Planar Kinetics of a Rigid Body

Impulse & Momentum

Rigid-body impulse–momentum connects forces and couples acting over time to changes in translation and rotation. It is the preferred kinetics tool for pulses, short events, and impacts, while kinematic constraints still determine compatible velocities.

Learning objectives

Compute linear and angular momentum about appropriate reference points.
Integrate force and couple histories to obtain vector and angular impulse.
Select useful systems and moment centers that eliminate unknown internal or support impulses.
Apply conservation laws only when the relevant external impulse or moment is negligible.
Analyze eccentric impact using contact-point restitution and momentum equations.
Verify signs, units, separation, energy loss, and contact admissibility.

19.1

Linear and Angular Momentum

Momentum state of a planar rigid body

For constant mass, linear momentum is \(\mathbf L=m\mathbf v_G\). Angular momentum about the mass center is \(H_G=I_G\omega\). About another fixed inertial point \(P\), include the moment of linear momentum: \(H_P=I_G\omega+\mathbf r_{G/P}\times m\mathbf v_G\).

Planar momentum

\[\mathbf L=m\mathbf v_G,\qquad H_G=I_G\omega,\qquad H_P=I_G\omega+(\mathbf r_{G/P}\times m\mathbf v_G)_z\]

Physical meaning

Linear momentum tracks mass-center translation and retains direction. Angular momentum measures rotational tendency about a chosen point; changing the point changes the translational moment term, so all moments and angular impulses in one equation must use the same reference.

FIG. 19-1 — Body momentum diagram. The linear momentum \(m\mathbf v_G\) acts at the mass center; the angular momentum \(I_G\omega\) is a free vector that may be placed anywhere.

Three types of planar motion

Motion	Momentum model	Judgment
Translation	\(\mathbf L=m\mathbf v_G\), \(H_G=0\)	Angular momentum about another point may still be nonzero.
Fixed-axis rotation	\(H_O=I_O\omega\)	Use only when \(O\) is truly fixed in an inertial frame.
General plane motion	\(\mathbf L=m\mathbf v_G\), \(H_G=I_G\omega\)	Include translation and rotation; impose velocity constraints separately.

19.2

Principle of Impulse and Momentum

Integrated equations of motion

For a rigid body in planar motion, \(m\mathbf v_{G1}+\sum\int_{t_1}^{t_2}\mathbf F\,dt=m\mathbf v_{G2}\) and \(H_{P1}+\sum\int_{t_1}^{t_2}M_P\,dt=H_{P2}\), provided angular momentum and moment are formed consistently about an allowable inertial point.

Mass-center scalar form

\[m(v_{Gx})_1+\sum\int F_xdt=m(v_{Gx})_2,\quad m(v_{Gy})_1+\sum\int F_ydt=m(v_{Gy})_2,\quad I_G\omega_1+\sum\int M_Gdt=I_G\omega_2\]

Impulse is the signed area under a force–time or moment–time curve. During a very short impact, bounded forces such as weight often have negligible impulse compared with contact and support impulses; this is an approximation that should be justified by time scale and magnitude.

FIG. 19-2 — Initial momentum + applied impulse = final momentum. Drawing all three keeps moment arms visible.

Procedure for analysis

Problem-solving strategy

Recognize time, impulse, collision, or sudden coupling cues; use work–energy instead when displacement and work dominate.
Select one body or a system, an inertial frame, positive velocity components, and positive rotation.
Draw external impulses and couples, then choose a moment center through troublesome unknown impulses when valid.
Write symbolic linear and angular impulse–momentum equations and add independent velocity constraints.
For impact, apply restitution only along the line of impact at the actual contact points.
Solve and audit units, directions, separation, energy loss, and friction or contact capacity.

19.3

Conservation of Momentum

Conservation is component- and point-specific

If net external impulse in one direction is negligible, that component of system linear momentum is conserved. If net external angular impulse about point \(P\) is negligible, angular momentum about \(P\) is conserved. One statement can hold while the other does not.

Conservation tests

\[\sum\int F_xdt\approx0\Rightarrow (L_x)_1=(L_x)_2,\qquad \sum\int M_Pdt\approx0\Rightarrow (H_P)_1=(H_P)_2\]

Physical meaning

Conservation does not mean forces vanish; internal impulses may be large and supports may exert impulses whose moments vanish about a selected pin. The system boundary and reference point determine which cancellation is legitimate.

19.4

Eccentric Impact

In eccentric impact the contact impulse does not pass through a body's mass center, so translation and rotation are coupled. Define the normal line of impact from contact geometry; smooth contact carries no tangential impulse, whereas rough contact requires a tangential impulse law and a stick–slip admissibility check.

Restitution at contact

\[e=\frac{(v_{B,c})_{n2}-(v_{A,c})_{n2}}{(v_{A,c})_{n1}-(v_{B,c})_{n1}},\qquad 0\le e\le1\]

Contact and energy checks

Use rigid-body kinematics to compute each contact-point normal velocity. After solving, the bodies must be separating or just together, the normal impulse must be compressive, and final kinetic energy must not exceed initial kinetic energy for a passive impact model.

Applications after method development

EXAMPLE 1Programmed torque impulse on a rotor

A 12.0 kg uniform disk of radius 0.350 m rotates clockwise at 3.00 rad/s. A counterclockwise couple \(M(t)=10.0+4.00t\) N m acts for 1.50 s. Find the disk's final angular velocity.

Recognition strategy

The applied couple is specified over a time interval and the target is a later angular velocity. Angular impulse–momentum about the fixed axis directly relates the torque-time area to angular momentum change.

Given information

Uniform disk: \(m=12.0\ \text{kg}\), \(r=0.350\ \text{m}\), \(I_O=\tfrac12mr^2\).
Counterclockwise is positive, so \(\omega_1=-3.00\ \text{rad/s}\).
Applied couple \(M(t)=10.0+4.00t\ \text{N m}\) for \(0\le t\le1.50\ \text{s}\); find \(\omega_2\).

Model and interpretation

The disk is rigid and rotates on an ideal fixed bearing. Bearing reactions pass through the moment center and create no angular impulse about it. This is kinetics; the given time history, not geometry alone, changes momentum.

Step-by-step solution

1. Compute rotational inertia. For the uniform disk, \(I_O=\tfrac12(12.0)(0.350)^2=0.735\ \text{kg m}^2\).

2. Integrate the signed couple. The counterclockwise angular impulse is \(\int_0^{1.50}(10.0+4.00t)dt=[10.0t+2.00t^2]_0^{1.50}=19.5\ \text{N m s}\).

3. Apply angular impulse–momentum. About the fixed axis, \(I_O\omega_1+\int Mdt=I_O\omega_2\), so \(\omega_2=\omega_1+(19.5)/(0.735)\).

4. Resolve magnitude and sense. Numerically, \(\omega_2=-3.00+26.53=23.5\ \text{rad/s}\); the positive result means the initial clockwise rotation reverses.

Verification

Units: Angular impulse has units \(\text{N m s}=\text{kg m}^2/\text{s}\), identical to \(I\omega\). Dividing impulse by inertia therefore yields rad/s.

Signs and direction: The initial momentum is negative and the applied angular impulse positive. Since 19.5 N m s exceeds the 2.205 N m s needed to stop the disk, reversal to counterclockwise rotation is required.

Plausibility: The stopping impulse occurs early, and the remaining impulse accelerates the disk positively. A constant 10 N m lower bound alone would already exceed the initial momentum over 1.50 s.

Final result: \(\boxed{\omega_2=23.5\ \text{rad/s counterclockwise}}\).

Engineering interpretation

Torque-time area, rather than peak torque alone, controls momentum change. A practical rotor drive also requires checks of reversal transients, shaft stress, bearing loads, controller limits, and overspeed protection.

EXAMPLE 2Impulse needed for no-slip spool launch

A 20.0 kg uniform solid cylinder of radius 0.200 m rests on a horizontal floor. A 24.0 N s horizontal impulse acts rightward through its center, while static contact prevents slip. Find its mass-center speed, angular speed, and floor impulse; verify a 30.0 N s static-impulse capacity is sufficient.

Recognition strategy

A short prescribed impulse changes both linear and angular momentum. The center impulse has no moment about the mass center, so an unknown floor impulse plus the no-slip velocity constraint closes the rigid-body system.

Given information

Cylinder \(m=20.0\ \text{kg}\), \(r=0.200\ \text{m}\), \(I_G=\tfrac12mr^2=0.400\ \text{kg m}^2\), initially at rest.
Applied center impulse \(J=+24.0\ \text{N s}\), with rightward translation positive and counterclockwise rotation positive.
No-slip final constraint \(v_G=-\omega r\); maximum sustainable static contact impulse is 30.0 N s.

Model and interpretation

Treat the cylinder as rigid and the impulsive interval as so short that weight and normal impulses do not affect horizontal or angular equations. Let floor impulse \(K\) be positive rightward; its sign will determine the actual direction.

Step-by-step solution

1. Write linear impulse–momentum. In the horizontal direction, \(J+K=mv_G\), or \(24.0+K=20.0v_G\).

2. Write angular impulse–momentum. About \(G\), the floor impulse gives \(-Kr=I_G\omega\). With \(v_G=-\omega r\), this becomes \(K=I_Gv_G/r^2=10.0v_G\).

3. Solve the coupled equations. Substitution gives \(24.0+10.0v_G=20.0v_G\), hence \(v_G=2.40\ \text{m/s}\) and \(K=24.0\ \text{N s}\).

4. Recover spin and test contact. The rolling constraint gives \(\omega=-v_G/r=-12.0\ \text{rad/s}\), clockwise. Since \(|K|=24.0<30.0\ \text{N s}\), the supplied static-impulse capacity supports no slip.

Verification

Units: Linear impulse N s equals kg m/s and matches \(mv_G\). Multiplying the floor impulse by radius gives N m s, matching \(I_G\omega\).

Signs and direction: The solved floor impulse is positive, so it acts rightward and creates clockwise angular impulse about the center. The negative angular velocity is consistent with rightward rolling.

Plausibility: Without floor impulse the cylinder would translate at 1.20 m/s without spinning. Static contact adds rightward momentum while generating spin; the stated capacity exceeds the required impulse.

Final result: \(\boxed{v_G=2.40\ \text{m/s right},\quad \omega=12.0\ \text{rad/s clockwise},\quad K=24.0\ \text{N s right}}\).

Engineering interpretation

An impulsive no-slip assumption must be supported by contact-impulse capacity, not an ordinary force comparison alone. Real launches also depend on pulse duration, compliance, pressure distribution, and surface damage.

EXAMPLE 3Eccentric puck impact on a pinned gate

A 2.00 kg puck moving at 6.00 m/s strikes normally and smoothly at the free end of a stationary 8.00 kg slender gate of length 1.20 m pinned at its other end. With restitution \(e=0.400\), find the puck speed and gate angular speed immediately after impact.

Recognition strategy

The line of impact misses the gate's mass center and fixed pin, so the gate gains rotation. Angular momentum about the pin eliminates the unknown pin impulse, while restitution must use contact-point velocities along the impact line.

Given information

Puck mass \(m=2.00\ \text{kg}\), incoming velocity \(+6.00\ \text{m/s}\) along the smooth line of impact.
Gate mass \(8.00\ \text{kg}\), length \(L=1.20\ \text{m}\), initially at rest, with \(I_O=\tfrac13m_gL^2=3.84\ \text{kg m}^2\).
Coefficient of restitution \(e=0.400\); positive puck velocity and gate-end velocity are in the incoming direction.

Model and interpretation

The impact duration is short, the pin is ideal, gravity impulse is negligible, and contact impulse acts only along the stated normal. Angular momentum of puck plus gate is conserved about the pin during impact, but kinetic energy is not assumed conserved.

Step-by-step solution

1. Form angular momentum about the pin. The pin impulse has zero moment, so \((2.00)(6.00)(1.20)=(2.00)u(1.20)+3.84\omega\), where \(u\) is puck velocity after impact.

2. Write restitution at contact. The gate-end speed is \(v_C=\omega L\). Separation relative speed equals \(e\) times approach speed: \(\omega(1.20)-u=0.400(6.00)=2.40\ \text{m/s}\).

3. Solve the simultaneous relations. Using \(u=1.20\omega-2.40\) in the angular-momentum equation gives \(14.4=2.40u+3.84\omega\), hence \(\omega=3.00\ \text{rad/s}\).

4. Recover velocity and impulse checks. Then \(u=1.20(3.00)-2.40=1.20\ \text{m/s}\). Puck impulse is \(2.00(1.20-6.00)=-9.60\ \text{N s}\), so the gate receives 9.60 N s at contact.

Verification

Units: Every angular-momentum term has kg m²/s, while restitution compares velocities in m/s. The contact impulse from mass times velocity change has the required N s units.

Signs and direction: Both outgoing quantities are positive in the declared direction; the gate end moves at 3.60 m/s and separates faster than the puck at 1.20 m/s, satisfying the restitution convention.

Plausibility: Initial kinetic energy is 36.0 J and final energy is \(1.44+17.28=18.72\) J, so 17.28 J is lost rather than created. The positive loss and separating contact velocities are admissible.

Final result: \(\boxed{u=1.20\ \text{m/s forward},\qquad \omega=3.00\ \text{rad/s in the impact-driven sense}}\).

Engineering interpretation

The offset contact converts projectile momentum into gate rotation and a substantial pin reaction impulse. Hardware assessment still needs contact stiffness, local stress, pin capacity, rebound variability, guarding, and repeated-impact fatigue review.

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Chapter recap

Impulse–momentum is the time-based kinetics method for changes in rigid-body velocity.
Choose the system and moment center to expose or eliminate unknown impulses legitimately.
Retain vector signs and include both translation and rotation for general plane motion.
Invoke conservation only after showing the corresponding external impulse is negligible.
For impact, use contact-point normal velocities and verify separation, compressive impulse, and nonnegative energy loss.

Before moving on

You should be able to integrate a force or couple history, construct consistent linear and angular momentum equations, close them with rigid-body kinematics, and defend the result using dimensional, directional, contact, and energy checks.