23. Some moment of inertia problems.
LAST TIME

• Defined the moment of inertia tensor
• Expressed the kinetic energy of a rigid body in terms of this quantity
• Gave some examples of how it can be calculated.

TODAY
Solve a few sample problems involving moments of inertia.

EXAMPLE
Moment of inertia of a sphere about its center
By symmetry the three principal moments of inertia are equal

$$I_{xx}=I_{yy}=I_{zz}=I$$

while all the off-diagonal products of inertia are zero. We have with $a$ the radius of the sphere

$$I_{xx}+I_{yy}+I_{zz}=3I=2\int\rho d^3r(x^2+y^2+z^2)$$

$$I=\frac{2}{3}\int_0^adr\; 4\pi r^2\;r^2=\frac{8\pi\rho a^5}{15}$$

The total mass is

$$M=\frac{4\pi\rho a^3}{3}$$

Hence

$$I=\frac{2Ma^2}{5}$$

EXAMPLE
A massive cylinder and a cylindrical shell roll down an inclined plane. Which is fastest?
The moment of inertia about its axis of a cylindrical shell of radius $a$ and thickness $\delta«a$ is

$$\rho 2\pi a^3b\delta=Ma^2$$

where $b$ is the length of the cylinder, $\rho$ the mass density and $M$ the mass.

The moment of inertia of a solid cylinder is

$$I_{zz}=b\rho\int_0^adr2\pi r r^2=\frac{\pi ba^4}{2}=\frac{Ma^2}{2}$$

If a cylinder has rolled down a vertical height drop $h$ the kinetic energy will be

$$Mgh=\frac{MV^2}{2}+\frac{I_{zz}\omega^2}{2}$$

where $V$ is the speed of the center of mass and $\omega$ is the angular velocity. The rolling constraint implies that $V=a\omega$. For the solid cylinder we find

$$Mgh=\frac{Ma^2\omega_{solid}^2}{2}+\frac{Ma^2\omega_{solid}^2}{4}$$

or

$$\omega^2_{solid}=\frac{4gh}{3a^2}$$

while we find for the shell

$$Mgh=\frac{Ma^2\omega_{shell}^2}{2}+\frac{Ma^2\omega_{shell}^2}{2}$$

or

$$\omega^2_{shell}=\frac{gh}{a^2}$$

i.e. the solid cylinder rolls faster!

EXAMPLE
The physical pendulum A rigid body that swings under the influence of gravity about a fixed horizontal axis is called a physical pendulum. We label that axis the $x-$axis. The constraint that the axis is fixed means that

$$\vec{\omega}=(\omega_x,0,0)$$

The kinetic energy is then

$$T=\frac{I_{xx}\omega_{x}^2}{2}$$

The radius of gyration is defined as

$$k=\sqrt{\frac{I_{xx}}{M}}$$

where $M$ is the mass of the pendulum. We let $\l$ be the perpendicular distance between the center of mass and the axis of rotation, and $\theta$ the angle whose angular velocity is $\omega_x$ with $\theta=0$ corresponding to the case where the center of mass is directly below the axis of rotation.

The Lagrangian of the pendulum is then

$${\cal L}=\frac{Mk^2\dot{\theta}^2}{2}+Mgl\cos\theta$$

with equation of motion

$$\frac{d}{dt}(\frac{\partial \cal L}{\partial \dot{\theta}})-\... ...al \cal L}{\partial \theta} =M(k^2\ddot{\theta}+g\l\sin\theta)$$

We conclude that the behavior of the physical pendulum is the same as that of a mathematical pendulum ( pendulum with a point mass $M$)with effective length

$$l_{eff}=\frac{k^2}{\l }$$

FINDING THE PRINCIPAL AXES
In general the moment of inertia tensor is non-diagonal, but since it is a real symmetric matrix it can always be diagonalized with real and orthogonal eigenvalues. If $\hat{e}_i$ is an eigenvector of $I$ with eigenvalue $I_i$

$$I\hat{e}_i=I_i\hat{e}_i,\; i=1,2,3$$

we refer to $\hat{e}_i$ as a principal axis of $I$, and $I_i$ as a principal moment. The Cartesian coordinate system with axes $\hat{e}_i$ is the principal axes frame. The orthogonal rotation matrix $R$ which diagonalizes the matrix $I$

$$RIR^{-1}=\left(\begin{array}{ccc} I_1&0&0\\ 0&I_2&0\\ 0&0&I_3\\ \end{array}\right)$$

is

$$R=\left(\begin{array}{ccc} e_{1x}&e_{1y}&e_{1z}\\ e_{2x}&e_{2y}&e_{2z}\\ e_{3x}&e_{3y}&e_{3z}\\ \end{array}\right)$$

Some general results:

• If all three principal moments are equal (as e.g. for the moments of inertia about the center of a sphere) the moment of inertia matrix is proportional to the unit matrix, and any Cartesian frame is a principal axis frame.
• If two principal moments are equal but different from the third we talk about a symmetric top. If e.g. $I_1=I_2\neq I_3$ we can choose the axes perpendicular to $e_3$ arbitrarily.
• If all three principal moments are different we talk about an asymmetric top.
• The principal moments satisfies the triangle inequality that the sum of two principal moments is always larger than or or equal to the third. If $I_1+I_2=I_3$ the body is a plane sheet with normal along $\hat{e}_3$.

Some comments about how to solve this numerically is given in the attached Maple worksheet.

Model problem 3 of 1999 problem set 6 with solution, question 1 of2001 problem set 7 with solution, question 2 of2001 problem set 8 with solution

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