Planar Rigid Body Kinetics
Planar rigid body kinetics examines the motion of rigid bodies in a two-dimensional plane, explicitly accounting for the forces and moments that cause this motion. It combines the principles of kinematics with Newton’s laws of motion and the equations of equilibrium to analyze translational and rotational dynamics.
By addressing parameters such as net forces, torques, mass, and moments of inertia, this section provides a framework to solve problems involving linear and angular acceleration, energy methods, impulse-momentum relationships, and dynamic equilibrium in planar motion.
The starting point is Euler’s second Law (also called the moment equation) which was derived for a system of particles here.
For a continuous rigid body, the summation is replaced with an integral, so for any continuous body and any point \mathbf P:
\sum \mathbf{M}_P = \int \mathbf{r} \times \mathbf{F} \, \mathrm{d}m = \frac{\mathrm{d}}{\mathrm{d}t} \left( \int \mathbf{r} \times \mathbf{v} \, \mathrm{d}m \right)
where \mathbf{F} is the external force density, and \mathbf{v} is the velocity field of the body.
Translational motion
In pure translation, every point on the rigid body has the same acceleration \mathbf{a}. Therefore, the acceleration can be factored out of the integral:
\sum \mathbf{M}_P = \int \mathbf{r} \times m \mathbf{a} \, \mathrm{d}m = \left( \int \mathbf{r} \, \mathrm{d}m \right) \times \mathbf{a}
The integral \int \mathbf{r} \, \mathrm{d}m represents the moment of mass about point \mathbf P:
\int \mathbf{r} \, \mathrm{d}m = M \mathbf{r}_{PC}
where M is the total mass of the body and \mathbf{r}_{PC} is the distance between \mathbf P and the center of mass C.
Substituting:
\sum \mathbf{M}_P = M \mathbf{r}_{PC} \times \mathbf a = \mathbf{r}_{PC} \times (M\mathbf a)
If we select point \mathbf P as the center of mass, so \mathbf{r}_{PC} = 0. This simplifies the equation:
\sum \mathbf{M}_C = 0
Angular momentum
For a system of particles, each particle has a momentum m \mathbf{v}. Taking the cross product of the position vector \mathbf{r} with m \mathbf{v} gives the moment of momentum, or angular momentum, denoted as \mathbf L.
To extend this to a rigid body, consider it as an infinite collection of particles. For a rigid body, angular momentum about a point \mathbf P is defined as:
\mathbf L_P = \int (\mathbf{r} \times \mathbf{v}) \, \mathrm{d}m
where \mathbf{r} is the position vector from \mathbf P to the differential mass element \mathrm{d}m, and \mathbf{v} is its velocity.
Restricting the motion to 2D planar motion about the z-axis, consider a rigid body rotating in the plane. Any point \mathbf Q in the plane, with coordinates (x, y), has the same linear velocity as any other point on the line parallel to the z-axis at \mathbf Q. The angular momentum about \mathbf P becomes:
\mathbf L_P = \int (\mathbf{r} \times \mathbf{v}_Q) \, \mathrm{d}m
Using relative velocity kinematics, the velocity \mathbf{v}_Q is:
\mathbf{v}_Q = \mathbf{v}_P + \omega\,\mathbf k \times (x\, \mathbf i + y\, \mathbf j)
where \mathbf{v}_P is the velocity of point \mathbf P, \omega \, \mathbf k is the angular velocity about the z-axis, and x\, \mathbf i + y\, \mathbf y is the relative position vector from \mathbf P to \mathbf Q. Substituting this into the expression for \mathbf L_P, and rearranging terms using vector identities:
\mathbf L_P = \mathbf{r}_{PC} \times m \mathbf{v}_P + \omega \, \mathbf k \left( \int (x^2 + y^2) \, \mathrm{d}m \right) - x\, \mathbf i \left( \int xz \, \mathrm{d}m \right) - y\, \mathbf j \left( \int yz \, \mathrm{d}m \right)
The mass moment of inertia about the z-axis through \mathbf P is defined as:
I_{zz} = \int (x^2 + y^2) \, \mathrm{d}m
The products of inertia with respect to x and y about the z-axis through \mathbf P are defined as:
\begin{aligned} I_{xz} & = - \int xz \, \mathrm{d}m\\ I_{yz} & = - \int yz \, \mathrm{d}m \end{aligned}
Substituting these into the expression for \mathbf L_P, the angular momentum simplifies to:
\mathbf L_P = \mathbf{r}_{PC} \times m \mathbf{v}_P + \omega \, \mathbf i\, I_{xz} + \omega \, \mathbf j\, I_{yz} + \omega \, \mathbf k\, I_{zz}
If \mathbf P is the center of mass, \mathbf{r}_{PC} = 0 If p has no velocity (\mathbf{v}_p = 0), the term \mathbf{r}_p \times m \mathbf{v}_p vanishes.
The mass moment of inertia represents how the mass of a body is distributed relative to the axis of rotation. It measures how far the mass elements of the body are from the z-axis, integrated over the entire body. For a physical intuition, consider a body rotating about the z-axis. If the mass elements are moved further away from the axis (in the x- or y-directions), the mass moment of inertia increases.
The products of inertia are mathematical measures of a body’s lack of symmetry. They can be zero under certain symmetry conditions. For example, if there is symmetry with respect to the z-axis or the xy-plane, the products of inertia are zero. A cone, for instance, has its z-axis as an axis of symmetry, and the xy-plane as a plane of symmetry. For such a shape, the products of inertia are zero.
As a practical example, consider a wheel spinning in planar motion. The wheel is symmetric about both the z-axis and the xy-plane, resulting in zero products of inertia. However, a body does not always need visible symmetry to have zero products of inertia, and symmetry might not be obvious in all cases. It is also worth noting that the choice of coordinate axes influences the products of inertia. If coordinates are not aligned with the inherent symmetry of the body, the products of inertia may not be zero for those axes.
The moment of inertia and the product of inertia can be retrieved from table or computed mathematically using the formulas.
For example to compute the mass moment of inertia (I_{zz}) for a homogeneous solid cylinder of radius r, length L, and constant density \rho, about the z-axis, we proceed as follows. Using \mathrm{d}m = \rho \, \mathrm{d}V, we have:
I_{zz} = \int (x^2 + y^2) \, \mathrm{d}m = \rho \int_V \left( x^2 + y^2 \right) \, \mathrm{d}V
In cylindrical coordinates the integral becomes:
I_{zz} = \rho \int_0^L \int_0^{2\pi} \int_0^R r^3 \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}z = \rho \int_0^R r^3 \mathrm{d}r \int_0^{2\pi} \mathrm{d}\theta \int_0^L \mathrm{d}z
This integral can be split in three independent integrals:
\begin{aligned} & \int_0^R r^3 \, \mathrm{d}r = \left[ \frac{r^4}{4} \right]_0^R = \frac{R^4}{4}\\ & \int_0^{2\pi} \mathrm{d}\theta = 2\pi \\ &\int_0^L \mathrm{d}z = L \end{aligned}
Therefore:
I_{zz} = \rho \cdot L \cdot 2\pi \cdot \frac{R^4}{4}
The total mass m of the cylinder is:
m = \rho \cdot V = \rho \cdot (\pi R^2 L)
Solving for \rho:
\rho = \frac{m}{\pi R^2 L}
Substituting \rho into I_z gives:
I_{zz} = \frac{m}{\pi R^2 L} \cdot L \cdot 2\pi \cdot \frac{R^4}{4} = \frac{m \cdot R^2}{2}
Equations of motion
Le’s consider a rigid body undergoing planar motion in two dimensions. The angular momentum \mathbf{L} about a point \mathbf P (either the center of mass or a point with zero velocity) is given by:
\mathbf{L} = \omega \, \mathbf i\, I_{xz} + \omega \, \mathbf j\, I_{yz} + \omega \, \mathbf k\, I_{zz}
Euler’s second law states that the sum of the moments about the center of mass C is equal to the time derivative of the angular momentum about C:
\sum \mathbf{M}_C = \frac{\mathrm{d}\mathbf{L}_C}{\mathrm{d}t}
Substituting the expression for angular momentum:
\sum \mathbf{M}_C = \frac{\mathrm{d}}{\mathrm{d}t} \left(\omega \, \mathbf i\, I_{xz} + \omega \, \mathbf j\, I_{yz} + \omega \, \mathbf k\, I_{zz}\right)
Let’s choose a body-fixed reference frame 2 welded into the body with axes \mathbf{i}_2, \mathbf{j}_2, \mathbf{k}_2 such that the moments and products of inertia remain constant over time. In planar motion, the body rotates about the z-axis, and the \mathbf{i}_2 and \mathbf{j}_2 axes become time-dependent with respect to the inertial frame 1.
The transformation between the inertial frame 1 and the body-fixed frame 2 involves rotation by an angle \theta about the z-axis. The time derivatives of the unit vectors in the inertial frame are:
\begin{aligned} \frac{\mathrm{d}\mathbf{i}_2}{\mathrm{d}t} & = \omega \mathbf{j}_2 \\[5pt] \frac{\mathrm{d}\mathbf{j}_2}{\mathrm{d}t} & = -\omega \mathbf{i}_2 \end{aligned}
where \omega = \frac{\mathrm{d}\theta}{\mathrm{d}t} is the angular velocity.
Applying the product rule to differentiate the angular momentum expression:
\begin{aligned} \frac{\mathrm{d}\mathbf{L}_C}{\mathrm{d}t} & = \frac{\mathrm{d}}{\mathrm{d}t} \left(\omega \, \mathbf i_2\, I_{xz} + \omega \, \mathbf j_2\, I_{yz} + \omega \, \mathbf k_2\, I_{zz}\right)\\ & = \frac{\mathrm{d} \omega}{\mathrm{d}t} \mathbf{i}_2 I_{xz} + \omega \frac{\mathrm{d} \mathbf{i}_2}{\mathrm{d}t} I_{xz} + \frac{\mathrm{d} \omega}{\mathrm{d}t} \mathbf{j}_2 I_{yz} + \omega \frac{\mathrm{d} \mathbf{j}_2}{\mathrm{d}t} I_{yz} + \frac{\mathrm{d} \omega}{\mathrm{d}t} \mathbf{k}_2 I_{zz} + \omega \frac{\mathrm{d} \mathbf{k}_2}{\mathrm{d}t} I_{zz} \\ & = \frac{\mathrm{d}\omega}{\mathrm{d}t} \mathbf{i}_2 I_{xz} + \omega (\omega \mathbf{j}_2) I_{xz} + \frac{\mathrm{d}\omega}{\mathrm{d}t} \mathbf{j}_2 I_{yz} + \omega (-\omega \mathbf{i}_2) I_{yz} + \frac{\mathrm{d}\omega}{\mathrm{d}t} \mathbf{k}_2 I_{zz} \\ & = \left(I_{xz} \frac{\mathrm{d}\omega}{\mathrm{d}t} - I_{yz} \omega^2\right) \mathbf{i}_2 + \left(I_{yz} \frac{\mathrm{d}\omega}{\mathrm{d}t} + I_{xz} \omega^2\right) \mathbf{j}_2 + I_{zz} \frac{\mathrm{d}\omega}{\mathrm{d}t} \mathbf{k}_2 \\ & = \left(I_{xz} \alpha - I_{yz} \omega^2\right) \mathbf{i}_2 + \left(I_{yz} \alpha + I_{xz} \omega^2\right) \mathbf{j}_2 + I_{zz} \alpha \mathbf{k}_2 \end{aligned}
where \alpha = \frac{\mathrm{d}\omega}{\mathrm{d}t} = \frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} is the angular acceleration.
By equating the components in the \mathbf{i}_2, \mathbf{j}_2, and \mathbf{k}_2 directions, we obtain three scalar equations. About the x_2-axis:
\sum M_{x_2} = I_{xz} \alpha - I_{yz} \omega^2
About the y_2-axis:
\sum M_{y_2} = I_{yz} \alpha + I_{xz} \omega^2
About the z-axis:
\sum M_{z} = I_{zz} \alpha
For purely two-dimensional planar motion about the z-axis, the moments about the x_2 and y_2-axes must balance:
\begin{aligned} \sum M_{x_2} & = 0 \\ \sum M_{y_2} & = 0 \end{aligned}
These conditions imply that the products of inertia I_{x_2 y_2} are zero, which is achieved when the body has symmetry about the x_2 z-plane and y_2 z-plane. This symmetry ensures that there are no gyroscopic moments that could cause the body to deviate out of the plane.
Additionally, the sum of forces in the z-direction must be zero to prevent motion out of the plane:
\sum F_z = 0
With the above conditions satisfied, the only non-trivial equation governing the rotational motion about the z-axis is:
\sum M_z = I_{zz} \alpha
where \alpha = \frac{\mathrm{d}\omega}{\mathrm{d}t} is the angular acceleration.
This equation is analogous to Newton’s second law F = ma for linear motion, where:
- moments (M_z) correspond to forces (F),
- mass moment of inertia (I_{zz}) corresponds to mass (m),
- angular acceleration (\alpha) corresponds to linear acceleration (a).
Consider an ice skaters who spin, they often start with their arms and legs extended, which increases their mass moment of inertia I_{zz}. According to the rotational equation of motion:
\sum M_z = I_{zz} \alpha
For the same torque \sum M_z applied, a larger I_{zz} results in a smaller angular acceleration \alpha:
\alpha = \frac{\sum M_z}{I_{zz}}
When the skater pulls their arms and legs closer to their body, the distribution of their mass moves closer to the axis of rotation (the z-axis). This decreases I_{zz}, allowing for a larger angular acceleration \alpha with the same applied torque. This increase in \alpha makes the skater spin faster, demonstrating the conservation of angular momentum:
L = I_{zz} \omega = \text{constant}
Since L is conserved, when I_{zz} decreases (by pulling arms in), \omega (angular velocity) must increase, leading to a faster spin.