gives us

$$\begin{array}{cc}\hfill \mathrm{and}& ={\displaystyle \underset{\text{\u2212}\mathrm{large}\text{/}2}{\overset{\mathrm{large}\text{/}2}{\int}}{X}^{2}\mathrm{Lift}\mathrm{Man}X=\mathrm{Lift}}\frac{{X}^{3}}{3}{|}_{\text{\u2212}\mathrm{large}\text{/}2}^{\mathrm{large}\text{/}2}=\mathrm{Lift}\left(\frac{1}{3}\right)\left[{\left(\frac{\mathrm{large}}{2}\right)}^{3}-{\left(\frac{\text{\u2212}\mathrm{large}}{2}\right)}^{3}\right]\hfill \\ & =\mathrm{Lift}\left(\frac{1}{3}\right)\frac{{\mathrm{large}}^{3}}{8}(2)=\frac{\mathrm{rice}}{\mathrm{large}}\left(\frac{1}{3}\right)\frac{{\mathrm{large}}^{3}}{8}(2)=\frac{1}{12}\mathrm{rice}{\mathrm{large}}^{2}.\hfill \end{array}$$

We then calculate the moment of inertia for the same uniform thin rod but with a different choice of axis so that we can compare the results. We want the moment of inertia to be smaller around the axis through the center of mass than around the endpoints, like the weight at the beginning of this section. This happens because more mass is distributed away from the axis of rotation.

#### an equally thin rod with a shaft at the end

Now consider the same uniform thin mass rod*rice*and length*large*, but this time we move the rotation axis to the end of the rod. We want to find the moment of inertia about this new axis (Figure 10.26).crowd*division of rice*Also defined as the small mass elements that make up the rod. As before, we receive

$$\mathrm{and}={\displaystyle \int {R}^{2}\mathrm{Man}\mathrm{rice}={\displaystyle \int {X}^{2}\mathrm{Man}\mathrm{rice}={\displaystyle \int {X}^{2}\mathrm{Lift}\mathrm{Man}X}}}.$$

However, this time we have a different point limit. bar extends from$X=0$again$X=\mathrm{large}$, since the axis is at the end of the rod$X=0$.that's why we think

$$\begin{array}{cc}\hfill \mathrm{and}& ={\displaystyle \underset{0}{\overset{\mathrm{large}}{\int}}{X}^{2}\mathrm{Lift}\mathrm{Man}X=\mathrm{Lift}}\frac{{X}^{3}}{3}{|}_{0}^{\mathrm{large}}=\mathrm{Lift}\left(\frac{1}{3}\right)[{(\mathrm{large})}^{3}-{(0)}^{3}]\hfill \\ & =\mathrm{Lift}\left(\frac{1}{3}\right){\mathrm{large}}^{3}=\frac{\mathrm{rice}}{\mathrm{large}}\left(\frac{1}{3}\right){\mathrm{large}}^{3}=\frac{1}{3}\mathrm{rice}{\mathrm{large}}^{2}.\hfill \end{array}$$

I like10.26 Calculation of moment of inertia*and*For uniform thin rods with shafts passing through the rod ends.

Note that the moment of inertia of the bar about the endpoints is four times the moment of inertia about the midpoint (consistent with the barbell example).

### set of parallel axes

The similarity between the procedures for calculating the moment of inertia of a rod about the central axis and the axis of the rod end is striking, suggesting that there may be simpler ways to determine the moment of inertia of a rod about an arbitrary axis parallel to the axis passing through the center of mass. Such an axis is calledparallel operating system. There is a saying calledParallel shaft group, which we specify here but not in detail in this paper.

### set of parallel axes

allow*rice*is the mass of the object and is*Man*is the distance from the axis passing through the object's center of mass to the new axis. then we have

$${\mathrm{and}}_{\text{parallel operating system}}={\mathrm{and}}_{\text{Marseille Center}}+\mathrm{rice}{\mathrm{Man}}^{2}.$$

10.20

Let's apply this to the bar example we solved above:

$${\mathrm{and}}_{\text{end}}={\mathrm{and}}_{\text{Marseille Center}}+\mathrm{rice}{\mathrm{Man}}^{2}=\frac{1}{12}\mathrm{rice}{\mathrm{large}}^{2}+\mathrm{rice}{\left(\frac{\mathrm{large}}{2}\right)}^{2}=\left(\frac{1}{12}+\frac{1}{4}\right)\mathrm{rice}{\mathrm{large}}^{2}=\frac{1}{3}\mathrm{rice}{\mathrm{large}}^{2}.$$

This result is consistent with our longer calculation above. This is a useful equation that we will use in some examples and problems.

### check your understanding 10.5

What is the moment of inertia of the roller radius?*R*I have time*rice*Around an axis passing through a point on the surface as shown below?

#### uniform thin disk around an axis through the center

The integral to determine the moment of inertia of a two-dimensional object is a bit more complicated, but at this level of investigation a shape is usually performed - a uniform thin disk around an axis passing through its center (Figure 10.27).

I like10.27 Calculates the moment of inertia of a thin coil about an axis passing through its center.

Because the disk is so thin, we can assume that the mass is completely distributed inside*xy*- airplane. we restart the relationshipsurface mass density, which is the mass per unit area. Since it is uniform, the surface density of the mass is$\mathrm{the\; strait}$is a constant:

$$\mathrm{the\; strait}=\frac{\mathrm{rice}}{A}\phantom{\rule{0ex}{0ex}}\text{or}\phantom{\rule{0ex}{0ex}}\mathrm{the\; strait}A=\mathrm{rice},\phantom{\rule{0ex}{0ex}}\text{so}\phantom{\rule{0ex}{0ex}}\mathrm{Man}\mathrm{rice}=\mathrm{the\; strait}(\mathrm{Man}A).$$

Now we use surface simplification. This region can be thought of as a series of thin rings, each ring representing an increment of mass*division of rice*radius*R*Equidistant from the axis, as shown in part (b) of the figure. The infinitesimal area of each ring*I*is thus given by the length of each ring ($2\mathrm{Pi}R$) times the infinitesimal width of each ring*DR*:

$$A=\mathrm{Pi}{R}^{2},\mathrm{Man}A=\mathrm{Man}(\mathrm{Pi}{R}^{2})=\mathrm{Pi}\mathrm{Man}{R}^{2}=2\mathrm{Pi}R\mathrm{Man}R.$$

Then, the entire area of the disk consists of radii ranging from 0 to*R*. This radius becomes the limit of our integral*DR*, that is, we start with$R=0$again$R=R$.If we put them together we have

$$\begin{array}{cc}\hfill \mathrm{and}& ={\displaystyle \underset{0}{\overset{R}{\int}}{R}^{2}\mathrm{the\; strait}(2\mathrm{Pi}R)\mathrm{Man}R=2\mathrm{Pi}\mathrm{the\; strait}{\displaystyle \underset{0}{\overset{R}{\int}}{R}^{3}}\mathrm{Man}R=}2\mathrm{Pi}\mathrm{the\; strait}\frac{{R}^{4}}{4}{|}_{0}^{R}=2\mathrm{Pi}\mathrm{the\; strait}\left(\frac{{R}^{4}}{4}-0\right)\hfill \\ & =2\mathrm{Pi}\frac{\mathrm{rice}}{A}\left(\frac{{R}^{4}}{4}\right)=2\mathrm{Pi}\frac{\mathrm{rice}}{\mathrm{Pi}{R}^{2}}\left(\frac{{R}^{4}}{4}\right)=\frac{1}{2}\mathrm{rice}{R}^{2}.\hfill \end{array}$$

Note that this matches the specified valueFigure 10.20.

#### Calculating the Moment of Inertia of Complex Objects

Now consider a complex object u like thisFigure 10.28, showing a thin disk at the end of a thin rod. It cannot be easily integrated to determine moments of inertia because it is not a uniformly shaped object. However, if we go back to the original definition of the moment of inertia as a sum, we can conclude that the moment of inertia of a complex object can be found from the sum of each part of the object:

$${\mathrm{and}}_{\text{In the summer}}={\displaystyle \sum _{\mathrm{and}}{\mathrm{and}}_{\mathrm{and}}}.$$

10.21

It is important to pay attention to the moment of inertia of the bodyofficial 10.21yes*around a common axis*.For the purposes of this article, this will be a stick with a length*large*around its ends and a radius of*R*Rotate around an axis by a certain distance from the center$\mathrm{large}+R$, millet*R*is the radius of the disk. Let's determine the mass of the rod, it should be${\mathrm{rice}}_{\text{R}}$and the mass of the disc${\mathrm{rice}}_{\text{Man}}.$

I like10.28 A composite object consisting of a disc at the end of a rod. including axis of rotation*A*.

The moment of inertia of the rod is simply$\frac{1}{3}{\mathrm{rice}}_{\text{R}}{\mathrm{large}}^{2}$, but we have to use the parallel axis theorem to find the moment of inertia of the disk about the axis shown. The moment of inertia of the disk about its center is$\frac{1}{2}{\mathrm{rice}}_{\text{Man}}{R}^{2}$We apply the parallel axis theorem${\mathrm{and}}_{\text{parallel operating system}}={\mathrm{and}}_{\text{Marseille Center}}+\mathrm{rice}{\mathrm{Man}}^{2}$Look for

$${\mathrm{and}}_{\text{parallel operating system}}=\frac{1}{2}{\mathrm{rice}}_{\text{Man}}{R}^{2}+{\mathrm{rice}}_{\text{Man}}{(\mathrm{large}+R)}^{2}.$$

Adding the moment of inertia of the rod and the moment of inertia of the disc with the axis of rotation offset, we get that the moment of inertia of the complex body is the same

$${\mathrm{and}}_{\text{In the summer}}=\frac{1}{3}{\mathrm{rice}}_{\text{R}}{\mathrm{large}}^{2}+\frac{1}{2}{\mathrm{rice}}_{\text{Man}}{R}^{2}+{\mathrm{rice}}_{\text{Man}}{(\mathrm{large}+R)}^{2}.$$

#### Application of Moment of Inertia Calculation in Problem Solving

Now let's look at some practical applications of moment of inertia calculations.

### example October 11th

#### people in carousel

A 25kg child stands in the distance$R=\mathrm{1,0}\phantom{\rule{0ex}{0ex}}\text{rice}$from the axis of the carousel (Figure 10.29). The carousel can be approximated as a monolithic solid disk with a mass of 500 kg and a radius of 2.0 m. Determine the moment of inertia of the system.

I like10.29 Calculate the child's moment of inertia on the merry-go-round.

#### strategy

This question is about calculating the moment of inertia. We get the mass and distance of the child's axis of rotation, and the mass and radius of the carousel. Since the child's mass and size are much smaller than the carousel, we can approximate the child as a point mass. The notation we use is${\mathrm{rice}}_{\text{C}}=25\phantom{\rule{0ex}{0ex}}\text{Kilogram},{R}_{\text{C}}=\mathrm{1,0}\phantom{\rule{0ex}{0ex}}\text{rice},{\mathrm{rice}}_{\text{rice}}=500\phantom{\rule{0ex}{0ex}}\text{Kilogram},{R}_{\text{rice}}=\mathrm{2,0}\phantom{\rule{0ex}{0ex}}\text{rice}$.

Our goal is to find${\mathrm{and}}_{\text{In the summer}}={\displaystyle \sum _{\mathrm{and}}{\mathrm{and}}_{\mathrm{and}}}$.

#### solution

type,${\mathrm{and}}_{\text{C}}={\mathrm{rice}}_{\text{C}}{R}^{2}$, for the carousel,${\mathrm{and}}_{\text{rice}}=\frac{1}{2}{\mathrm{rice}}_{\text{rice}}{R}^{2}$.for this reason

$${\mathrm{and}}_{\text{In the summer}}=25{(1)}^{2}+\frac{1}{2}(500){(2)}^{2}=25+1000=1025\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{rice}}^{2}.$$

#### significance

This value should be close to the moment of inertia of the carousel itself, since it has much more mass distributed from its axis than the child.

### example 10.12

#### a stick and a medicine ball

Determine the moments of inertia about the two axes for the combination of the rod and the fixed sphere as shown below. The stick has a length of 0.5 m and a mass of 2.0 kg. A sphere has a radius of 20.0 cm and a mass of 1.0 kg.

#### strategy

Since we have a complex body in both cases, we can use the parallel axis theorem to find the moment of inertia about each axis. In (a), the center of mass of the sphere is farther away$\mathrm{large}+R$from the axis of rotation. In (b), the center of mass of the sphere is farther away*R*from the axis of rotation. In both cases the moment of inertia of the rod is about the axis at one end. RefersTable 10.4The moment of inertia for a single object.

- ${\mathrm{and}}_{\text{In the summer}}={\displaystyle \sum _{\mathrm{and}}{\mathrm{and}}_{\mathrm{and}}}={\mathrm{and}}_{\text{in stock}}+{\mathrm{and}}_{\text{ball}}$;

${\mathrm{and}}_{\text{ball}}={\mathrm{and}}_{\text{Marseille Center}}+{\mathrm{rice}}_{\text{ball}}{(\mathrm{large}+R)}^{2}=\frac{2}{5}{\mathrm{rice}}_{\text{ball}}{R}^{2}+{\mathrm{rice}}_{\text{ball}}{(\mathrm{large}+R)}^{2}$;

${\mathrm{and}}_{\text{In the summer}}={\mathrm{and}}_{\text{in stock}}+{\mathrm{and}}_{\text{ball}}=\frac{1}{3}{\mathrm{rice}}_{\text{in stock}}{\mathrm{large}}^{2}+\frac{2}{5}{\mathrm{rice}}_{\text{ball}}{R}^{2}+{\mathrm{rice}}_{\text{ball}}{(\mathrm{large}+R)}^{2};$

${\mathrm{and}}_{\text{In the summer}}=\frac{1}{3}(\mathrm{2,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram}){(\mathrm{0,5}\phantom{\rule{0ex}{0ex}}\text{rice})}^{2}+\frac{2}{5}(\mathrm{1,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram})(\mathrm{0,2}\phantom{\rule{0ex}{0ex}}{\text{rice})}^{2}+(\mathrm{1,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram}){(\mathrm{0,5}\phantom{\rule{0ex}{0ex}}\text{rice}+\mathrm{0,2}\phantom{\rule{0ex}{0ex}}\text{rice})}^{2};$

${\mathrm{and}}_{\text{In the summer}}=(\mathrm{0,167}+\mathrm{0,016}+\mathrm{0,490})\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{rice}}^{2}=\mathrm{0,673}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{rice}}^{2}.$ - ${\mathrm{and}}_{\text{ball}}=\frac{2}{5}{\mathrm{rice}}_{\text{ball}}{R}^{2}+{\mathrm{rice}}_{\text{ball}}{R}^{2}$;

${\mathrm{and}}_{\text{In the summer}}={\mathrm{and}}_{\text{in stock}}+{\mathrm{and}}_{\text{ball}}=\frac{1}{3}{\mathrm{rice}}_{\text{in stock}}{\mathrm{large}}^{2}+\frac{2}{5}{\mathrm{rice}}_{\text{ball}}{R}^{2}+{\mathrm{rice}}_{\text{ball}}{R}^{2}$;

${\mathrm{and}}_{\text{In the summer}}=\frac{1}{3}(\mathrm{2,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram}){(\mathrm{0,5}\phantom{\rule{0ex}{0ex}}\text{rice})}^{2}+\frac{2}{5}(\mathrm{1,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram})(\mathrm{0,2}\phantom{\rule{0ex}{0ex}}{\text{rice})}^{2}+(\mathrm{1,0}\phantom{\rule{0ex}{0ex}}\text{Kilogram}){(\mathrm{0,2}\phantom{\rule{0ex}{0ex}}\text{rice})}^{2}$;

${\mathrm{and}}_{\text{In the summer}}=(\mathrm{0,167}+\mathrm{0,016}+\mathrm{0,04})\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{rice}}^{2}=\mathrm{0,223}\phantom{\rule{0ex}{0ex}}\text{Kilogram}\xb7{\text{rice}}^{2}.$

#### significance

The moment of inertia of complex objects can be calculated more easily using the parallel axis theorem. We see that the moment of inertia in (a) is greater than that in (b). This is because the axis of rotation in (b) is closer to the center of mass of the system. A simple analogy is employees. The moment of inertia about the endpoints is$\frac{1}{3}\mathrm{rice}{\mathrm{large}}^{2}$, but the moment of inertia passing through the center of mass along its length is$\frac{1}{12}\mathrm{rice}{\mathrm{large}}^{2}$.

### example October 13th

#### pendulum shopping speed

rod pendulum (Figure 10.30) released from rest at an angle of$30\text{\xb0}$.It is 30cm long and weighs 300g. What is its angular velocity at its lowest point?

I like10:30 a.m. The rod pendulum is released at an angle from the rest position$30\text{\xb0}.$

#### strategy

Use energy saving to solve the problem. When triggered, the pendulum has a gravitational potential energy determined by the height of the center of mass above the lowest point of swing. At the bottom of the swing, all the gravitational potential energy is converted into rotational kinetic energy.

#### solution

A change in potential energy is equal to a change in rotational kinetic energy.$\mathrm{Man}\xfc+\mathrm{Man}\mathrm{Potassium}=0$.

On the swing:$\xfc=\mathrm{rice}G{H}_{\text{centimeter}}=\mathrm{rice}G\frac{\mathrm{large}}{2}(\text{cosine}\phantom{\rule{0ex}{0ex}}\mathrm{and})$.at the bottom of the swing$\xfc=\mathrm{rice}G\frac{\mathrm{large}}{2}$The rocker swings relative to the lowest point.

The top of the swing is rotational kinetic energy$\mathrm{Potassium}=0$.at the bottom of the swing$\mathrm{Potassium}=\frac{1}{2}\mathrm{and}{\mathrm{oh}}^{2}$.for this reason:

$$\text{Man}\xfc+\text{Man}\mathrm{Potassium}=0\Rightarrow (\mathrm{rice}G\frac{\mathrm{large}}{2}(1-\text{cosine}\phantom{\rule{0ex}{0ex}}\mathrm{and})-0)+(0-\frac{1}{2}\mathrm{and}{\mathrm{oh}}^{2})=0$$

or

$$\frac{1}{2}\mathrm{and}{\mathrm{oh}}^{2}=\mathrm{rice}G\frac{\mathrm{large}}{2}(1-\text{cosine}\phantom{\rule{0ex}{0ex}}\mathrm{and}).$$

after dissolution$\mathrm{oh}$, We have

$$\mathrm{oh}=\sqrt{\mathrm{rice}G\frac{\mathrm{large}}{\mathrm{and}}(1-\text{cosine}\phantom{\rule{0ex}{0ex}}\mathrm{and})}=\sqrt{\mathrm{rice}G\frac{\mathrm{large}}{1\text{/}3\mathrm{rice}{\mathrm{large}}^{2}}(1-\text{cosine}\phantom{\rule{0ex}{0ex}}\mathrm{and})}=\sqrt{G\frac{3}{\mathrm{large}}(1-\text{cosine}\phantom{\rule{0ex}{0ex}}\mathrm{and})}.$$

By interpolating the values we have

$$\mathrm{oh}=\sqrt{9.8\phantom{\rule{0ex}{0ex}}\text{rice}\text{/}{\text{small}}^{2}\frac{3}{\mathrm{0,3}\phantom{\rule{0ex}{0ex}}\text{rice}}(1-\text{cosine}\phantom{\rule{0ex}{0ex}}30)}=3.6\phantom{\rule{0ex}{0ex}}\text{radian}\text{/}\text{small}.$$

#### significance

Note that the angular velocity of a pendulum is independent of its mass.

## FAQs

### What is the formula for moment of inertia in physics 1? ›

The formula for the moment of inertia is the “sum of the product of mass” of each particle with the “square of its distance from the axis of the rotation”. The formula of Moment of Inertia is expressed as **I = Σ m _{i}r_{i}^{2}**.

**What is the formula for the moment of inertia in Openstax? ›**

This quantity is called the moment of inertia I, with units of kg · m 2 kg · m 2 : **I = ∑ j m j r j 2** . I = ∑ j m j r j 2 . For now, we leave the expression in summation form, representing the moment of inertia of a system of point particles rotating about a fixed axis.

**How do you solve moment of inertia problems? ›**

The moment of inertia of a system of particles is given by **I=∑imir2i**. I = ∑ i m i r i 2 . The moment of inertia of a body having continuous mass distribution is given by I=∫bodyr2dm, I = ∫ body r 2 d m , where r is perpendicular distance of the mass element dm from the axis.

**What is the moment of inertia for volume? ›**

Volume moment of inertia has units [m^5] , which you can think of as **[m^3*m^2]** , the first one for the volume, the second for the rest. Mass moment of inertial has units [kg*m^2] , the kg for the mass, the second is the same as the units above ( [m^2] ), So you need to go from [m^3] to [kg] .

**How do you calculate moment of inertia? ›**

Moments of inertia can be found by summing or integrating over every 'piece of mass' that makes up an object, multiplied by the square of the distance of each 'piece of mass' to the axis. In integral form the moment of inertia is **I=∫r2dm** I = ∫ r 2 d m .

**How do you calculate inertia force? ›**

The law states that **F = m a** , where is the force applied to an object, is the inertial mass of the object, and is the accleration of the object.

**What is moment of inertia for beginners? ›**

The moment of inertia is **a physical quantity which describes how easily a body can be rotated about a given axis**. It is a rotational analogue of mass, which describes an object's resistance to translational motion. Inertia is the property of matter which resists change in its state of motion.

**What is the moment of inertia in college physics? ›**

moment of inertia, in physics, **quantitative measure of the rotational inertia of a body**—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.

**Why do we calculate moment of inertia? ›**

The Importance of Moment of Inertia

The MOI of an object **determines how much torque an object needs to reach a specific angular acceleration**. When calculating torque, or rotational force, you need to know the mass MOI.

**What is the formula for moment of inertia for different objects? ›**

The formula for the moment of inertia is **I=m ×r²**. This is a general equation. The moment of inertia of different shapes and objects as well as the moment of inertia formula of different shapes and objects does not remain the same.

### What is moment of inertia examples? ›

Real life examples of moment of inertia

**FLYWHEEL of an automobile**: Flywheel is a heavy mass mounted on the crankshaft of an engine. The magnitude of MOI of the flywheel is very high and helps in storing the energy. Hollow shaft- An hollow shaft transmits more power compared to that of a solid shaft(both of same mass).

**What is moment of inertia 11? ›**

The moment of inertia of a body about a given axis is **equal to the sum of the products of the masses of its constituent particles and the square of their respective distances from the axis of rotation**. Its unit is kg. m ^{2} and its dimensional formula is [ML ^{2}].

**Why is moment of inertia in 4? ›**

This is important because **it specifies the area's resistance to bending**. The equation for planar moment of inertia takes the second integral of the distance to the reference plane, multiplied by the differential element of area. The result is expressed in units of length to the fourth power: m^{4} or in^{4}.

**What is ML 2 moment of inertia? ›**

Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, it is the rotational analogue to mass (which determines an object's resistance to linear acceleration). The moments of inertia of a mass have units of dimension **ML ^{2}([mass] × [length]^{2})**.

**What is law of inertia in physics example? ›**

Examples of Law of Inertia in Everyday Life (Inertia of Motion) **When the bus stops suddenly, people fall forward**. When the driver of a bus brakes suddenly, the lower part of the body comes to rest as the bus comes to rest but the upper part of the body continues to move forward due to inertia of motion.

**What is the force of inertia? ›**

The force of inertia is **the property common to all bodies that remain in their state, either at rest or in motion, unless some external cause is introduced to make them alter this state**. Bodies do not display this force except when one alters their state; at which point it is called resistance or action.

**What is the formula for 2nd law of inertia? ›**

Newton's Second Law, **F = ma** is a vector equation. It says that the net force (a vector) acting on a mass m (a number) causes an acceleration (a vector) of the object in the same direction as the net force. The net force is the combined force of all individual forces acting on an object.

**Why do we calculate moments? ›**

Introduction. Statistical Moments plays a crucial role while we specify our probability distribution to work with since, with the help of moments, **we can describe the properties of statistical distribution**. Therefore, they are helpful to describe the distribution.

**What is first moment formula? ›**

The first moment of area of a shape, about a certain axis, **equals the sum over all the infinitesimal parts of the shape of the area of that part times its distance from the axis [Σad]**.

**What is moment of a force in physics? ›**

**The turning effect of force** is known as moment of force. It is the product of the force multiplied by the perpendicular distance from the line of action of the force to the pivot or point where the object will turn. the S.I unit is newton metre (Nm).

### What is inertia for dummies? ›

The law states that, **if a body is at rest, it will remain at rest unless it is acted upon by a force**. Alternatively, if a body is moving at a constant speed in a straight line, it will keep moving in a straight line at constant speed unless acted upon by a force.

**What is inertia simplified? ›**

: **a property of matter by which it remains at rest or in unchanging motion unless acted on by some external force**.

**What is the inertia table? ›**

Widely used for experiments and calculations, we are supplying Inertia Table at affordable market rates. This apparatus is **designed to determine the moment of inertia of an irregular body about an axis passing through its centre of gravity and perpendicular to its plane by dynamical method**.

**Which law is moment of inertia? ›**

law of inertia, also called **Newton's first law**, postulate in physics that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

**Is moment of inertia always constant? ›**

**Moment of inertia is always constant**.

**What is difference between inertia and moment of inertia? ›**

**Momentum depends on mass and velocity.** **Inertia depends only on mass**. The total momentum is conserved. Inertia has nothing to do with the conservation of energy.

**What is the formula in physics? ›**

F | Force |
---|---|

m | Mass of the body |

a | Acceleration in velocity available |

**What is the formula for moment of an object? ›**

Moment is calculated using the moment equation. This equation is: **moment force = F sin(x) r**, where F is the applied force, x is the angle between the force vector and lever arm, and r is the distance between the axis and the applied force point.

**What is moment of inertia with formula and unit? ›**

Formula of the moment of inertia: Moment of Inertia, **I = ∑ m i r i 2** , where is the mass of the i t h particle in a body and is the distance of the i t h particle from the axis of rotation. S.I. unit of moment of inertia: I = k g × m 2 I = k g m 2. Thus, the S.I. unit of the moment of inertia is k g m 2 .

**What are 5 examples of law of inertia? ›**

**10 examples of law on inertia in our daily life**

- You tend to move forward when a sudden break is applied.
- You feel a backward force when the bus moves quickly from rest.
- Dusting bed with a broom removes dust due to inertia of rest.
- when you shake a branch the leaves get detached.
- Experiencing jerk when lift suddenly starts.

### Can inertia be negative? ›

Negative inertia is an unusual and counter-intuitive property of matter, extensively investigated in some of the most exotic branches of physics and engineering at both macroscopic and microscopic levels. Such an exotic property promises a wide range of applications, from Alcubierre drive to acoustic wave manipulation.

**Is moment of inertia zero? ›**

As we know, I=mr2 where r is the perpendicular distance between axis of rotation and centre of mass. **if r is 0 then moment of inertia is also zero**.

**What is inertia 8th? ›**

Inertia is defined as **a property of matter by which it remains at the state of rest or in uniform motion in the same straight line unless acted upon by some external force**.

**Can moment of inertia change? ›**

BUT... the moment of inertia is a property of the body. **It will not change as you change the position of your force**. In much the same way that the mass of a body will not change as you change the direction or magnitude of a force acting on it.

**What is a first moment? ›**

Definitions of first moment. **the sum of the values of a random variable divided by the number of values**. synonyms: arithmetic mean, expectation, expected value.

**What is the unit of inertia of mass? ›**

From this, the unit of mass is Kg and the unit of r, i.e., distance is m. Therefore, the moment of the Inertia unit is **Kg.** **m2**.

**Is moment of inertia equal to 2 kinetic energy? ›**

The kinetic energy of the rotating rigid body is equal to the sum of the kinetic energies of all the particles. It shows that **moment of inertia of a body is equal to twice the kinetic energy of a rotating body whose angular velocity is one radian per second**.

**What is the symbol for inertia? ›**

The Moment of Inertia is often given the symbol I. It is the rotational analogue of mass. In Newtonian physics the acceleration of a body is inversely proportional to mass.

**Why is r2 in moment of inertia? ›**

tl;dr Basically, the reason that inertia is proportional to r2 is because **one of the r's comes from the fact that aT=rα, where aT is the tangential acceleration**. The other one comes from the fact that s=rθ.

**What is law of inertia 1? ›**

Newton's First Law: Inertia

Newton's first law states that **every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force**. This tendency to resist changes in a state of motion is inertia.

### What is the formula of 1st moment? ›

The first moment of area of a shape, about a certain axis, **equals the sum over all the infinitesimal parts of the shape of the area of that part times its distance from the axis [Σad]**.

**What is moment of inertia Class 1? ›**

Moment of Inertia: It is defined as **sum of product of mass and square of distance from the axis of rotation**. the point from the axis of rotation is called radius of gyration.

**What is inertia answers? ›**

Inertia can be defined as **a property of matter by which it remains at the state of rest or in uniform motion in the same straight line unless acted upon by some external force**. Inertia is a property of a body that resists changing its state of motion or state of rest.

**What is an example of a moment of inertia? ›**

Real life examples of moment of inertia

**FLYWHEEL of an automobile**: Flywheel is a heavy mass mounted on the crankshaft of an engine. The magnitude of MOI of the flywheel is very high and helps in storing the energy. Hollow shaft- An hollow shaft transmits more power compared to that of a solid shaft(both of same mass).

**Is inertia a force in physics? ›**

**Inertia is the force that holds the universe together**. Literally. Without it, matter would lack the electric forces necessary to form its current arrangement. Inertia is counteracted by the heat and kinetic energy produced by moving particles.

**What is Newton's law of inertia? ›**

law of inertia, also called Newton's first law, postulate in physics that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

**What causes inertia? ›**

The inertia is the result of **the initial resistance between the moving bodies against the static vacuum**. The moving body drags the resisting vacuum during acceleration, till the point that the vacuum travels with the moving body and has the same velocity.

**What is Newton's 2nd law called? ›**

To understand this we must use Newton's second law - **the law of acceleration** (acceleration = force/mass). Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

**How do I calculate a moment? ›**

The moment of a force about a point is **(the magnitude of the force) × (the perpendicular distance of the line of action of the force from the point)**.

**What is 1st order moment? ›**

The first-order moment (a statistical moment in mechanics) of a random variable X is **the mathematical expectation EX**. The value E(X−a)k is called the moment of order k relative to a, E(X−EX)k is the central moment of order k.

### Is moment of inertia easy? ›

It requires more efforts for larger mass to set up into the rotation. Thus, moment of inertia depends upon mass. **It is not equally easy to rotate both of them about the same axis of rotation**. More efforts are required for the object at a greater distance to accelerate to the same angular velocity.

**Is torque a vector? ›**

**Torque is a vector quantity**. The direction of the torque vector depends on the direction of the force on the axis.