Rotsional.inertis if You Dont Know Mass

5.two: Rotational Inertia

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Rotational Kinetic Free energy and Rotational Inertia

While our first arroyo to studying dynamics for linear motility was Newton's laws (forces cause accelerations), we will detect it easier to examine rotational dynamics from a standpoint of energy beginning. Consider an object that is rotating around a stationary center of mass. Does such an object possess kinetic free energy? We might be inclined to say that information technology does, but with the middle of mass not moving, its momentum is cypher, which would make the quantity \(\frac{p^2}{2m}\) too equal to zero.

We must non be such slaves to memorized equations! This equation (past itself) never applied to a system of multiple particles, which tin can easily have a nil full momentum and nevertheless withal have a nonzero kinetic energy. Well, rigid objects are systems of multiple particles, and when they are rotating, all those particles (except those right at the pivot point) are moving, which means they all have kinetic energy. At any given moment, in that location are particles moving in opposite directions, and if the middle of mass of the object is stationary, these reverse momenta (which are vectors) abolish, Their kinetic energies, on the other paw, are non vectors, and are all positive numbers, so they tin never cancel out.

In some sense, the particles comprising a rotating object can be thought of as contributing to the "internal" free energy of the object as we discussed dorsum in Section 4.4. But doing this runs contrary to the primary reason for the introduction of the mechanical/internal energy idea, which was to separate the kinetic energy of the system that we tin can clearly see from the kinetic energy that is concealed from u.s.a. inside the confines of the organisation. We can clearly see rotational motion of an object, so nosotros choose to include rotational kinetic energy in the category of "mechanical energy."

Okay, then a rotating object does possess kinetic energy. Our chore now is to express that kinetic energy in terms of the rotation variables nosotros accept already defined, merely all we know well-nigh kinetic energy is the linear version. In Figure v.2.i we consider the movement of a unmarried particle inside a rigid rotating object.

Effigy five.2.one – Motion of a Single Particle in a Rotating Rigid Body

This is particle #ane – one of many inside the rigid object. We can write down its kinetic free energy, and in fact we can limited it in terms of a rotational variable and the particle'due south distance from the pivot:

\[ KE_1 = \frac{i}{2}m_1v_1^2 = \frac{1}{2}m_1\left(R_1\omega\correct)^2 = \frac{1}{2}m_1R_1^2\omega^ii \]

If we want the full kinetic energy of the object, we need to add up the kinetic energy of all the particles. Thank you to our definition of angular velocity, we can cistron that function out of all the terms:

\[ KE_{whole\;object} = \frac{ane}{ii}m_1v_1^2 + \frac{one}{2}m_2v_2^ii + \dots = \frac{1}{two}m_1R_1^2\omega^2 + \frac{ane}{2}m_2R_2^2\omega^2 + \dots = \frac{i}{ii} \left[ m_1R_1^2 + m_2R_2^2 + \dots \right] \omega^2 \]

Discover that the quantity in brackets in the terminal equality is adamant past the distribution of mass throughout the object. That is, it is an intrinsic property of the object, not dependent upon how it is moving. We by and large abbreviate this quantity with an \(I\), which gives us a familiar form for the kinetic energy formula:

\[ KE_{rotating\;rigid\;body} = \frac{1}{2}I\omega^2, \;\;\;\;\;\;\;\;\; where:\;\; I \equiv m_1R_1^two + m_2R_2^2 + \dots \]

This looks just like the linear kinetic free energy formula, with the athwart speed replacing the linear speed, and \(I\) replacing the mass. This quantity certainly contains some information about the mass of the object, merely information technology is more than complicated than but the mass, and is called the rotational inertia , or more commonly (and less descriptively), the moment of inertia . Notice that this "inertia" depends not but upon the amount of stuff (mass), only likewise where that mass is. This means that ii different objects can actually weigh exactly the aforementioned corporeality, but when they are rotated at equal speeds, one of them has more KE than the other. As you might guess, this occurs when more of the mass is concentrated farther from the pivot for the erstwhile object than the latter.

Warning

It is important to note that we will only be because rotations around axes, not points. In our two-dimensional figures, an centrality that is perpendicular to the plane of the figure is duplicate from a single point, but we volition not discuss motion that involves an object'due south motion changing its plane of rotation. And so rotational inertia for three-dimensional objects involves the distances of the tiny masses from a common axis, not a common indicate.

Calculating Rotational Inertia for Continuous Objects

Our task is to compute the rotational inertia, for which the formula in terms of masses and their positions is different from the one for centre of mass (see Section 4.2), but the procedure is exactly the same. We start with the same picture (Figure iv.2.three, which is reproduced below), and convert the sums into integrals, as before.

Figure 5.ii.ii – Setup Diagram for Computations Involving Mass Density of a Sparse Rod

\[ I = dm_1x_1^two + dm_2x_2^2 + \dots = \int\limits_{x=0}^{x=L} dm\;ten^2 \]

Annotation that the rotational inertia is calculated around a specific pivot point, which we have chosen to exist our origin for the calculation.

Every bit before, we supercede the \(dm\) with \(\lambda \left(x\correct) dx\), and we have our formula for the rotational inertia along the \(x\)-centrality around the pin point at the origin:

\[ I = \int\limits_{x=0}^{ten=L} \lambda\left(10\right)x^2dx \]

Allow's return to the cases for which we computed the centers of mass in Department 4.two – the compatible and non-uniform rod. Different the case of center of mass, where the respond is a location on the rod, the terminal answer for the rotational inertia will accept units of \(kg \cdot m^ii\), and the formula for it will involve the total mass of the rod and its length. Besides information technology is of import to remember that while the center of mass is a location that doesn't depend upon where we put our coordinate arrangement to calculate it, the rotational inertia is only defined relative to a specific pivot point.

A Uniform Rod of Mass M and Length Fifty, Pivoted Virtually an End

Plugging the constant \(\lambda\) into Equation five.2.v and performing the integral gives:

\[ I = \int\limits_{10=0}^{10=Fifty} \lambda x^2dx = \lambda \left[\frac{1}{3}x^3\correct]_0^Fifty =\frac{ane}{3} \lambda L^3 \]

We are non finished yet, considering this reply is not in terms of the rod's mass. Since this rod is uniform, the mass is simply the (constant) density multiplied past its length, which gives:

\[ I =\frac{1}{iii} \left(\dfrac{M}{L}\right) L^3 = \frac{1}{3} 1000 50^2 \]

We will discover that every rotational inertia we come across has this bones grade: A abiding (usually written as a fraction) multiplied by the mass of the object and the square of some natural length dimension of the object. In this case information technology is the length of the rod, merely it may also be something similar the radius of a disk or sphere.

A Non-Uniform Rod of Mass M and Length Fifty, Pivoted About Its Lighter Stop

Now we repeat the process for the not-compatible density function for which we computed the center of mass in Department iv.2:

\[ \lambda\left(ten\correct) = \lambda_o\left(\dfrac{x}{L}+1\right)\]

Notation that unlike the uniform example, the results should not come up out the aforementioned for both ends of the rod, since more than of the mass is concentrated nigh the end at \(x=L\). We are calculating this rotational inertia about the lighter end, since all of the \(x\) values in the integral are measured from that terminate.

\[ I = \int\limits_{x=0}^{x=50} \lambda_o\left(\dfrac{ten}{L}+1\right)x^2dx = \lambda_o \left[\dfrac{one}{4L}x^4 +\frac{1}{3} x^iii\right]_0^50 = \frac{seven}{12} \lambda_o L^3\]

Nosotros are not done yet, because we are given the mass of the rod, not the constant \(\lambda_o\). We therefore need to compute the total mass in terms of this constant. We do this by integrating density function over the length of the rod:

\[ K = \int\limits_{x=0}^{10=Fifty} \lambda_o\left(\dfrac{10}{L}+i\correct)dx = \lambda_o \left[\dfrac{1}{2L}x^2 + 10\right]_0^L = \frac{3}{2} \lambda_o L \;\;\; \Rightarrow \;\;\; \lambda_o = \frac{2}{3} \dfrac{M}{50} \]

Plugging this back in above gives our answer:

\[ I = \frac{vii}{12} \left(\frac{2}{3} \dfrac{1000}{Fifty}\right) L^iii = \frac{vii}{18}ML^two\]

Instance \(\PageIndex{1}\)

Discover the rotational inertia of the non-uniform rod of mass \(M\) and length \(L\) whose mass density function is given by Equation 5.2.8, when rotated well-nigh its heavier end (\(10=Fifty\)).

Solution

The difference betwixt this calculation and the i above is that the variable \(x\) that appears in Equation v.three.five doesn't friction match the \(x\) that appears in the density formula. The density formula is referenced to our coordinate organization, only the \(x\) in the rotational inertia integral represents the distance of each tiny slice of mass \(dm\) from the pivot bespeak at \(x=L\). So nosotros need to make a change in the integral so that the \(x\) variable that appears in information technology matches the \(x\) in the density function. Making the substitution \(x \rightarrow L-ten\) (so \(dx \rightarrow -dx\)), into the integral does the trick, because then the integrand is zero at the pivot point (\(10=L\)) as it should be:

\[ \begin{assortment}{l} I_{heavy\;end} & = \int\limits_{x=L}^{10=0} dm\;\left(10-L\right)^2 = \int\limits_{x=L}^{x=0} \lambda\left(x\correct)\left(x-L\right)^2\left(-dx\right)\ = \int\limits_{x=0}^{ten=L} \lambda_o\left(\dfrac{ten}{50}+i\correct)\left(10-L\right)^2dx \\ & = \lambda_o \int\limits_{ten=0}^{x=L} \left(\dfrac{x^3}{50}-x^2-xL+L^two\correct)dx = \lambda_o\left[\dfrac{ten^4}{4L}-\dfrac{x^three}{iii}-\dfrac{x^2L}{2}+twoscore^2\right]_0^L \\ & = \frac{5}{12}\lambda_o L^three \finish{array}\nonumber \]

We need to plug in for \(\lambda_o\) (which was computed higher up) to go our final answer:

\[ I_{heavy\;end} = \frac{5}{12}\left(\frac{ii}{three} \dfrac{M}{L}\right) L^3 = \boxed{\frac{5}{eighteen}ML^ii} \nonumber \]

Principal Axes

It's clear that the choice of the pin is important to the calculation of the rotational inertia, but then is the axis. Existent objects are 3-dimensional, then they really take three independent rotation axes, each of which has its own rotational inertia around it. These axes are chosen the master axes . The origin of these axes is located at – what else? – the centre of mass of the object. The principal axes are only like shooting fish in a barrel to identify for objects with some degree of symmetry. Some objects are so symmetric that more than one gear up of axes volition work. For case, a uniform sphere has so much symmetry that whatever prepare of three mutually perpendicular axes whose origin coincides with the heart of the sphere volition piece of work, and of class the rotational inertias around all these axes are the same.

The reason it is natural to ascertain the origin of the chief axes to be at the eye of mass is that if an object is rotating freely in space with no forces on it, its centrality of rotation must pass through its center of mass (though it doesn't need to exist around one of the principal axes). This is actually surprisingly easy to evidence. Suppose an object was rotating around an axis that does not laissez passer through the center of mass. This would mean that the center of mass is moving in a circle effectually the axis of rotation. But circular move is accelerated motion. Co-ordinate to Newton's second police force, the heart of mass cannot exist accelerating if there are no forces on the object, which contradicts our supposition.

Computing Rotational Inertia Without Integration

Throughout our report of mechanics, our goal has been to develop shortcut tools to help us deal with concrete systems in simpler ways. We developed work-energy so that we could solve problems that pay no attention to direction or time without slogging through Newton'southward laws (such as speed at a given height on a loop-de-loop). We developed impulse-momentum so that nosotros could more hands solve problems involving systems in which the internal forces are complicated (such every bit collisions). At present nosotros are developing a tools related to rigid torso rotations so that nosotros don't have to rail the linear motions of all the particles in the system. With this very practical mindset, information technology is non surprising that physicists have developed tools for computing rotational inertia that avoid the ugliness of e'er having to perform integrals. The first such shortcut is just a collection of rotational inertias that are associated with mutual symmetric geometries, such as rods, disks, and spheres. Our collection is given at the end of the section. There are 2 tools that we can combine with our collection of rotational inertias that will allow us to "bootstrap" our way to determining many more.

Additivity Around a Mutual Axis

Suppose we know the rotational inertias of 2 separate objects around a common axis. If these two objects are attached so that they rotate together rigidly effectually that common axis, then the rotational inertia of the combined object is only the sum of their rotational inertias. This is evident from the formula for rotational inertia: Each object has its ain sum of \(mx^2\) terms, and when the objects are combined such that their \(x\) axes are common, then the new sum of \(mx^2\) terms is simply the combination of the two individual sums. To summarize:

\[I = I_1 +I_2 \]

Instance \(\PageIndex{2}\)

Use the condiment holding of rotational inertia and the issue given by Equation 5.ii.7 to find the rotational inertia of a compatible thin rod of mass \(M\) and length \(Fifty\) about its heart of mass.

Solution

We can treat a rod rotated around an centrality through its center equally if it is two separate one-half-rods of one-half the mass and half the length, fastened at their ends. The centrality that passes though the eye of the rod passes through the ends of these two half-rods, and we know the rotational inertia of each half-rod. The additivity property then gives us the rotational inertia of the whole rod about its centre:

\[ I_{uniform\;thin\;rod\;nearly\;its\;eye} = 2I_{half-rod\;about\;terminate} = 2 \left[ \frac{1}{3} \left( \dfrac{Grand}{2} \right) \left( \dfrac{Fifty}{2} \right)^2 \correct] = \boxed{ \frac{i}{12}ML^2} \nonumber \]

Parallel Axis Theorem

Equally we have seen multiple times already, just changing the axis effectually which an object is rotated will result in a different rotational inertia. Suppose nosotros summate the rotational inertia of an object about an centrality, then slide that axis in a parallel style on the object, and calculate the new rotational inertia, and so do it over and over, recording the new values each time. I might ask, "Where is the axis (parallel to the original one) for which the rotational inertia is the smallest?" Is there any way to estimate where this might be, and is it unique, or might there exist multiple places where the rotational inertia hits a minimum?

To answer this question, permit'southward await at a one-dimensional object that lies forth the \(ten\)-axis, and consider its rotational inertia effectually the \(y\)-axis. Writing it as a sum rather than an integral, it is:

\[ I = m_1x_1^ii + m_2x_2^2 + \dots\]

Now let's suppose nosotros decide to change where nosotros place the origin, moving it a distance \(+x\) forth the \(x\)-axis. When we do this, the distance from the centrality to mass \(m_1\) changes from \(x_1\) to \(x_1-x\). Too, since the original axis went through the origin, this new axis is no longer the \(y\)-axis – at present it intersects the \(x\)-axis at \(10\). The new rotational inertia is, therefore:

\[ I = m_1\left(x_1-ten\right)^2 + m_2\left(x_2-x\right)^2 + \dots\]

We can consider this to exist a function of \(x\). That is, this formula provides the rotational inertia of the object nigh the axis located at \(10\). We can now respond our question virtually where the rotational inertia is a minimum by using calculus. The value of \(x\) for which the part \(I\left(x\right)\) is a minimum satisfies:

\[ 0 = \dfrac{dI}{dx} = -2m_1\left(x_1-x\right) - 2m_2\left(x_2-x\right) + \dots\]

Solving for \(x\) here provides a familiar result:

\[ x = \dfrac{m_1x_1 + m_2x_2 + \dots}{m_1 + m_2 + \dots} \]

The rotational inertia of an object for all axes parallel to each other is a minimum for the axis that passes through the center of mass! Actually, this should non be besides surprising. The rotational inertia of an object volition be minimized effectually an centrality that is as close equally possible to equally much of the object'south mass equally possible, and the center of mass is the "average location of mass," so it makes sense that this would be "as close to as much of the object's mass equally possible."

Given this information, we tin can write the rotational inertia of an object effectually an centrality parallel to an axis passing through the centre of mass a positive-valued "adjustment" to the rotational inertia effectually the center of mass. It turns out (we will not prove it here) that this adjustment is quite simple – it is only the mass of the object multiplied by the square of the starting time distance between the new axis and the axis through the center of mass. This is called the parallel axis theorem :

\[ I_{new} = I_{cm} + Md^2, \]

where \(d\) is the altitude separating the new axis and the center of mass.

Example \(\PageIndex{3}\)

Use the parallel axis theorem and the outcome given by Equation 5.2.7 to find the rotational inertia of a compatible thin rod of mass \(G\) and length \(L\) well-nigh its center of mass.

Solution

The distance that the terminate of the rod is separated from the rod's center of mass is \(d=L/ii\). Plugging this into the parallel centrality theorem gives our answer, which agrees with what nosotros got in Example 5.2.2:

\[ I_{new} = I_{cm} + Md^2 \;\;\; \Rightarrow \;\;\; I_{cm} = I_{new} - Md^two = \frac{1}{3}ML^two - M\left(\dfrac{Fifty}{ii}\right)^2 = \left(\frac{ane}{iii}-\frac{1}{four}\right)ML^2 = \boxed{\frac{1}{12}ML^2} \nonumber \]

Rotational Inertias of Some Mutual Geometries

In all of the cases indicated below, the mass of the object is \(1000\), and the fabric making upwardly the object has uniform density. The reader is encouraged (equally an exercise) to navigate their way betwixt various relations using the additivity and parallel axes theorem tools. [Note: When it comes to rotating two-dimensional objects such as rings and disks, we will confine our studies to axes perpendicular to the two-dimensional planes in which these objects lie. For rotations around axes parallel to this plane, 1 would need yet another useful tool, known every bit the perpendicular axes theorem.]

Sparse Rods

Effigy 5.2.iii – Sparse Directly Rod Rotated Nigh I stop

\[I = \frac{1}{3}ML^2\]

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Figure 5.2.4 – Thin Directly Rod Rotated Nearly Centre

\[I = \frac{one}{12}ML^2\]

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Effigy five.2.5 – Sparse Round Ring (or Sparse Cylindrical Beat) Rotated Nearly Center

\[I = MR^2\]

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Figure 5.2.6 – Sparse Circular Ring (or Sparse Cylindrical Shell) Rotated About Edge

\[I = 2MR^2\]

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Disks (or Cylinders)

Figure 5.2.7 – Solid Deejay (or Cylinder) Rotated About Center

\[I = \frac{i}{2}MR^2\]

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Figure 5.2.8 – Solid Deejay (or Cylinder) Rotated About Border

\[I = \frac{3}{two}MR^2\]

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Effigy 5.2.9 – Hollow Disk (or Cylinder) Rotated Well-nigh Center

\[I = \frac{1}{two}M\left(R_1^2+R_2^ii\right)\]

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Spheres

Figure 5.2.10 – Solid Sphere Rotated About Center

\[I = \frac{2}{five}MR^2\]

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F igure five.2.11 – Solid Sphere Rotated About Edge

\[I = \frac{seven}{5}MR^2\]

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F igure five.2.12 – Sparse Spherical Beat out Rotated About Center

\[I = \frac{2}{3}MR^2\]

Instance \(\PageIndex{4}\)

The frame of a badminton racquet is constructed from 2 identical thin aluminum rods of compatible density, mass M, and length Fifty. One of the rods is aptitude into a circle and is welded to the end of the other rod. Notice the rotational inertia of this racquet around the axis perpendicular to the aeroplane of the circle that passes through the bespeak where the loop connects to the rod.

Solution

Commencement with a diagram of the racquet:

The points labeled \(cm_1\) and \(cm_2\) are the centers of mass of the ii rods, respectively. Nosotros accept the rotational inertias of the two rods about their respective centers of mass, and using the fact that the length of the bent rod is its circumference gives:

\[ \begin{assortment}{50} I_{cm_1} = \dfrac{1}{12} ML^2 \\ I_{cm_2} = MR^2 = Thou\left(\dfrac{L}{2\pi}\right)^2 = \dfrac{ane}{4\pi^2}ML^ii \stop{array} \nonumber \]

Now we demand to utilise our 2 tools. First, we need to change the axes for each of these rotational inertias from \(cm_1\) and \(cm_2\) to the pivot. We do this using the parallel axis theorem. The distance that the new axis is from \(cm_1\) is \frac{L}{2} , and the distance betwixt \(cm_2\) and the pin is \(R=\frac{50}{2\pi}\), and then:

\[ \begin{array}{fifty} I_1 = I_{cm_1} + Md_1^2 = \dfrac{1}{12} ML^2 + G\left(\dfrac{L}{two}\right)^two = \dfrac{ane}{iii} ML^2 \\ I_2 = I_{cm_2} + Md_2^2 = \dfrac{i}{4\pi^ii}ML^2 + Thou\left(\dfrac{L}{2\pi}\right)^ii = \dfrac{1}{2\pi^two} ML^2 \end{array} \nonumber \]

And at present that we take the 2 rotational inertias about the same axis, we can use the additive property to get the full rotational inertia:

\[ I = I_1 + I_2 = \boxed{\left(\dfrac{i}{3} + \dfrac{1}{2\pi^ii}\right)ML^2} \nonumber \]

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Source: https://phys.libretexts.org/Courses/University_of_California_Davis/UCD%3A_Physics_9A__Classical_Mechanics/5%3A_Rotations_and_Rigid_Bodies/5.2%3A_Rotational_Inertia

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