When rigid objects rotate they will contain kinetic energy and associated angular momentum. What we find, however, is that rather than depending only on mass, as both [math]K=\tfrac{1}{2}mv^2[/math] and [math]\vec{p}=m\vec{v}[/math] do, the rotational quantities will depend on two things: 1) How much mass is rotating, and 2) How that mass is distributed around the axis of rotation.[br][br]Together, these two things are accounted for by using the moment of inertia. It is often derived by considering the kinetic energy of a system of N particles (where N is just some integer). The kinetic energy for a system of N particles is just the sum of the individual particle kinetic energies. Then the derivation proceeds with simplifications that are possible if those individual particles are locked together such that they form a rigid object in rotation. For instance, all of the parts must have the same angular velocity (and corresponding period of rotation). The derivation for a system of N particles goes like this: [br][br][center][math]K_{system}=\Sigma_{i=1}^N K_i \\[br]K_{system}=\Sigma_{i=1}^N \tfrac{1}{2}m_iv_i^2 \\[br]\text{Using $v_{tan}=r\omega$, and understanding that the parts of a rigid body can only have $v_{tan}$, } \\[br]K_{system}=\Sigma_{i=1}^N \tfrac{1}{2}m_i(r_i\omega_i)^2 \\[br]\text{Since all parts of a rigid body have the same angular velocity \omega, it doesn't need a subscript. } \\[br]K_{system}=\tfrac{1}{2}(\Sigma_{i=1}^N m_ir_i^2)\omega^2 \\[br]K_{system}=\tfrac{1}{2}I\omega^2 \\[br]\text{Here we define the moment of inertia: } \\[br]I\equiv\sum_{i=1}^N m_ir_i^2.[br][/math][/center]Notice that in this derivation we also found the kinetic energy associated with a rotating object. We will use this result later in this chapter. It is [math]K_{rot}=\tfrac{1}{2}I\omega^2.[/math]

[color=#1e84cc]EXAMPLE: As an example, consider something like a diatomic molecule of nitrogen. What is the moment of inertia of the molecule about its center of mass? The bond length (separation of atoms) is [math]L=1.1\AA,[/math] and the mass of nitrogen is [math]m_{N_{14}}=\tfrac{14 g}{mole}.[/math] By definition, [math]1\AA\equiv 1\times10^{-10}m.[/math] This unit is called the Ångstrom, usually written with a little circle above the A.[br]ANSWER: We will view it like two masses on the end of a stick, where the mass-less stick represents the bond and the masses are the atoms. The mass of each atom is just the mass in kilograms given above divided by Avogadro's number, or the number of atoms in one mole of a substance. Avogadro's number is [math]N_A=6.02\times10^{23}\tfrac{atoms}{mole}.[/math] The inertia measured about the center of mass would be [math]I=m(\tfrac{L}{2})^2+m(\tfrac{L}{2})^2=2m(\tfrac{L}{2})^2=\tfrac{1}{2}mL^2.[/math][/color]

Given an object like a plank of wood or a disk, how would we find the moment of inertia? We will need a continuous version of the sum. The inertia above is a sum of masses times the square of the distance of each mass away from the rotation axis. The way to turn the sum to an integral is just to have the masses go to infinitesimally small masses, so we get:[br][center][math][br]I=\int r^2\;dm.[/math][/center] [br]Besides thinking about solving the integral, we need to discuss how to find a proper expression for 'dm', since it has something to do with geometry of the object.[br][br]Let's assume we have something like a meter stick of length L and mass m made of some uniformly dense material like plastic. If we chop the stick into smaller and smaller pieces, there will always be a quantity that is unchanging about each piece - the ratio of mass to length. The reason for this is that while a stick cut in half has half of its original mass, it also has half its original length. Therefore the ratio of mass to length remains constant. This quantity is called[b] linear density[/b], and is denoted with a Greek lambda: [math]\lambda=\tfrac{M}{L}.[/math][br][br]In the event that a stick is made of some material that doesn't have constant density, we would define the linear density in a more general way based on infinitesimal masses and lengths. This, therefore, is the general definition of linear density:[br][center][math]\lambda = \tfrac{dM}{dL}[/math][/center]

EXAMPLE: Calculate the rotational inertia of a rod of length L about both its center and its end around an axis that is perpendicular to the rod's length. [br][br]SOLUTION: The limits of the integral depend on the location of the center of rotation. The idea is that r=0 at the axis of rotation. If we treat the rod as if it lies along the x-axis, then when we rotate the rod about its center, the integration limits are x=-L/2 to x=L/2. When we rotate the rod about its end, the limits are x=0 to L. We get: [br][br][center][math]I=\int_{-L/2}^{L/2}x^2 \;dm \\[br]I=\int_{-L/2}^{L/2}\lambda x^2 \;dx \\[br]I=\lambda\frac{x^3}{3}|_{-L/2}^{L/2} \\[br]\text{Using \lambda=m/L gives: } \\[br]I=\tfrac{1}{12}mL^2. \\[br]\text{About the end point we get: } \\[br]I=\int_{0}^{L}\lambda x^2 \;dx \\[br]I=\lambda\frac{x^3}{3}|_{0}^{L} \\[br]\text{Using \lambda=m/L gives: } \\[br]I=\tfrac{1}{3}mL^2[br].[/math][/center][br][br]This result tells us that it is four times more difficult to get a rod rotating if we wish to rotate it about its end point than about its center. Holding a meter stick in your hand and attempting to rotate it, this fact is readily apparent. [br][br]For objects of other common shapes - spheres, cylinders, rings - the integrals are all either 2D or 3D. We will not do such calculations. It is common to refer to the results of such calculations and to refer to tables of inertias as seen in the Wikipedia link below. Notice when you look at that link that the moments of inertia are always in the form of some fraction times mass times a linear dimension squared.