DoAssignment.ca

9.5 · Apply introductory planar rigid-body equations of motion

Learn to apply introductory planar rigid-body equations of motion through clear examples and targeted practice.

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Introduction to Rigid-Body Dynamics

Connecting external forces and moments to translation and rotation

A rigid body can translate and rotate at the same time. In planar motion, its points move in a plane and its rotation axis is perpendicular to that plane. Choose the body as the system and describe its motion relative to an inertial observer, such as someone standing on the ground. Use fixed horizontal and vertical axes unless the geometry suggests more convenient axes. The basic equations connect the net external force to the acceleration of the mass center, and the net external moment about the mass center to angular acceleration. Constraints such as a pin or rolling contact add relationships between motion and geometry. State your axes and positive rotation direction before assigning signs.

What you will learn

  • Describe planar rigid-body motion using the mass center's acceleration and the body's angular acceleration.
  • Draw a free-body diagram and choose consistent positive directions.
  • Apply force and moment equations to find unknown accelerations or forces.
  • Use the no-slip relation for a body rolling on a fixed surface.

1. Describe the motion and draw the forces

A rigid body is an idealized object whose parts keep the same distances from one another. Its planar motion can be described by the position of its mass center GG and its angular position. The mass center represents the body's translation in the equations; it need not be a contact point or a visible mark.
Start by naming the system: the body or bodies whose motion you are studying. State the observer and reference frame; in this lesson, use a stationary ground observer and fixed axes. Draw a free-body diagram of the isolated body, showing all external forces, such as weight, applied forces, and support reactions. Mark where forces act, because their locations determine moments.
Choose positive coordinate directions and a positive rotation direction. Counterclockwise is a common positive choice, but another choice works if used consistently. A force's moment about a point depends on its perpendicular distance from that point. If its line of action passes through the point, its moment there is zero.
rG/O=xGi+yGj\mathbf{r}_{G/O}=x_G\mathbf{i}+y_G\mathbf{j}
  • Identify the body, observer, reference frame, and axes.
  • Show external forces and their application points.
  • Choose positive rotation before assigning moment signs.

2. Apply the planar equations of motion

For translation, add the external force components. Newton's second law for a rigid body relates the resultant force to the mass times the acceleration of the mass center. Resolve the vector equation into components along your selected axes.
For rotation about the mass center, add the moments of the external forces about GG. The result equals the mass moment of inertia about the axis through GG perpendicular to the plane, multiplied by angular acceleration. This inertia measures how strongly the body's mass distribution resists angular acceleration.
The force and moment equations work together for planar motion. A pin, rolling contact, or other constraint may provide additional equations. If taking moments about a fixed pin, use the body's mass moment of inertia about that pin, not its inertia about GG.
∑F=maG,∑MG=IGα\sum \mathbf{F}=m\mathbf{a}_G,\qquad \sum M_G=I_G\alpha
  • Use force equations for mass-center translation.
  • Use moments about GG for angular acceleration, or use a suitable fixed point with its matching inertia.
  • Unknown reaction forces are included as external forces.

3. Add constraints, solve, and check

Write equations symbolically before substituting values. Include a kinematic constraint only when it is stated or follows from the contact geometry. For a rigid body rolling without slipping on a stationary straight surface, the contact point has zero instantaneous velocity relative to the surface. Along the surface, this gives a relation between the mass center's tangential acceleration and angular acceleration.
Use SI units consistently: mass in kilograms, length in metres, force in newtons, moment in newton-metres, and angular acceleration in radians per second squared. Radians are dimensionless in the equations, but writing them identifies an angular quantity. Check that both sides of a force equation have force units and both sides of a moment equation have moment units.
Interpret signs after solving. A negative acceleration means the actual direction is opposite the positive direction you chose. Check that the result agrees with the diagram and constraints. If the net force is zero, the mass center has zero acceleration; if the net moment about GG is zero, the angular acceleration is zero.
aG,t=Rαa_{G,t}=R\alpha
  • Use only justified motion constraints and keep their signs consistent.
  • Check units, directions, and limiting cases.
  • A negative result reports direction relative to your chosen positive direction.

Worked example

A bar released from a pin

A uniform slender bar of mass 3.0 kg3.0\,\mathrm{kg} and length 1.2 m1.2\,\mathrm{m} is horizontal and pinned at its left end. It is released from rest. Find its initial angular acceleration. Use g=9.81 m/s2g=9.81\,\mathrm{m/s^2}.
Bar and weight
Bar and weightbar directionupuniform barpinmgL

Weight acts at the bar's mass center, halfway from the pin.

  1. Set the system and signs
    Take the bar as the system, viewed from the stationary ground frame. Let counterclockwise be positive. At release, the bar has zero angular velocity, but it can have nonzero angular acceleration.
  2. Take moments about the pin
    The pin reactions have zero moment about the pin. The weight acts downward at distance L/2L/2, creating a clockwise, negative moment. For rotation about a fixed pin, use the bar's inertia about that pin.
    −mgL2=IPα,IP=13mL2-mg\frac{L}{2}=I_P\alpha,\qquad I_P=\frac{1}{3}mL^2
  3. Solve and interpret
    The negative result means the bar initially rotates clockwise. The mass cancels, as expected for a uniform bar under its own weight.
    α=−3g2L=−12.3 rad/s2\alpha=-\frac{3g}{2L}=-12.3\,\mathrm{rad/s^2}
Answer: The initial angular acceleration is 12.3 rad/s212.3\,\mathrm{rad/s^2} clockwise.
Check: The units of g/Lg/L are s−2\mathrm{s^{-2}}. The clockwise sign agrees with the downward weight acting to the right of the pin.

Worked example

A force applied off the mass center

A thin rectangular plate has mass 4.0 kg4.0\,\mathrm{kg}, width 0.60 m0.60\,\mathrm{m}, and height 0.40 m0.40\,\mathrm{m}. It moves in a horizontal plane on ideal guides that supply an upward force equal to its weight and prevent vertical motion. A horizontal 12 N12\,\mathrm{N} force to the right is applied at the midpoint of its lower edge. Find the mass-center acceleration and angular acceleration at that instant. The guides exert no moment about the mass center.
Plate with guide reactions
Plate with guide reactionsxyplate12 Nmgguide force

The horizontal force acts below the mass center; guide forces balance weight and create no moment about G.

  1. Define the system and axes
    Take the plate as the system in the stationary ground frame. Let +x+x point right, +y+y point up, and counterclockwise rotation be positive. The upward guide force balances weight, so the net vertical force is zero.
  2. Find translation
    The only unbalanced force is the 12 N12\,\mathrm{N} horizontal force. Apply the force equation to the mass center.
    aGx=124.0=3.0 m/s2,aGy=0a_{Gx}=\frac{12}{4.0}=3.0\,\mathrm{m/s^2},\qquad a_{Gy}=0
  3. Find rotation
    The horizontal force line is 0.20 m0.20\,\mathrm{m} below GG, so it creates a counterclockwise positive moment. For a thin rectangular plate, use its planar mass moment of inertia about GG.
    IG=m12(w2+h2)=0.173 kg m2,α=(12)(0.20)IG=13.8 rad/s2I_G=\frac{m}{12}(w^2+h^2)=0.173\,\mathrm{kg\,m^2},\qquad \alpha=\frac{(12)(0.20)}{I_G}=13.8\,\mathrm{rad/s^2}
Answer: The mass center accelerates at 3.0 m/s23.0\,\mathrm{m/s^2} to the right, and the plate's angular acceleration is 13.8 rad/s213.8\,\mathrm{rad/s^2} counterclockwise.
Check: The moment has units N m\mathrm{N\,m}, and dividing by IGI_G gives s−2\mathrm{s^{-2}}. A force applied through GG would produce the same translation but zero angular acceleration.

Worked example

A solid cylinder rolling down an incline

A uniform solid cylinder of mass 2.0 kg2.0\,\mathrm{kg} and radius 0.25 m0.25\,\mathrm{m} rolls without slipping down a fixed incline at 30∘30^\circ to the horizontal. Find its mass-center acceleration and the friction force. It is released from rest; use g=9.81 m/s2g=9.81\,\mathrm{m/s^2}.
Cylinder on an incline
Cylinder on an inclinecylindermgnormalfriction

The incline rises 30 degrees to the right. Normal force acts from the contact point toward the cylinder center; friction acts up the incline at contact.

  1. Choose incline-aligned coordinates
    Take the cylinder as the system in the ground frame. Let positive translation point down the incline and positive rotation be clockwise, the direction that accompanies rolling downhill. The initial speed is zero; find the accelerations and contact friction.
  2. Write translation and rotation equations
    Along the incline, gravity pulls downhill and static friction acts uphill. The normal force acts at the contact point along the radius toward the center, so its line of action passes through the center and it has no moment about GG. Use the solid-cylinder inertia and the no-slip acceleration relation.
    mgsin⁡30∘−f=ma,fR=IGα,a=Rα,IG=12mR2mg\sin 30^\circ-f=ma,\qquad fR=I_G\alpha,\qquad a=R\alpha,\qquad I_G=\frac{1}{2}mR^2
  3. Solve for acceleration and friction
    Substituting the rolling relation into the moment equation gives the friction needed for rotation. Combining it with translation yields the acceleration and then the friction magnitude.
    a=23gsin⁡30∘=3.27 m/s2,f=mgsin⁡30∘−ma=3.27 Na=\frac{2}{3}g\sin 30^\circ=3.27\,\mathrm{m/s^2},\qquad f=mg\sin 30^\circ-ma=3.27\,\mathrm{N}
Answer: The cylinder's mass center accelerates down the incline at 3.27 m/s23.27\,\mathrm{m/s^2}. Friction has magnitude 3.27 N3.27\,\mathrm{N} and acts up the incline.
Check: The acceleration is less than gsin⁡30∘=4.91 m/s2g\sin 30^\circ=4.91\,\mathrm{m/s^2} because some of gravity's effect produces rotation. The friction moment has the positive sign for the stated clockwise rotation.

Common mistakes and how to avoid them

Using the inertia about the mass center in a moment equation taken about a pin.
Correction: Match the inertia to the moment reference point. The fixed-pin bar equation uses IPI_P.
Treating a force applied away from the mass center as causing translation only.
Correction: A force with a nonzero moment about GG also causes angular acceleration.
Assuming friction always points opposite the body's motion.
Correction: Friction opposes relative slipping at contact. For the rolling cylinder here, it points up the incline and enables the required rotation.
Changing the positive rotation direction midway through a calculation.
Correction: Keep one sign convention throughout; interpret a negative result as a direction opposite the chosen positive direction.

Lesson summary

  • For planar rigid-body motion, net force determines mass-center acceleration and net moment determines angular acceleration.
  • Draw the body, external forces, force locations, axes, and positive rotation direction before writing equations.
  • Use an inertia value that matches the moment reference point and include only justified constraints.
  • Check units, signs, physical direction, and limiting cases.

Check your understanding

Question 1

A body's net moment about its mass center is zero. What does the planar moment equation imply about its angular acceleration?
  1. It is zero.
  2. It must equal the mass-center acceleration.
  3. It is always clockwise.
  4. It cannot be determined even if the inertia is known.
Show answer and explanation
It is zero.
Since IGI_G is positive, a zero net moment gives α=0\alpha=0.

Question 2

A horizontal force acts through a body's mass center. What is its moment about that center?
  1. Zero.
  2. The force magnitude multiplied by the body's mass.
  3. The force magnitude divided by the mass.
  4. Always counterclockwise.
Show answer and explanation
Zero.
The perpendicular distance from the mass center to the force's line of action is zero.

Question 3

For the rolling cylinder, what is the direction of static friction?
  1. Up the incline.
  2. Down the incline.
  3. Perpendicular to the incline and away from the surface.
  4. Zero, because the cylinder is rolling.
Show answer and explanation
Up the incline.
With positive translation downhill, uphill friction supplies the moment for the stated clockwise rotation.

Key terms

Mass center
The point used to represent a body's translation in the force equation.
Mass moment of inertia
A measure of a body's resistance to angular acceleration about a specified axis.
Angular acceleration
The rate of change of angular velocity, measured here in radians per second squared.
Rolling without slipping
Rolling contact with no relative motion between the contacting surfaces at the instant of contact.

Continue through EN PH 131

View the complete EN PH 131 University of Alberta EN PH 131: Engineering Mechanics: Dynamics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows University of Alberta EN PH 131: Engineering Mechanics: Dynamics, study topic 9.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question