10-rigid_Bodies

In Newton’s second law of motion, we saw that the change in velocity depends on a force acting on the object and the object’s mass:

\[\ddot{p} = \frac{1}{m}f \qquad \qquad [10.1] \]

For rotation we have a very similar law. The change in angular velocity depends on two things: we have torque \(\tau\) rather than force, and the moment of inertia \(I\) rather than mass:

\[\ddot{\theta} = I^{-1}\tau \qquad \qquad [10.2] \]

Let’s look at these two in more depth.

Torque

Torque (also sometimes called “moments”) can be thought of as a twisting force.The equation that links force and torque is

\[\tau = p_{f}\times f \qquad \qquad [10.3] \]

where \(f\) is the force being applied, and \(p_{f}\) is the point at which the force is being
applied, relative to the origin of the object.

Every force that applies to an object will generate a corresponding torque.Whenever we apply a force to a rigid body, we need to use it in the way we have so far: to perform a linear acceleration.We will additionally need to use it to generate a torque.If you look at equation 10.3, you may notice that any force applied so that \(f\) and \(p_{f}\) are in the same direction will have zero torque Geometrically this is equivalent to saying that if the extended line of the force passes through the center of mass, then no torque is generated.In three dimensions it is important to notice that a torque needs to have an axis.We give torques in a scaled axis representation:

\[\tau = a \widehat{d} \qquad \qquad [10.4] \]

where \(a\) is the magnitude of the torque and \(d\) is a unit-length vector in the axis around which the torque applies.We always consider that torques act clockwise when looking in the direction of their axis.The torque is the vector product of the force

The moment of inertia

The moment of inertia of an object is ameasure of how difficult it is to change that object’s rotation speed. The moment of inertia depends on the mass of the object and the distance of that mass from the axis of rotation. Imagine the stick being made up of lots of particles; twirling the stick in the manner of a majorette involves accelerating particles that lie a long way from the axis of rotation. In comparison to twirling the stick lengthwise, the particles of the stick are a long way from the axis. The inertia will therefore be greater, and the stick will be more difficult to rotate. We can calculate the moment of inertia about an axis in terms of a set of particles in the object:

\[I_{a} = \sum\limits_{i=1}^{n} m_{i}d^2_{p_{i} \rightarrow a} \]

where n is the number of particles, \(d_{p_{i}}\rightarrow a\) is the distance of particle \(i\) from the axis of rotation \(a\), and \(I_{a}\) is the moment of inertia about that axis.

Clearly we can’t use a single value for the moment of inertia as we did for mass. It depends completely on the axis we choose. About any particular axis, we have only onemoment of inertia, but there are any number of axes we could choose. Fortunately the physics of rigid bodies means we don’t need to have an unlimited number of different values either.We can compactly represent all the different values in a matrix called the “inertia tensor.”

The inertia tensor in three dimensions is a \(3 × 3\) matrix that is characteristic of a rigid body (in other words, we keep an inertia tensor for each body, just as each body has its own mass). Along the leading diagonals the tensor has the moment of inertia about each of its axes—X, Y, and Z:

\[\left[ \begin{array}{ccc} I_{x} & & \\ & I_{y} & \\ & & I_{z}\end{array} \right] \]

where \(I_{x}\) is the moment of inertia of the object about its X axis through its center of mass; similarly for \(I_{y}\) and \(I_{z}\) . The remaining entries don’t hold moments of inertia. They are called “products of inertia” and are defined in this way:

\[I_{ab} = \sum\limits_{i = 1}^{n}m_{i}a_{pi}b_{pi} \]

It is difficult to visualize what the product of inertia means either in geometrical or mathematical terms. It represents the tendency of an object to rotate in a direction different from the direction in which the torque is being applied.

where \(a_{pi}\) is the distance of particle i from the center of mass of the object, in the direction of a. We use this to calculate \(I_{xy}\), \(I_{xz}\), and \(I_{yz}\) . In the case of \(I_{xy}\), we get

\[I_{xy} = \sum\limits_{i = 1}^{n} m_{i}x_{pi}y_{pi} \]

where \(x_{pi}\) is the distance of the particle from the center of mass in the X axis direction; similarly for \(y_{pi}\) in the Y axis direction. Using the scalar products of vectors we get

\[I_{xy} = \sum\limits_{i =1}^{n}m_{i}(p_{i} * x)(p_{i} * y) \]

We place the products of inertia into our inertia tensor to give the final structure:

\[\left[ \begin{array}{ccc} I_{x} & -I_{xy} & -I_{xz} \\ -I_{xy} & I_{y} & -I_{yz} \\ -I_{xz} & -I_{yz} & I_{z}\end{array} \right] \qquad \qquad [10.2] \]

The mathematician Euler gave the rotational version of Newton’s second law of motion in terms of this structures

\[\tau = I \ddot{\theta } \qquad \qquad [10.3] \]

Note that because of the presence of the products of inertia, the direction of the torque vector \(\tau\) is not necessarily the same as the angular acceleration vector \(\theta\) . If the products of inertia are all zero

\[I = \left[ \begin{array}{ccc} I_{x} & 0 & 0\\ 0 & I_{y} & 0\\ 0 & 0& I_{z}\end{array} \right] \]

and the torque vector is in one of the principal axis directions—X, Y, or Z—then the acceleration will be in the direction of the torque.

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