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Rotating rigid bodies#
Rigid bodies and the moment of inertia#
A rigid body is a body that has some finite extended shape—so not a point particle, but a shape that does not deform under reasonable forces. When we want to describe the motion of such bodies, particularly rotations, their shape becomes important. As an example, consider a uniform solid cylinder with height \(h\), radius \(R\), and mass \(M\). We can set it to rotate with angular velocity \(\omega\) around its symmetry axis, namely the axis through the centre perpendicular to the circular cross section. This requires a torque, and the rotating cylinder has some angular momentum \(L\) relative to the origin at the centre of mass that lies on the rotation axis. The direction of \(L\) is along the rotation axis.
Next, consider a hollow cylinder with otherwise identical specs: height \(h\), radius \(R\), and mass \(M\). Now all the mass is concentrated in the outside layer. For simplicity we assume that the end caps are also hollow, so the object is a tube or pipe. When we set this object spinning around the same axis with the same angular velocity \(\omega\), does this object gain the same angular momentum as the solid cylinder? You may already intuit that the answer has to be `no’. The solid cylinder has some of its mass closer to the rotation axis, requiring a smaller \(\r\) in \(\L=\r\times\p\), and a correspondingly lower tangential velocity \(\v\). We need to find out how we can calculate this.
We can think of a rigid body as a collection of small masses put together. Each of these masses \(m_j\) has its own angular momentum
For simplicity, assume that \( \vec{\omega}\) is pointing in the \(z\)-direction. Then \(L_j = m_j r_j^2 \omega\). Note that \(\omega\) does not depend on \(j\); all elements move at the same angular velocity. This is what makes it a rigid body. When we put all the small masses together, the total angular momentum is the sum of all the individual angular momenta:
Here we defined a new quantity, the moment of inertia \(I\) that captures the effect of the shape of the rotating object on the angular momentum.
Example 1: The figure skater
You have likely seen the effect of a spinning ice skater who pulls in her arms to spin faster. We can easily explain this effect using the changing moment of inertia. In its most basic form, the angular momentum of the skater is conserved, since there are no torques acting on her. When she has her arms spread wide, the angular momentum is
(75)#\[\begin{align} L = I_{\rm wide} \omega_{\rm wide}\, , \end{align}\]and when she has her arms tucked in close, her angular momentum is
(76)#\[\begin{align} L = I_{\rm close} \omega_{\rm close}\, . \end{align}\]Conservation of angular momentum then determines that
(77)#\[\begin{align} I_{\rm wide} \omega_{\rm wide} = I_{\rm close} \omega_{\rm close} \qquad\text{or}\qquad \omega_{\rm close} = \frac{I_{\rm wide}}{I_{\rm close}} \omega_{\rm wide}\, . \end{align}\]When \(I_{\rm wide} > I_{\rm close}\), the angular velocity \( \omega_{\rm close}\) increases.
Before we calculate \(I\) for a variety of shapes, consider the parallels between \(\p = m\v\) and \(\L = I \vec{\omega}\). Both \(\p\) and \(\L\) measure the amount of motion of a body, linear and rotational, respectively. The linear and angular velocities \(\v\) and \(\vec{\omega}\) also clearly play analogous roles. This suggests an interpretation of \(I\) by analogy. Recall that \(m\) is the inertial mass, i.e., the resistance of a body to change its (linear) motion. Similarly, \(I\) is the resistance of a rigid body to a change in rotational motion. Hence the name `moment of inertia’.
Calculating the moment of inertia#
For a collection of particles with mass \(m_j\) that all rotate around the same axis with angular velocity \(\omega\), the moment of inertia is immediately calculated by summing the masses times the distance-squared to the rotation axis:
Things become a bit more involved when we have a solid body like a disk, cylinder or sphere. Let’s assume that the density of the object at a position \(\r\) is given by \(\rho(\r)\). Then a `small mass’ can be put in infinitesimal form as
where \(\d V\) is an infinitesimal volume around the position \(\r\). Summation now becomes integration (the summing of tiny bits!), and we can write for the moment of inertia:
Note that we take \(r\) as the distance to the rotation axis, not the actual distance from \(\d m\) to the origin (so cylindrical coordinates would be convenient here), since \(\L = \r\times\p\) means that we need to take only the part of \(\r\) that is perpendicular to \(\L\). And given that \(\L\) lies along the rotation axis, only the component perpendicular to the rotation axis contributes to \(I\).
Example 2: The moment of inertia for a ring
Consider a thin ring of radius \(a\) and uniform mass distribution, rotating around its symmetry axis. Since the mass is concentrated in a one-dimensional circle, we can write \(\rho = M/2\pi a\). The sum over all masses then becomes the integral around the ring over an angle between 0 and \(2\pi\), where all radii are equal to \(a\):
(81)#\[\begin{align} I_{\rm ring} = \int_0^{2\pi} \rho a^2\, a\, \d \theta = \int_0^{2\pi} \frac{M a^2}{2\pi}\, \d \theta = Ma^2\, . \end{align}\]Note that the integral around the ring is equivalent to summing over infinitesimal lengths \(a\d\theta\). Also, this is the same moment of inertia as for a point mass \(M\) a distance \(a\) from the rotation axis. Therefore, how the mass is distributed around the circle does not seem to matter. Can you prove this mathematically with a general mass distribution \(\rho(\theta)\)?
Example 3: The moment of inertia for a disk
We can use the moment of inertia of a ring \(I_{\rm ring} = Ma^2\) to calculate the moment of inertia of a disk rotating around its symmetry axis. Remember that calculating the moment of inertia is like summing lots of small parts. We can think of the disk as a collection of thin rings that we need to sum. Each ring has an area \(2\pi r \d r\), and the mass in this ring is the area times the mass density per area, \(\sigma = M/\pi a^2\). Summing these rings together then amounts to the integral
(82)#\[\begin{align} I_{\rm disk} = \int_0^a r^2 \d m(r) = \int_0^a r^2 \frac{M}{\pi a^2} 2\pi r\, \d r = \frac{2M}{a^2} \int_0^a r^3\, \d r = \frac12 Ma^2\, . \end{align}\]From this you see that the moment of inertia of a ring is larger than the moment of inertia of a disk, if both have the same mass.
Example 4: Circling bodies
Consider two masses, a distance \(d\) apart, rotating around their centre of mass in circular orbits. This could for example be a planet of mass \(m\) orbiting a star of mass \(M\), or a binary star system. We make no assumptions about which mass is bigger. The distances of \(m\) and \(M\) to the centre of mass are:
(83)#\[\begin{align} r_m = \frac{Md}{m+M} \qquad\text{and}\qquad r_M = \frac{md}{m+M} \, . \end{align}\]We consider only the magnitudes of \(\r_m\) and \(\r_M\) here. The moment of inertia for this system is
(84)#\[\begin{align} I = m r_m^2 + M r_M^2 = \frac{m M^2 d^2}{(m+M)^2} + \frac{M m^2 d^2}{(m+M)^2} = \frac{mM d^2}{m+M} \equiv \mu d^2\, , \end{align}\]where we again encountered the reduced mass \(\mu\) for two masses. It is like a single mass \(\mu\) rotating around a circle with radius \(d\).