Dynamics: Work-Energy for Rigid Bodies

Analyzing the motion of complex mechanical systems using Newton's second law or momentum principles often requires solving for internal forces you don't care about. The work-energy principle for rigid bodies provides a powerful scalar alternative, letting you relate changes in a system's speed to the action of forces over a distance. By focusing on energy accounting, you can efficiently solve for velocities, angular velocities, and displacements in systems involving translation, rotation, and rolling, bypassing complex vector equations.

Work Done on a Rigid Body

For a particle, work is defined as the dot product of force and displacement. For a rigid body—an idealized object where the distance between any two points remains constant—work calculations must account for both the movement of the body's center of mass and its rotation.

The work done by a force acting on a rigid body as it moves from position 1 to position 2 is given by: $$U_{1\to 2}={\int }_1^2\vec{F}\cdot d{\vec{r}}_A$$ where $d{\vec{r}}_A$ is the differential displacement of the point of application of the force $\vec{F}$. Crucially, if the point of application does not move (e.g., at a frictionless pin support), that force does no work.

For a pure couple (a moment applied to the body), the work done is calculated from the angular displacement. The work of a constant couple moment $M$ is: $$U_{1\to 2}={\int }_{{\theta }_1}^{{\theta }_2}Md\theta =M({\theta }_2-{\theta }_1)$$ This holds true regardless of the rigid body's translation; only its rotation matters. Forces like the weight of a body perform work based on the vertical displacement of its center of gravity: U_{\text{weight}} = -mg (y_G_2 - y_G_1).

Kinetic Energy of a Rigid Body

The kinetic energy ($T$) of a rigid body is the sum of the kinetic energies of all its particles. This sum resolves into two distinct, separable parts: one due to translation and one due to rotation.

The general formula for the kinetic energy of a rigid body is: $$T=\frac{1}{2}m{v}_G^2+\frac{1}{2}{I}_G{\omega }^2$$ Here, $m$ is the total mass, $v_G$ is the speed of the center of mass $G$, $I_G$ is the mass moment of inertia about an axis through $G$, and $\omega$ is the angular speed of the body.

This reveals a major simplification:

• Translation: If the body is only translating ($\omega =0$), then $T=\frac{1}{2}m{v}_G^2$.
• Rotation about a Fixed Axis: If the body rotates about a fixed point $O$, you can also use $T=\frac{1}{2}{I}_O{\omega }^2$, where $I_O$ is the moment of inertia about $O$. This is derived from the general formula using the parallel-axis theorem.

The Work-Energy Theorem and Conservation of Energy

The work-energy theorem for rigid bodies formalizes the relationship between work and kinetic energy. It states that the total work done by all external forces and couples acting on a rigid body as it moves between two states is equal to the change in its kinetic energy. $$\sum U_{1\to 2}=T_2-T_1=\Delta T$$ This is a scalar equation. You must account for the work done by all external agents: applied forces, couples, friction, and weights. Internal forces between connected parts of a rigid body do no work.

A direct and powerful corollary is the principle of conservation of energy. If only conservative forces (like gravity or spring forces) do work on the body, then the total mechanical energy (kinetic + potential) is constant. $$T_1+V_1=T_2+V_2$$ Here, $V$ represents the potential energy (e.g., gravitational $V_g=mg{y}_G$, elastic $V_e=\frac{1}{2}k{x}^2$). Using conservation of energy is often the fastest solution path, as it requires analyzing only the initial and final states, not the path between them.

Applying the Method to Connected Systems and Rolling

The true utility of the work-energy approach shines in solving complex, connected systems. The procedure is methodical:

1. Define the System: Clearly identify the rigid body or assembly of connected bodies you are analyzing.
2. Calculate Initial and Final Kinetic Energy ($T_1$, $T_2$): For systems, $T$ is the sum of the kinetic energies of all moving parts. You must relate velocities using kinematics (e.g., constrained motion).
3. Calculate the Total Work ($\sum U_{1\to 2}$): Sum the work of all external forces/couples acting on your defined system. Forces at frictionless pins or non-moving points of application do zero work.
4. Apply the Theorem: Set $\sum U_{1\to 2}=T_2-T_1$ and solve for the unknown (usually a velocity or displacement).

Consider these classic applications:

• Rolling Without Slipping: For a wheel or disk rolling on a fixed surface, the friction force at the contact point does no work if there is no slip, because the point of application is instantaneously at rest. The kinetic energy is $T=\frac{1}{2}m{v}_G^2+\frac{1}{2}{I}_G{\omega }^2$, with the kinematic constraint $v_G=\omega r$.
• Pulley Systems: For a pulley with mass (a rigid body), its kinetic energy includes rotational energy: $T_{\text{pulley}}=\frac{1}{2}{I}_O{\omega }^2$. The angular velocity $\omega$ is kinematically linked to the linear velocity of the attached cable ($v=\omega R$). The work done by cable tensions on the pulley is zero if the cable doesn't slip, as the point of application moves perpendicular to the tension direction.
• Connected Rigid Bodies: For a system like a linkage, you must write the kinetic energy of each link separately, using kinematics to relate the motion of their centers of mass to a single variable (like an angular velocity).

Common Pitfalls

1. Incorrect Work Calculation for a Force: A common error is using the displacement of the body's center of mass to calculate the work of a force that is applied at a different point. Always use the displacement of the force's point of application. Conversely, for weight, you must use the vertical displacement of the center of mass.

1. Misapplying Conservation of Energy: Conservation ($T_1+V_1=T_2+V_2$) only holds if all forces doing work are conservative. If friction that does work (like sliding friction) or an applied external force/couple is present, you must use the full work-energy theorem ($\sum U=\Delta T$). Do not ignore the work done by non-conservative forces.

1. Kinematic Errors in System Kinetic Energy: The most frequent mistake in system problems is an incorrect kinematic relationship. For example, stating $v_B={\omega }_A{r}_A$ without verifying the cable doesn't slip or the gears are properly meshed. Always double-check your velocity relationships before plugging them into the energy equation.

1. Forgetting the Work of a Couple: When an external couple or moment acts on a body, it contributes work equal to $M\Delta \theta$. This is easy to overlook if you are only summing forces. Similarly, remember that a pure couple does work even if the center of mass does not translate.

Summary

• The work-energy theorem for a rigid body states that the total work of external forces and couples equals the change in its kinetic energy: $\sum U_{1\to 2}=\Delta T$.
• The kinetic energy of a rigid body is the sum of its translational and rotational components: $T=\frac{1}{2}m{v}_G^2+\frac{1}{2}{I}_G{\omega }^2$.
• Work calculations must use the displacement of a force's point of application, while a couple's work depends on angular displacement.
• Mechanical energy is conserved ($T+V=\text{constant}$) only when all work is done by conservative forces (e.g., gravity, ideal springs).
• This scalar method is exceptionally powerful for solving problems involving velocity and displacement in systems with connected bodies, rolling without slip, and pulleys with mass, as it often avoids solving for unknown internal or constraint forces.