Planar Rigid Body Work Energy Principle
It is possible extend the work-energy principle already discussed for particles here to rigid bodies in planar (2D) motion.
Bar Rotating About a Fixed Axis
Kinetic energy for rigid body
For a small mass element \mathrm dm, its kinetic energy is:
T = \frac{1}{2} \mathrm dm \, v^2
Where v is the velocity of the differential mass element. To calculate the total kinetic energy of the rigid body, we integrate over all \mathrm dm in the body:
T = \frac{1}{2} \int (\mathbf{v} \cdot \mathbf{v}) \, \mathrm dm
The velocity of any point \mathbf P in the body can be expressed using the relative velocity equation as the sum of the velocity of the center of mass (\mathbf{v}_C) and the rotational velocity component due to angular velocity (\mathbf{\omega}) about the center of mass.
If the body is in planar motion, it is not necessary to compute the distance of the point \mathbf P, but is sufficient to compute the distance with the companion point \mathbf Q, which refers to a point on the rigid body that lies in the plane of rotation and has the same instantaneous velocity as any other point going from that point in the z direction.
\mathbf{v}_P = \mathbf{v}_C + \mathbf{\omega} \times \mathbf{r}_{QC} = \mathbf v_C + \omega (-y \mathbf{i} + x \mathbf{j})
The kinetic energy involves the dot product of \mathbf{v}_P:
\begin{aligned} \mathbf{v}_P \cdot \mathbf{v}_P & = (\mathbf{v}_C + \omega (-y \mathbf{i} + x \mathbf{j})) \cdot (\mathbf{v}_C + \omega (-y \mathbf{i} + x \mathbf{j})) \\ & = \mathbf{v}_C \cdot \mathbf{v}_C + \omega^2 (x^2 + y^2) + 2 \omega \mathbf{v}_C \cdot (-y\mathbf{i} + x\mathbf{j}) \end{aligned}
The kinetic energy of the rigid body becomes:
\begin{aligned} T &= \frac{1}{2} \int \big[ \mathbf{v}_C \cdot \mathbf{v}_C + \omega^2 (x^2 + y^2) + 2 \omega \mathbf{v}_C \cdot (-y\mathbf{i} + x\mathbf{j}) \big] \mathrm dm \end{aligned}
Breaking this into terms:
\begin{aligned} T & = \frac{1}{2} \mathbf{v}_C \cdot \mathbf{v}_C \int \mathrm dm + \frac{1}{2} \omega^2 \int (x^2 + y^2) \mathrm dm + \omega \mathbf{v}_C \cdot \int (-y\mathbf{i} + x\mathbf{j}) \mathrm dm \\ & = \frac{1}{2} m v_C^2 + \frac{1}{2} I_C \omega^2 \end{aligned}
where:
- m = \int \mathrm dm is the total mass of the body,
- I_{zz}^C = \int (x^2 + y^2) \mathrm dm is the moment of inertia about the center of mass and for brevity can be written as I_C.
The last term involves \int (-y\mathbf{i} + x\mathbf{j}) \mathrm dm and by definition of the center of mass:
\int x \, \mathrm dm = \int y \, \mathrm dm = 0 and therefore this term vanishes:
\omega \mathbf{v}_C \cdot \int (-y\mathbf{i} + x\mathbf{j}) \mathrm dm = 0 The kinetic energy:
T = \frac{1}{2} m v_C^2 + \frac{1}{2} I_C \omega^2
is the sum of two terms: the translational kinetic energy \frac{1}{2} m v_C^2 which is the same as in the case of a system of particles and a rotational kinetic energy \frac{1}{2} I_C \omega^2.
No-slip wheel
When analyzing the motion of a wheel rolling without slipping on a flat surface, it is essential to consider both its translational and rotational dynamics. The wheel’s center of mass undergoes translational motion, while simultaneously, the wheel itself rotates about its axis. Consequently, the total kinetic energy of the system comprises both translational kinetic energy and rotational kinetic energy.
The translational kinetic energy is given by:
For a wheel rolling without slipping, the no-slip condition relates the translational velocity v_C to the angular velocity \omega through the radius R of the wheel:
v_C = \omega R
Substituting v_C = \omega R into the expression for translational kinetic energy, we obtain:
\frac{1}{2} m (\omega R)^2 = \frac{1}{2} m R^2 \omega^2
To determine the moment of inertia I for a solid cylinder (a common model for a uniform wheel), we perform an integration over its mass distribution. The moment of inertia about the central axis (the z-axis) for a solid cylinder is derived as follows:
Consider a solid cylinder of mass M, radius R, and height h. The mass element \mathrm dm at a distance r from the central axis contributes to the moment of inertia as \mathrm dI = r^2 \mathrm dm. In cylindrical coordinates, the mass element can be expressed as \mathrm dm = \rho \, r \, \mathrm dr \, \mathrm d\theta \, \mathrm dz, where \rho is the mass density.
Integrating over the entire volume of the cylinder:
I = \int dI = \int_0^h \int_0^{2\pi} \int_0^R r^2 \rho \, r \,\mathrm dr \, \mathrm d\theta \, \mathrm dz
Simplifying the integrals:
I = \rho \int_0^h \mathrm dz \int_0^{2\pi} \mathrm d\theta \int_0^R r^3 \mathrm dr = \rho h (2\pi) \left[ \frac{R^4}{4} \right] = \frac{1}{2} m R^2
Here, m = \rho \pi R^2 h, and after performing the integration, we find that the moment of inertia of a solid cylinder about its central axis is:
I = \frac{1}{2} m R^2
Substituting this expression for I back into the rotational kinetic energy:
\frac{1}{2} \left( \frac{1}{2} m R^2 \right) \omega^2 = \frac{1}{4} m R^2 \omega^2
Combining both the translational and rotational kinetic energies:
T = \frac{1}{2} m R^2 \omega^2 + \frac{1}{4} m R^2 \omega^2 = \frac{3}{4} m R^2 \omega^2
Unbalanced wheel
When dealing with an unbalanced rolling wheel or cylinder, the analysis requires considering both translational and rotational motion while accounting for the fact that the mass center is offset from the geometric center. Unlike a uniform cylinder with its mass distributed symmetrically about its geometric center, the unbalanced wheel has a shift in the mass center.
For a rolling wheel, the velocity of the center of mass \mathbf{v}_C can be expressed in terms of the velocity \mathbf{v}_Q of the geometric center and the rotational effects due to the angular velocity \mathbf{\omega}. Using the relative velocity equation:
\mathbf{v}_C = \mathbf{v}_Q + \mathbf{\omega} \times \mathbf{r}
Here, \mathbf{r} represents the vector from the geometric center to the mass center. Substituting \mathbf{v}_Q = R\omega \mathbf{i} and expanding the cross product \mathbf{\omega} \times \mathbf{r}, we find:
\mathbf{v}_C = R\omega \mathbf{i} + \left[ (\omega \mathbf{k}) \times (r \sin\theta \mathbf{i} + r \cos\theta \mathbf{j}) \right]
Simplifying further, the velocity of the mass center becomes:
\mathbf{v}_C = (R\omega - r\omega \cos\theta) \mathbf{i} + (r\omega \sin\theta) \mathbf{j}
The magnitude of \mathbf{v}_C, representing the velocity of the mass center, is calculated as:
\mathbf{v}_C \cdot \mathbf{v}_C = |\mathbf{v}_C|^2 = v_C^2
Expanding and simplifying:
v_C^2 = R^2\omega^2 + r^2\omega^2 - 2Rr\omega^2\cos\theta
For the kinetic energy of the unbalanced rolling wheel, we include both translational and rotational components. The rotational kinetic energy is based on the mass moment of inertia I_C, which is now expressed in terms of the radius of gyration k_C. The mass moment of inertia becomes:
I_C = m k_C^2
where k_C is defined as the distance from the mass center C at which a point mass would have the same moment of inertia. For a uniform cylinder, k_C = R / \sqrt{2}, resulting in I_C = \frac{1}{2} m R^2.
In the case of the unbalanced wheel, the calculation of k_C and the adjustment to the moment of inertia reflect the offset mass center. Using the relative velocity equation, v_C is determined, allowing substitution into the expressions for kinetic energy. The total kinetic energy is the sum of translational and rotational components:
T = \frac{1}{2} m v_C^2 + \frac{1}{2} I_C \omega^2
Substituting v_C^2 and I_C = m k_C^2:
\begin{aligned} T & = \frac{1}{2} m \left( R^2\omega^2 + r^2\omega^2 - 2Rr\omega^2 \cos\theta \right) + \frac{1}{2} m k_C^2 \omega^2 \\ & = \frac{1}{2} m R^2 \omega^2 \left[ 1 - \frac{2r}{R}\cos\theta \left(\frac{r}{R} \right)^2 + \left( \frac{k_C}{R} \right)^2 \right] \end{aligned}
Bar rotating about a fixed axis
When analyzing the motion of a slender bar undergoing rotation about a fixed axis, we consider the bar’s kinetic energy, which is influenced by its translational motion and rotational motion.
For a uniform bar, the mass center is located at its midpoint. If we analyze the bar’s motion from the perspective of its mass center, it will exhibit both translational kinetic energy, due to the velocity of the mass center, and rotational kinetic energy, due to the rotation about this mass center.
Using kinematics, the velocity of the mass center \mathbf{v}_C is given as \mathbf{v}_C = \frac{L}{2} \omega, where L is the length of the bar, and \omega is its angular velocity. The mass moment of inertia of the slender bar about its mass center C is obtained by integration of:
I_C = \int r^2 \, \mathrm{d}m
where r is the perpendicular distance from the axis of rotation to the mass element \mathrm{d}m. For a slender bar of length L, uniform mass m, and mass per unit length \lambda = \frac{m}{L}, the mass element is given as \mathrm{d}m = \lambda \, \mathrm{d}x.
We position the coordinate system such that the origin is at the mass center of the bar, with the bar extending from -\frac{L}{2} to \frac{L}{2}. The distance r from the axis is simply x, so:
\begin{aligned} I_C & = \int_{-\frac{L}{2}}^{\frac{L}{2}} x^2 \, \mathrm{d}m = \int_{-\frac{L}{2}}^{\frac{L}{2}} x^2 \lambda \, \mathrm{d}x = \frac{m}{L} \int_{-\frac{L}{2}}^{\frac{L}{2}} x^2 \, \mathrm{d}x \\ & = \frac{m}{L} \left[ \frac{x^3}{3} \right]_{-\frac{L}{2}}^{\frac{L}{2}} = \frac{m}{L}\frac{1}{3}\left[ \left(\frac{L}{2}\right)^3 - \left(-\frac{L}{2}\right)^3 \right] \\ & = \frac{m}{L}\frac{1}{3}\left( \frac{L^3}{8} + \frac{L^3}{8} \right) = \frac{1}{3}\frac{m}{L}\frac{L^2}{4} = \frac{1}{12}mL^2 \end{aligned}
where m is the mass of the bar. The total kinetic energy T of the bar, substituting v_C^2 = \left(\frac{L}{2} \omega\right)^2 and I_C = \frac{1}{12} m L^2, the expression for the kinetic energy becomes:
\begin{aligned} T & = \frac{1}{2} m \left(\frac{L}{2} \omega\right)^2 + \frac{1}{2} \left(\frac{1}{12} m L^2\right) \omega^2 \\ & = \frac{1}{8} m L^2 \omega^2 + \frac{1}{24} m L^2 \omega^2 \\ & = \frac{1}{6} m L^2 \omega^2 \end{aligned}
Now, if we analyze the same problem from the perspective of point A, located at one end of the bar, we recognize that this is a case of pure rotation about point A. In this scenario, the velocity of point A is zero (\mathbf{v}_A = 0).
The mass moment of inertia about point A, denoted as I_A, can be obtained from references as:
To compute the moment of inertia about the endpoint A, we adjust the limits of integration. The bar now extends from 0 to L, and the distance r remains x. The moment of inertia is:
I_A = \int_0^L x^2 \, \mathrm{d}m
Substituting \mathrm{d}m = \lambda \, \mathrm{d}x and \lambda = \frac{m}{L}:
\begin{aligned} I_A & = \frac{m}{L} \int_0^L x^2 \, \mathrm{d}x = \frac{m}{L} \left[ \frac{x^3}{3} \right]_0^L = \frac{m}{L}\frac{L^3}{3} \\ & = \frac{1}{3} m L^2 \end{aligned}
Using the rotational kinetic energy formula for pure rotation:
T = \frac{1}{2} I_A \omega^2
Substituting I_A = \frac{1}{3} m L^2, we find:
T = \frac{1}{2} \left(\frac{1}{3} m L^2\right) \omega^2 = \frac{1}{6} m L^2 \omega^2
This result matches the kinetic energy derived from the perspective of the mass center.
The equivalence of these results confirms that the choice of reference point, whether the mass center or an endpoint, does not affect the total kinetic energy.
This approach provides flexibility in solving problems involving rigid bodies with rotational motion, allowing the use of the reference frame most convenient for the problem being analyzed.
Work energy principle
The work energy principle was derived already for particles here.
The results is that the difference of the work of the external forces on the body is equal to a change in translational kinetic energy:
W = \Delta T = \left( \frac{1}{2}m v_C^2\right) \bigg|_{t_1}^{t_2}
When dealing with rigid bodies, the internal forces between particles cancel each other out. This means you it is not needed to consider the work done by these internal forces and can focus solely on external forces.
However, for rigid bodies, work involves more than just translation; rotation also plays a role. Therefore, work is performed not only by external forces but also by moments from couples or other forces acting on the body. Consequently, the kinetic energy of a rigid body includes both translational and rotational components, unlike the purely translational kinetic energy considered for individual particles.
The work-energy principle for two-dimensional planar rigid body motion states that the work done by external forces and moments from couples equals the change in the body’s kinetic energy, which now includes both translational and rotational energy, over a given time period:
W = \Delta T = \left( \frac{1}{2}m v_C^2 + \frac{1}{2}I_C \omega^2 \right) \bigg|_{t_1}^{t_2}
This principle accounts for the combined effects of translation and rotation in the motion of rigid bodies.
Work of a constant force
Consider a constant force \mathbf{F}_1. Since the force is constant, it can be factored out of the integral when calculating work. The work done is then the dot product of \mathbf{F}_1 with the integral of velocity over the time interval from t_1 to t_2:
W = \mathbf{F}_1 \cdot \int_{t_1}^{t_2} \mathbf{v} \, dt
Work of a variable force
For a variable force acting on a single point throughout its motion, the work done is given by:
W = \int \mathbf{F}(s) \cdot \mathrm d\mathbf{s}
Suppose the force varies with position as \mathbf{F}(s) = 3s^2. If the object moves from position s = 0 to s = d, the work done is:
W = \int_{0}^{d} s^2 \, \mathrm ds = \frac{1}{3}d^3
This example illustrates how to compute work for a position-dependent force.
Work done by gravity
The work done by gravity is a conservative force. The work done by gravity can be treated similarly to a constant force.
Consider a rigid body with a weight mg moving vertically through a distance \Delta h. The work done by gravity is:
W = mg \Delta h
Here, \Delta h represents the change in the height of the mass center.
For the sign convention, the work is positive when the force aids the motion (e.g., gravity acting downward while the object moves downward) and it is negative when the force opposes the motion (e.g., gravity acting downward while the object moves upward).
Work of a spring
It was previously derived for particles here:
W = -\frac{k}{2} \left[ \delta(t_2)^2 - \delta(t_1)^2\right]
The work is negative because the force of the spring hinders the motion.
Work of a moment
The work done by a moment due to a couple is analogous to the work done by a force, but it involves rotational motion instead of translational motion.
For translational motion, the work done by a force \mathbf{F}_1 is given as:
W = \int_{t_1}^{t_2} \mathbf{F}_1 \cdot \mathbf{v}_1 \, \mathrm{d}t
For planar rotational motion, the work done by a couple \mathbf{C} is expressed as:
W = \int_{t_1}^{t_2} \mathbf{C} \cdot \boldsymbol{\omega} \, \mathrm{d}t
In planar motion, where the rotation is about the \mathbf{k}-axis, the angular velocity \boldsymbol{\omega} becomes:
\boldsymbol{\omega} = \frac{\mathrm{d}\theta}{\mathrm{d}t} \mathbf{k}
Similarly, the couple \mathbf{C} acts about the \mathbf{k}-axis. Substituting these into the work expression gives:
W = \int_{t_1}^{t_2} C \mathbf{k} \cdot \frac{\mathrm{d}\theta}{\mathrm{d}t} \mathbf{k} \, \mathrm{d}t
Since \mathbf{k} \cdot \mathbf{k} = 1, this simplifies to:
W = \int_{t_1}^{t_2} C \frac{\mathrm{d}\theta}{\mathrm{d}t} \, \mathrm{d}t
The time differentials \mathrm{d}t cancel out, leading to the integral with respect to angular displacement \theta:
W = \int_{\theta_1}^{\theta_2} C \, \mathrm{d}\theta
The sign convention is as follows: positive work occurs when the couple C and the angular displacement \theta are in the same direction, aiding the motion. Negative work occurs when C and \theta are in opposite directions, hindering the motion.