Dynamics · Chapter 17
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Dynamics · Chapter 17

Planar Kinetics of a Rigid Body:
Force and Acceleration

A force system applied to a body of definite size and shape can make it both move along a path and spin. This chapter links those two effects: translation is governed by force and the acceleration of the mass center, while rotation is governed by moment and a new quantity — the mass moment of inertia. Together they give three scalar equations that describe any planar motion of a symmetric rigid body.

Chapter Objectives

  • Introduce the methods used to compute the mass moment of inertia of a body.
  • Develop the planar kinetic equations of motion for a symmetric rigid body.
  • Apply these equations to bodies undergoing translation, fixed-axis rotation, and general plane motion.
  • Select a useful moment center and connect the kinetic equations to independent kinematic constraints.
  • Verify units, signs, contact admissibility, and static-friction capacity.
Instructor Note

Almost every problem in this chapter is solved the same way: draw a clean free-body diagram, draw the kinetic diagram beside it, then write at most three equations (ΣFx, ΣFy, ΣM). If you ever have more unknowns than equations, the missing one is almost always a kinematics relation such as aG = αr. Master that habit here and the rest of dynamics gets much easier.

17.1

Mass Moment of Inertia

Just as mass measures a body's resistance to linear acceleration in F = ma, the mass moment of inertia measures its resistance to angular acceleration in the rotational equation M = Iα. A large flywheel, for instance, is hard to start and hard to stop precisely because its moment of inertia is large.

Mass Moment of Inertia

The second moment of a body's mass about an axis, formed by summing each mass element multiplied by the square of its perpendicular distance from that axis. Because distance is squared, the result is always positive, and its value changes with the chosen axis. Common units are kg·m².

Moment of Inertia — Definition
\[ I = \int_m r^2\,dm \]
\(r\) = perpendicular distance from the axis to the mass element \(dm\).

When density varies through the body, the element is written as dm = ρ dV and the integral is taken over volume. If density is constant it factors out, leaving a purely geometric integral.

Volume Form (variable density)
\[ I = \int_V r^2\,\rho\,dV \qquad\Longrightarrow\qquad I = \rho\int_V r^2\,dV \;\;(\rho\text{ constant}) \]

Shell vs. Disk Elements

For a body of revolution, a single integration suffices if the element is thin in only one direction. A shell element has every part at the same radius \(r\), so it can be used directly. A disk element spans a range of radii, so the inertia of the element itself must be found first and then integrated.

z r Shell element z y dz Disk element
Fig 17.1Two differential elements for a body of revolution: a thin shell (left) sits entirely at one radius; a thin disk (right) spans many radii. Original schematic redrawn for MechForge.
THEOREMParallel-Axis Theorem

Condition. The moment of inertia \(I_G\) about an axis through the mass center \(G\) is known, and a second axis is parallel to it at perpendicular distance \(d\).

Conclusion. The moment of inertia about the second axis is

\[ I = I_G + m d^2 \]
  • IGmoment of inertia about the parallel axis through \(G\)
  • mmass of the body
  • dperpendicular distance between the two parallel axes

Radius of Gyration

A geometric length \(k\) that packages the moment of inertia: if the entire mass were concentrated at distance \(k\) from the axis, it would give the same inertia. Handbooks often list \(k\) instead of \(I\).

Radius of Gyration
\[ I = m k^2 \qquad\Longleftrightarrow\qquad k = \sqrt{\tfrac{I}{m}} \]

Composite Bodies

For a body made of simple shapes, find each part's inertia about the common axis (using the parallel-axis theorem and tabulated values) and add them. A region that has been removed, such as a hole, is added as a negative contribution.

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17.2

Planar Kinetic Equations of Motion

The analysis is restricted to bodies that are symmetric about a fixed reference plane, so every force and couple can be projected onto that plane. Within it, motion is described by two translational equations and one rotational equation.

Physical meaning

The resultant external force controls acceleration of the mass center, while the resultant moment about G controls angular acceleration. Neither equation alone describes the full motion of a finite body, and kinematics supplies any connection imposed by pins, guides, cords, or rolling contact.

Translational Equation of Motion

The vector sum of all external forces equals the body's mass times the acceleration of its mass center \(G\). In the plane it splits into two independent scalar equations.

Translation
\[ \sum \mathbf{F} = m\,\mathbf{a}_G \quad\Rightarrow\quad \sum F_x = m(a_G)_x,\ \ \sum F_y = m(a_G)_y \]

Rotational Equation of Motion

Summing moments of the external forces about the mass center gives the product of the body's moment of inertia about \(G\) and its angular acceleration. Internal forces cancel in collinear pairs, so only external effects remain.

Rotation About G
\[ \sum M_G = I_G\,\alpha \]

Kinetic Moments About an Arbitrary Point

Moments may instead be summed about any point \(P\). The external moments then equal the moment of the "kinetic" vectors — the components of \(m\mathbf{a}_G\) plus \(I_G\boldsymbol{\alpha}\) — taken about \(P\). This is written compactly as \(\sum M_P = \sum (\mathcal{M}_k)_P\).

General Planar Set
\[ \sum F_x = m(a_G)_x,\quad \sum F_y = m(a_G)_y,\quad \sum M_G = I_G\alpha \;\;\text{or}\;\; \sum M_P = \textstyle\sum(\mathcal{M}_k)_P \]
Instructor Note

The kinetic diagram is not optional decoration. Sketch \(m(a_G)_x\), \(m(a_G)_y\), and \(I_G\alpha\) next to the free-body diagram. When you sum moments about a clever point \(P\) (one that kills unknown reactions), the kinetic diagram tells you the moment arm of each kinetic vector — that is where sign errors usually creep in.

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17.3

Equations of Motion: Translation

In pure translation every particle has the same acceleration and the angular acceleration is zero, so \(\sum M_G = 0\). Two cases arise depending on the shape of the path.

Rectilinear Translation

All particles travel along parallel straight lines. Since \(I_G\alpha = 0\), only \(m\mathbf{a}_G\) appears on the kinetic diagram.

Rectilinear Translation
\[ \sum F_x = m(a_G)_x,\quad \sum F_y = m(a_G)_y,\quad \sum M_G = 0 \]

Curvilinear Translation

All particles follow congruent curved paths. It is convenient to use normal–tangential axes centered at \(G\). A moment sum about another point \(B\) must account for the moment of both acceleration components.

Curvilinear Translation
\[ \sum F_n = m(a_G)_n,\quad \sum F_t = m(a_G)_t,\quad \sum M_G = 0 \]
G W Free-body = m a_G Kinetic
Fig 17.2For translation the kinetic diagram shows only the single vector \(m\mathbf{a}_G\); there is no \(I_G\alpha\) term. Original schematic redrawn for MechForge.
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17.4

Equations of Motion: Rotation About a Fixed Axis

When a body rotates about a fixed pin \(O\), its mass center \(G\) travels a circular path, so the acceleration of \(G\) is best resolved into normal and tangential components. The normal component always points from \(G\) toward \(O\); the tangential component follows the sense of \(\alpha\).

Acceleration Components of G
\[ (a_G)_n = \omega^2 r_G,\qquad (a_G)_t = \alpha r_G \]
Fixed-Axis Rotation (about G)
\[ \sum F_n = m\omega^2 r_G,\quad \sum F_t = m\alpha r_G,\quad \sum M_G = I_G\alpha \]

Moment Equation About the Pin O

Summing moments about the pin eliminates the unknown pin reaction. Because the normal acceleration component passes through \(O\), it produces no moment, and the parallel-axis theorem collapses the result to a single clean equation using \(I_O\).

Moment About the Pin
\[ \sum M_O = I_O\,\alpha \qquad\text{where}\qquad I_O = I_G + m r_G^{\,2} \]
O G r_G (a_G)_n (a_G)_t α
Fig 17.3Body rotating about fixed pin \(O\): the mass center's acceleration splits into a normal component aimed at \(O\) and a tangential component along the sense of \(\alpha\). Original schematic redrawn for MechForge.
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17.5

Equations of Motion: General Plane Motion

In general plane motion a body both translates and rotates. The same three scalar equations apply, now with both translational components and \(I_G\alpha\) generally nonzero. Moments may be summed about \(G\) or about any convenient point \(P\).

General Plane Motion
\[ \sum F_x = m(a_G)_x,\quad \sum F_y = m(a_G)_y,\quad \sum M_G = I_G\alpha \;\;\text{or}\;\; \sum M_P = \textstyle\sum(\mathcal{M}_k)_P \]

Rolling Without Slipping

When a disk, cylinder, or wheel rolls without slipping, the contact point is the instantaneous center of zero velocity, and the mass-center acceleration ties to the angular acceleration through \(a_G = \alpha r\). Summing moments about that contact point gives a one-equation shortcut.

Moment About the Instantaneous Center
\[ \sum M_{IC} = I_{IC}\,\alpha \qquad (\text{rolling without slipping}) \]
Compare with \(\sum M_O = I_O\alpha\) for a pinned body.
Instructor Note

With friction present, never assume "rolling without slipping" — test it. Solve assuming no slip, then check \(F \le \mu_s N\). If the required friction exceeds that limit, throw out the kinematic constraint \(a_G = \alpha r\), set \(F = \mu_k N\) instead, and resolve. A valid solution must report this capacity check explicitly rather than assume that contact can supply any required friction.

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Problem-solving strategy

  1. Isolate each rigid body and draw every external force, couple, and contact reaction on an FBD.
  2. Declare coordinates and angular sign, classify the motion, and obtain the mass-center acceleration from kinematics.
  3. Use \(\sum\mathbf F=m\mathbf a_G\) and \(\sum M_G=I_G\alpha\), or a justified fixed-pivot moment equation.
  4. Add independent constraints such as rolling without slip, then count equations and unknowns before solving.
  5. Test unilateral contact and static-friction inequalities; if an assumption fails, rebuild the model rather than clipping the answer.

Integrated applications

EXAMPLE 1Released access-bar acceleration
A uniform \(18.0 \mathrm{kg}\), \(1.20 \mathrm m\) slender access bar is horizontal and pinned at its left end O. At release from rest, a \(40.0 \mathrm N\) downward actuator force acts at the free end. Determine angular acceleration and pin-reaction components.

Recognition strategy

A rigid body with a nonaccelerating pin undergoes fixed-axis kinetics; a moment about O removes both unknown pin reactions before force balance recovers them.

Given information

  • Take \(+x\) right, \(+y\) up, and counterclockwise positive; \(m=18.0 \mathrm{kg}\), \(L=1.20 \mathrm m\), \(g=9.81 \mathrm{m/s^2}\), and initially \(\omega=0\).

Model and interpretation

The bar is slender and rigid, the pin is fixed and frictionless, and gravity acts at G. This is kinetics because external forces cause acceleration.

Step-by-step solution
1. Evaluate inertia. About the fixed end, \(I_O= \tfrac13mL^2= \tfrac13(18.0)(1.20)^2=8.64 \mathrm{kg\,m^2}\).
2. Sum moments at O. Both downward loads are clockwise, so \(\sum M_O=-40.0(1.20)-18.0(9.81)(0.600)=I_O\alpha\).
3. Solve angular motion. The moment is \(-153.95 \mathrm{N\,m}\), hence \(\alpha=-17.82 \mathrm{rad/s^2}\); at release, \(\mathbf a_G=\alpha(0.600)\mathbf j=-10.69\mathbf j \mathrm{m/s^2}\).
4. Recover pin forces. Horizontal balance gives \(O_x=0\). Vertical balance \(O_y-40.0-176.58=18.0(-10.69)\) gives \(O_y=24.1 \mathrm N\) upward.

Verification

Units: Moment divided by mass moment of inertia gives inverse seconds squared, and multiplying angular acceleration by radius gives metres per second squared.

Signs and direction: Downward forces to the right of O create a clockwise, negative angular acceleration; the positive solved \(O_y\) acts upward as drawn.

Plausibility: With \(\omega=0\), G has no normal acceleration at release. Substituting the reaction into vertical force balance reproduces \(m a_{Gy}=-192.4 \mathrm N\).

Final result: \(\boxed{\alpha=-17.82 \mathrm{rad/s^2},\quad O_x=0,\quad O_y=24.1 \mathrm N \text{upward}}\)

Engineering interpretation

The pin load is small only at this release instant because G accelerates nearly at gravitational magnitude; later normal acceleration can substantially increase the resultant bearing load.

EXAMPLE 2No-slip check for an inspection roller
A uniform solid cylindrical inspection roller of mass \(32.0 \mathrm{kg}\) and radius \(0.250 \mathrm m\) is released from rest on an \(18.0^\circ\) incline. If \(\mu_s=0.160\), determine its acceleration and verify whether rolling without slipping is admissible.

Recognition strategy

Forces and instantaneous acceleration make this a kinetics problem; assume rolling, combine translation and rotation with \(a_G=\alpha r\), then test static friction.

Given information

  • Take positive down the incline and positive angular acceleration clockwise; \(I_G= \tfrac12mr^2\), \(g=9.81 \mathrm{m/s^2}\), and the normal direction is outward.

Model and interpretation

The roller and incline are rigid; contact is unilateral and static if the required friction does not exceed \(\mu_sN\). Air and rolling resistance are neglected.

Step-by-step solution
1. Draw equations. Let friction \(f\) act uphill. Translation gives \(mg\sin18^\circ-f=ma_G\), while normal balance gives \(N=mg\cos18^\circ\).
2. Add rotation and constraint. The friction moment gives \(fr=I_G\alpha\); rolling requires \(a_G=\alpha r\), so \(f=I_Ga_G/r^2= \tfrac12ma_G\).
3. Solve acceleration. Substitution yields \(a_G=g\sin18^\circ/(1+1/2)=2.021 \mathrm{m/s^2}\) down the incline and \(\alpha=a_G/r=8.084 \mathrm{rad/s^2}\) clockwise.
4. Test contact friction. Required \(f= \tfrac12m a_G=32.34 \mathrm N\); \(N=32.0(9.81)\cos18^\circ=298.6 \mathrm N\), so \(\mu_{s,\mathrm{req}}=0.108<0.160\).

Verification

Units: The translational result has metres per second squared, dividing by radius gives radians per second squared, and both friction calculations yield newtons.

Signs and direction: The center accelerates downhill while static friction acts uphill, providing the clockwise moment needed to spin up the roller.

Plausibility: The result is below the frictionless sliding value \(g\sin18^\circ=3.031 \mathrm{m/s^2}\), and the strict friction inequality confirms no slip is physically admissible.

Final result: \(\boxed{a_G=2.021 \mathrm{m/s^2} \text{down slope},\quad \alpha=8.084 \mathrm{rad/s^2} \text{clockwise},\quad f=32.34 \mathrm N}\)

Engineering interpretation

The available friction margin is modest but adequate for the ideal surface. Contamination, wear, slope tolerance, and startup vibration would need allowance before specifying a production roller system.

EXAMPLE 3Traction audit for a driven service wheel
A \(120 \mathrm{kg}\) uniform solid service wheel of radius \(0.400 \mathrm m\) rolls on level concrete. A \(180 \mathrm{N\,m}\) clockwise drive couple acts while a \(90.0 \mathrm N\) resistance acts left through its center. For \(\mu_s=0.350\), find the acceleration and verify traction.

Recognition strategy

The drive couple, resistance, and requested instantaneous acceleration call for general-plane force-acceleration equations plus a provisional rolling constraint and friction-capacity check.

Given information

  • Take \(+x\) right and clockwise angular quantities positive; \(m=120 \mathrm{kg}\), \(r=0.400 \mathrm m\), \(I_G= \tfrac12mr^2=9.60 \mathrm{kg\,m^2}\), and \(g=9.81 \mathrm{m/s^2}\).

Model and interpretation

The ground is stationary and level, the wheel is rigid, and static friction \(f\) is initially assumed rightward. Normal contact must remain compressive and \(|f|\le\mu_sN\).

Step-by-step solution
1. Write translation. Horizontal force balance is \(f-90.0=120a_G\); vertical acceleration is zero, so \(N=mg=1177.2 \mathrm N>0\).
2. Write rotation. Taking clockwise moments positive, \(180-fr=I_G\alpha\). The rightward ground friction creates a counterclockwise moment, explaining its negative sign.
3. Apply rolling compatibility. No slip gives \(a_G=\alpha r\). Substituting \(f=120a_G+90\) and \(\alpha=a_G/0.400\) gives \(180-0.400(120a_G+90)=24a_G\).
4. Solve and audit traction. The equation gives \(a_G=2.00 \mathrm{m/s^2}\), \(\alpha=5.00 \mathrm{rad/s^2}\), and \(f=330 \mathrm N\). Required \(\mu_s=330/1177.2=0.280<0.350\).

Verification

Units: Force balance remains in newtons, moment balance in newton metres, and \(a_G/r\) produces radians per second squared.

Signs and direction: Positive results confirm rightward acceleration and clockwise spin; positive friction confirms the initially assumed rightward traction direction rather than opposing center motion.

Plausibility: Replacing \(f=330 \mathrm N\) in translation gives \(240 \mathrm N=120(2.00)\), while moment balance gives \(48.0 \mathrm{N\,m}=9.60(5.00)\).

Final result: \(\boxed{a_G=2.00 \mathrm{m/s^2} \text{right},\quad \alpha=5.00 \mathrm{rad/s^2} \text{clockwise},\quad f=330 \mathrm N}\)

Engineering interpretation

The traction demand uses about 80% of the stated static capacity, leaving limited margin for wet surfaces or load transfer; tire data and duty-cycle safety review remain necessary.

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Chapter Summary

Moment of Inertia

A measure of resistance to angular acceleration, \(I = \int r^2\,dm\), different for every axis. Shift between parallel axes with \(I = I_G + md^2\); build composite bodies by adding parts and subtracting holes.

Translation
\[ \begin{aligned} \text{Rectilinear:}&\quad \sum F_x = m(a_G)_x,\ \sum F_y = m(a_G)_y,\ \sum M_G = 0 \\ \text{Curvilinear:}&\quad \sum F_n = m(a_G)_n,\ \sum F_t = m(a_G)_t,\ \sum M_G = 0 \end{aligned} \]
Rotation About a Fixed Axis
\[ \sum F_n = m\omega^2 r_G,\quad \sum F_t = m\alpha r_G,\quad \sum M_G = I_G\alpha\ \ \text{or}\ \ \sum M_O = I_O\alpha \]
General Plane Motion
\[ \sum F_x = m(a_G)_x,\quad \sum F_y = m(a_G)_y,\quad \sum M_G = I_G\alpha\ \ \text{or}\ \ \sum M_P = \textstyle\sum(\mathcal{M}_k)_P \]
Instructor Note

One mental model unifies the whole chapter: force ↔ linear acceleration and moment ↔ angular acceleration, connected by kinematics. Translation is the case where \(\alpha = 0\); fixed-axis rotation is the case where the path of \(G\) is a known circle; general plane motion is everything else. Pick your moment point to delete the unknowns you don't care about, and you'll rarely solve more than three equations at once.

Before moving on

  • Can you produce separate free-body and kinetic diagrams with consistent signs?
  • Can you justify the selected moment equation and inertia axis?
  • Can you test contact force, normal reaction, and no-slip admissibility after solving?