Engineering Mechanics · Dynamics

Planar Kinetics of a Rigid Body: Work and Energy

Rigid-body work–energy relates motion at two configurations without resolving acceleration throughout the path. It is a scalar kinetics method: forces and couples matter through their work, while rigid-body kinematics supplies relations such as rolling without slip.

Learning objectives

Classify translation, fixed-axis rotation, and general plane motion before writing kinetic energy.
Evaluate work at the actual point of force application and work of a couple through angular displacement.
Select work–energy or conservation of energy from the force and displacement information.
Apply rolling and mechanism constraints while summing the energy of every body.
Audit units, work signs, contact admissibility, and practical operating limits.

18.1Kinetic Energy

A rigid body's kinetic energy is the sum of the translational energy of its mass center and rotational energy about its mass center.

Motion classification controls the formula

For constant mass and a rigid planar body, \(T=\tfrac12mv_G^2+\tfrac12I_G\omega^2\). Pure translation has \(\omega=0\). For rotation about a fixed point \(O\), the parallel-axis theorem combines the terms as \(T=\tfrac12I_O\omega^2\). General plane motion requires both mass-center and spin terms.

Rigid-body kinetic energy

\[T=\tfrac12mv_G^2+\tfrac12I_G\omega^2,\qquad T_{\text{fixed }O}=\tfrac12I_O\omega^2\]

Physical meaning

The first term measures motion of the body as a whole; the second measures motion relative to its mass center. Energy is scalar and nonnegative, so velocity direction is not preserved and must come from kinematics or a stated motion sense.

Fig. 18-1An arbitrary particle i at position r from reference point P, with body angular velocity ω. (Original redrawn schematic.)

For connected bodies, sum one correctly classified kinetic-energy expression per body and use cord, gear, or rolling constraints to express all speeds with a common independent variable. An instantaneous center can give \(T=\tfrac12I_{IC}\omega^2\), but only for the instantaneous velocity state.

18.2The Work of a Force

Follow the point of application

The differential work of a force is \(dU=\mathbf F\cdot d\mathbf r_A\), where \(A\) is the point where that force acts. Using the mass-center displacement for an off-center cable or actuator is generally wrong.

Work contributions

\[U_F=\int_1^2\mathbf F\cdot d\mathbf r_A,\quad U_W=-mg(y_{G2}-y_{G1}),\quad U_s=-\tfrac12k(s_2^2-s_1^2)\]

Spring deformation is measured from natural length. A fixed pin reaction does zero work because its application point does not move. An ideal normal force does zero work when perpendicular to motion. Static friction at an ideal no-slip rolling contact also has zero instantaneous power, but kinetic friction during slip usually dissipates energy and must be included.

Physical meaning

Positive work adds kinetic energy, negative work removes it, and zero work can still accompany a dynamically essential constraint force. “No work” does not mean “no force.”

Fig. 18-2The spring force opposes the body's displacement as the stretch grows from s₁ to s₂, so its work is negative. (Original redrawn schematic.)

18.3The Work of a Couple Moment

A couple is a free vector and does work through body rotation, not translation.

Couple work and power

\[U_M=\int_{\theta_1}^{\theta_2}M\,d\theta,\qquad P_M=M\omega\]

Sign and applicability

Choose a positive angular sense before integrating. A couple in that sense contributes positive work for positive \(d\theta\); an opposing brake contributes negative work. Angles must be in radians. If moment varies with angle, integrate its signed function rather than multiplying endpoint values.

18.4Principle of Work & Energy

Finite-motion balance

For a rigid body or selected system in an inertial frame, \(T_1+\sum U_{1\to2}=T_2\). Only external work appears for a whole system when ideal internal connection forces cancel.

Problem-solving strategy

Recognize a displacement-based speed problem; choose impulse–momentum instead if the key data are force over time.
Define the body or system, coordinates, positive rotation, initial and final configurations, and all constraints.
Classify each body's motion and express every speed through independent generalized speeds.
Inventory work at each force application point and each couple; omit a force only after justifying zero work.
Write the symbolic energy balance, substitute values with units, then recover direction from the assumed motion.
Verify rolling, contact, cable tension, friction capacity, and operating plausibility separately.

18.5Conservation of Energy

Mechanical-energy form

\[T_1+V_1+U_{NC}=T_2+V_2,\qquad V_g=mgy_G,\quad V_s=\tfrac12ks^2\]

Use conservation of mechanical energy when gravity and ideal springs are represented by potential energy and all remaining nonconservative work is zero. Datum choice is arbitrary but fixed. Include motor work, braking, sliding friction, and other nonconservative effects explicitly as \(U_{NC}\).

Applications after method development

EXAMPLE 1Flywheel accelerated through a known angle

An 18 kg uniform flywheel of radius 0.300 m rotates at 4.00 rad/s when a 24.0 N m driving couple begins acting through 5.00 rad; a constant 4.00 N m bearing couple opposes the motion. Find its final angular speed.

Recognition strategy

The data connect two angular positions rather than two times, so fixed-axis rigid-body work–energy is the direct kinetics method. Couple work changes rotational kinetic energy; angular acceleration history is unnecessary.

Given information

Uniform disk: \(m=18.0\ \text{kg}\), \(r=0.300\ \text{m}\), and \(I_O=\tfrac12mr^2\).
Initial speed \(\omega_1=4.00\ \text{rad/s}\), positive counterclockwise; rotation \(\Delta\theta=+5.00\ \text{rad}\).
Driving couple is \(+24.0\ \text{N m}\); bearing resistance is \(-4.00\ \text{N m}\); find \(\omega_2\).

Model and interpretation

Treat the wheel as a rigid uniform disk on a fixed, ideal axis. Counterclockwise rotation and couples are positive. Pin reactions act at the stationary axis and do no work; this is kinetics because applied couples cause a speed change.

Step-by-step solution

1. Select the kinetic-energy form. Because the axis through \(O\) is fixed, \(T=\tfrac12I_O\omega^2\), with \(I_O=\tfrac12(18.0)(0.300)^2=0.810\ \text{kg m}^2\).

2. Evaluate all work. The two constant couples act through the same positive angle, so \(U_{1\to2}=(24.0-4.00)(5.00)=100\ \text{J}\).

3. Apply work–energy symbolically. Writing \(\tfrac12I_O\omega_1^2+U_{1\to2}=\tfrac12I_O\omega_2^2\) gives \(\omega_2^2=\omega_1^2+2U_{1\to2}/I_O\).

4. Calculate the final state. Substitution gives \(\omega_2=\sqrt{(4.00)^2+2(100)/(0.810)}=16.2\ \text{rad/s}\), retaining the positive sense.

Verification

Units: Couple work has \(\text{N m}\) times radians, hence joules; \(I\omega^2\) also has \(\text{kg m}^2/\text{s}^2=\text{J}\), so the energy equation is dimensionally consistent.

Signs and direction: The driving couple does positive work and resistance negative work under the declared counterclockwise convention. Their positive net work requires a larger speed without reversing the wheel.

Plausibility: Initial energy is 6.48 J and final energy is 106.48 J, a rise of exactly 100 J. Removing bearing resistance would produce a slightly larger result, as the limiting behavior requires.

Final result: \(\boxed{\omega_2=16.2\ \text{rad/s counterclockwise}}\).

Engineering interpretation

The large speed increase reflects the flywheel's modest inertia and 100 J net input. A real flywheel assessment must also check bearing ratings, rotor stress, overspeed containment, and torque transients.

EXAMPLE 2Rolling cylinder descending a ramp

A 40.0 kg uniform solid cylinder of radius 0.250 m starts from rest and rolls without slipping down a ramp through a vertical drop of 1.20 m. The ramp angle satisfies \(\sin\beta=0.400\), and \(\mu_s=0.220\). Find the speed and verify rolling is admissible.

Recognition strategy

Gravity and static contact are present, but ideal rolling friction does no work because the instantaneous contact point has zero velocity. Conservation of mechanical energy gives speed; a separate force check tests the assumed constraint.

Given information

Cylinder mass \(40.0\ \text{kg}\), radius \(0.250\ \text{m}\), \(I_G=\tfrac12mr^2\), and initial rest.
Vertical drop \(h=1.20\ \text{m}\), \(\sin\beta=0.400\), \(g=9.81\ \text{m/s}^2\).
No-slip constraint \(v_G=\omega r\); available static coefficient \(\mu_s=0.220\).

Model and interpretation

Choose positive motion down the incline and clockwise angular speed as positive for this calculation. The cylinder is rigid, the ramp is stationary, rolling resistance is neglected, and static friction is checked after the energy solution.

Step-by-step solution

1. Write energy with the rolling constraint. From rest, \(mgh=\tfrac12mv_G^2+\tfrac12I_G\omega^2\), and substituting \(I_G=\tfrac12mr^2\) and \(\omega=v_G/r\) gives \(mgh=\tfrac34mv_G^2\).

2. Solve for speed and spin. Thus \(v_G=\sqrt{4gh/3}=\sqrt{4(9.81)(1.20)/3}=3.96\ \text{m/s}\), and \(\omega=v_G/r=15.8\ \text{rad/s}\).

3. Determine required static friction. For a solid cylinder rolling down a constant slope, translation and rotation give \(f=(1/3)mg\sin\beta=(1/3)(40.0)(9.81)(0.400)=52.3\ \text{N}\), acting uphill.

4. Check contact admissibility. The normal force is \(N=mg\cos\beta=(40.0)(9.81)\sqrt{1-0.400^2}=359.6\ \text{N}\), so \(f/N=0.145<\mu_s=0.220\); no slip is feasible.

Verification

Units: Both \(mgh\) and the translational and rotational kinetic terms reduce to joules. The resulting square root has units m/s, while division by radius produces rad/s.

Signs and direction: Gravity decreases potential energy and increases kinetic energy. Although the cylinder moves downhill, static friction acts uphill to create the clockwise angular acceleration required by rolling.

Plausibility: The result is below the frictionless sliding value \(\sqrt{2gh}=4.85\ \text{m/s}\) because energy is shared with rotation. The positive normal force and friction margin confirm maintained contact and no slip.

Final result: \(\boxed{v_G=3.96\ \text{m/s downhill},\quad \omega=15.8\ \text{rad/s clockwise}}\), with \(f/N=0.145<0.220\).

Engineering interpretation

Static friction redistributes gravitational energy into rotation without dissipating it. Surface contamination or compliance could invalidate ideal rolling, so practical equipment needs measured traction and rolling-loss data.

EXAMPLE 3Payload hoist energy and operating speed

A 120 kg hoist drum has radius of gyration 0.320 m and cable radius 0.250 m. It raises a 300 kg payload 4.00 m while a 1200 N m motor couple drives and an 80.0 N m bearing couple resists. Starting at 2.00 rad/s, find final drum and payload speeds.

Recognition strategy

The requested speeds follow a finite payload displacement, and motor, bearing, and gravitational work are known. A system work–energy balance avoids solving cable tension and acceleration as primary unknowns.

Given information

Drum: \(m_d=120\ \text{kg}\), \(k_G=0.320\ \text{m}\), \(r=0.250\ \text{m}\), so \(I= m_dk_G^2\).
Payload \(m_p=300\ \text{kg}\) rises \(4.00\ \text{m}\); no cable slip gives \(v_p=r\omega\) and \(\Delta\theta=4.00/r\).
Couples: motor \(+1200\ \text{N m}\), bearing \(-80.0\ \text{N m}\); \(\omega_1=2.00\ \text{rad/s}\).

Model and interpretation

The system includes drum and payload, making cable tension internal. Positive drum rotation raises the load. The drum is rigid, cable mass and elasticity are neglected, the cable remains taut, and motor torque is constant over 16.0 rad.

Step-by-step solution

1. Reduce the system inertia. Compute \(I=120(0.320)^2=12.288\ \text{kg m}^2\). With \(v_p=r\omega\), total kinetic energy is \(T=\tfrac12(I+m_pr^2)\omega^2\).

2. Evaluate external work. The drum turns \(\Delta\theta=4.00/0.250=16.0\ \text{rad}\). Hence \(U=(1200-80.0)(16.0)-(300)(9.81)(4.00)=6148\ \text{J}\).

3. Apply the system balance. Using \(T_1+U=T_2\), where \(I+m_pr^2=12.288+300(0.250)^2=31.038\ \text{kg m}^2\), yields \(\omega_2^2=\omega_1^2+2U/31.038\).

4. Resolve speeds and admissibility. The result is \(\omega_2=20.0\ \text{rad/s}\) and \(v_{p2}=r\omega_2=5.00\ \text{m/s}\). The implied \(\alpha=12.38\ \text{rad/s}^2\) gives upward acceleration \(3.10\ \text{m/s}^2\), so cable tension remains positive.

Verification

Units: Couple-angle and weight-height products are joules. The equivalent inertia times \(\omega^2\) is also joules, and multiplying angular speed by cable radius correctly gives m/s.

Signs and direction: Motor work is positive, bearing and gravity work are negative, and the remaining net work is positive. Thus the upward-moving payload accelerates and the taut-cable check does not indicate reversal.

Plausibility: The independent constant-acceleration check \(\omega_2^2=\omega_1^2+2\alpha\Delta\theta\) gives approximately 400.2, agreeing with the energy result; all contact constraints remain admissible.

Final result: \(\boxed{\omega_2=20.0\ \text{rad/s},\qquad v_{p2}=5.00\ \text{m/s upward}}\).

Engineering interpretation

The predicted payload speed is high for many hoists, flagging a need for drive controls and braking analysis. Final design requires cable, drum, gearbox, motor thermal, overspeed, code, and safety-factor review.

★Chapter recap

Choose kinetic energy from the body's actual motion and sum every moving component.
Evaluate force work at its application point and couple work through signed angular displacement.
Use work–energy for finite displacement with nonconservative work; use conservation only when its assumptions hold.
Energy predicts speed magnitude, while kinematics and sign conventions establish direction.
Always test no-slip, contact, tension, friction, and operating limits after solving.

Before moving on

You should be able to build a complete work inventory, reduce a constrained multibody system to independent speeds, solve symbolically before substitution, and defend the result with units, signs, limiting behavior, and admissibility checks.