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2.4 · Use radial and transverse components for planar motion

Learn to use radial and transverse components for planar motion through clear examples and targeted practice.

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Curvilinear and Relative Motion

Resolving velocity, acceleration, and net force along directions that rotate with a particle

When a particle’s position is naturally described by its distance and direction from a fixed point, radial and transverse components can make its motion easier to analyze. The radial direction points from the origin to the particle; the transverse direction is perpendicular to it and points toward increasing angle. These directions turn as the particle moves. That changing orientation is why the acceleration formulas include terms beyond the time derivatives of the distance and angle alone.

What you will learn

  • Define radial and transverse directions for planar motion.
  • Use polar-coordinate rates to find velocity and acceleration components.
  • Interpret component signs using the chosen positive directions.
  • Apply Newton’s second law in radial and transverse directions.

1. Define the position and moving directions

Choose the particle as the system and use a stationary observer in a fixed, inertial reference frame. An inertial frame is one in which Newton’s laws can be used in their usual form. Place the origin at a fixed point in the plane. Describe the particle’s position by its distance rr from the origin and its angle θ\theta, measured counterclockwise from a fixed positive horizontal axis.
The radial unit vector er\mathbf e_r points outward from the origin to the particle. The transverse unit vector eθ\mathbf e_\theta is perpendicular to er\mathbf e_r and points toward increasing θ\theta. These are the positive directions. Thus a negative radial component points inward, and a negative transverse component points toward decreasing angle.
Position is a vector from the origin. Path length is the distance travelled along the route, while displacement is the change in position. Velocity and acceleration are vectors too; their components are measured along the radial and transverse directions at the particle’s current position.
r=rer\mathbf r=r\mathbf e_r
  • Use a fixed origin and a stationary inertial frame.
  • Positive radial is outward; positive transverse is toward increasing angle.
  • The radial and transverse directions rotate as the particle moves.

2. Obtain velocity and acceleration

A dot denotes a time derivative. The rate r˙\dot r describes how quickly the distance from the origin changes, and θ˙\dot\theta describes how quickly the angle changes. Use radians for θ\theta and its rates. Because the unit directions rotate with the particle’s angle, differentiating position requires differentiating both the scalar distance and the direction vector.
For planar motion, the unit-vector rates are e˙r=θ˙eθ\dot{\mathbf e}_r=\dot\theta\mathbf e_\theta and e˙θ=−θ˙er\dot{\mathbf e}_\theta=-\dot\theta\mathbf e_r. Applying the product rule to position gives velocity; differentiating velocity gives acceleration. The term −rθ˙2-r\dot\theta^2 is radial and points inward when r>0r>0. The term 2r˙θ˙2\dot r\dot\theta contributes to transverse acceleration when both radial and angular motion occur.
The velocity components are vr=r˙v_r=\dot r and vθ=rθ˙v_\theta=r\dot\theta. The acceleration components are ar=r¨−rθ˙2a_r=\ddot r-r\dot\theta^2 and aθ=rθ¨+2r˙θ˙a_\theta=r\ddot\theta+2\dot r\dot\theta. With rr in metres and angular rates in radians per second, each velocity term has units of metres per second and each acceleration term has units of metres per second squared.
v=r˙er+rθ˙eθ,a=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ\mathbf v=\dot r\mathbf e_r+r\dot\theta\mathbf e_\theta,\qquad \mathbf a=(\ddot r-r\dot\theta^2)\mathbf e_r+(r\ddot\theta+2\dot r\dot\theta)\mathbf e_\theta
  • Differentiate the rotating unit directions as well as the scalar rates.
  • A negative radial acceleration component points toward the origin.
  • The transverse acceleration includes the contribution 2r˙θ˙2\dot r\dot\theta.

3. Relate acceleration to net force

When a problem asks for the net force, use Newton’s second law along the instantaneous radial and transverse directions. First draw a free-body diagram if individual forces are given or needed. Resolve the actual forces along the positive radial and transverse directions; then match each net-force component to the corresponding acceleration component.
Keep the signs from the chosen axes. A negative radial net force points inward, and a negative transverse net force points toward decreasing angle. The force components are not the same thing as the acceleration components: each is related to its matching acceleration by the particle’s mass.
∑Fr=m(r¨−rθ˙2),∑Fθ=m(rθ¨+2r˙θ˙)\sum F_r=m(\ddot r-r\dot\theta^2),\qquad \sum F_\theta=m(r\ddot\theta+2\dot r\dot\theta)
  • Resolve actual forces along the current radial and transverse directions.
  • Apply Newton’s second law separately in each direction.
  • Mass is measured in kilograms and force in newtons.

4. Choose signs and check special cases

Before calculating, state the particle being analysed, the observer and fixed origin, the instant of interest, and the known rates. Mark positive radial outward and positive transverse toward increasing angle. If forces are involved, make the diagram consistent with the particle’s position and these directions.
A useful check is motion at constant radius: r˙=0\dot r=0 and r¨=0\ddot r=0. Then radial acceleration is inward, while transverse acceleration depends on whether angular speed is changing. If angular speed is also constant, transverse acceleration is zero. These results check both the signs and the interpretation of the components.
Keep component calculations in the rotating directions unless the question asks for fixed horizontal and vertical components. Attach SI units to numerical results, state negative directions in words, and check that each term in a component equation has matching units.
r˙=0,  r¨=0  ⟹  ar=−rθ˙2,aθ=rθ¨\dot r=0,\;\ddot r=0\;\Longrightarrow\;a_r=-r\dot\theta^2,\quad a_\theta=r\ddot\theta
  • Write down the state at the instant being analysed before substituting.
  • Use a simple special case to check the acceleration signs.
  • Report negative components with their physical directions.

Worked example

Circular motion with changing angular speed

A particle moves on a circle of radius 2.00 m2.00\,\mathrm{m} about a fixed origin. At one instant its angular velocity is 3.00 rad/s3.00\,\mathrm{rad/s} counterclockwise and its angular acceleration is 0.500 rad/s20.500\,\mathrm{rad/s^2} counterclockwise. Find its velocity and acceleration components.
Circular particle motion
Circular particle motionfixed xfixed yr = 2.00 mparticlev

At the shown position, positive transverse is upward and positive radial is outward.

  1. Set the coordinates
    The system is the particle, observed from a stationary frame with the origin at the circle’s centre. Its radius is constant, so r˙=0\dot r=0 and r¨=0\ddot r=0. Positive angle is counterclockwise.
  2. Find velocity
    There is no radial velocity because the radius is constant. The positive transverse component points toward increasing angle.
    vr=0,vθ=rθ˙=(2.00 m)(3.00 rad/s)=6.00 m/sv_r=0,\qquad v_\theta=r\dot\theta=(2.00\,\mathrm{m})(3.00\,\mathrm{rad/s})=6.00\,\mathrm{m/s}
  3. Find acceleration
    The radial component points inward because the direction of travel changes around the circle. The positive transverse component reflects the increasing angular speed.
    ar=−rθ˙2=−18.0 m/s2,aθ=rθ¨=1.00 m/s2a_r=-r\dot\theta^2=-18.0\,\mathrm{m/s^2},\qquad a_\theta=r\ddot\theta=1.00\,\mathrm{m/s^2}
Answer: The velocity is 6.00 m/s6.00\,\mathrm{m/s} in the positive transverse direction. The acceleration is 18.0 m/s218.0\,\mathrm{m/s^2} inward radially and 1.00 m/s21.00\,\mathrm{m/s^2} in the positive transverse direction.
Check: The product rθ˙r\dot\theta has units of metres per second, and rθ˙2r\dot\theta^2 has units of metres per second squared. If angular acceleration were zero, the transverse acceleration would vanish, as expected for constant speed on a circle.

Worked example

A particle moving inward while turning

At an instant, a particle has r=0.800 mr=0.800\,\mathrm{m}, r˙=−1.20 m/s\dot r=-1.20\,\mathrm{m/s}, r¨=0.400 m/s2\ddot r=0.400\,\mathrm{m/s^2}, θ˙=2.00 rad/s\dot\theta=2.00\,\mathrm{rad/s}, and θ¨=−0.500 rad/s2\ddot\theta=-0.500\,\mathrm{rad/s^2}. Find its radial and transverse acceleration components.
  1. Define directions and known state
    The system is the particle in a stationary frame with a fixed origin. Positive radial is outward and positive transverse is counterclockwise. The negative r˙\dot r means the particle is moving inward; the negative θ¨\ddot\theta means angular acceleration toward decreasing angle.
  2. Calculate radial acceleration
    Use the radial acceleration relation. The angular-motion contribution is inward and combines algebraically with the given positive radial acceleration r¨\ddot r.
    ar=r¨−rθ˙2=0.400−(0.800)(2.00)2=−2.80 m/s2a_r=\ddot r-r\dot\theta^2=0.400-(0.800)(2.00)^2=-2.80\,\mathrm{m/s^2}
  3. Calculate transverse acceleration
    The angular-acceleration term is negative. The product 2r˙θ˙2\dot r\dot\theta is also negative because the radial rate is negative and the angular rate is positive.
    aθ=rθ¨+2r˙θ˙=(0.800)(−0.500)+2(−1.20)(2.00)=−5.20 m/s2a_\theta=r\ddot\theta+2\dot r\dot\theta=(0.800)(-0.500)+2(-1.20)(2.00)=-5.20\,\mathrm{m/s^2}
Answer: The radial acceleration is 2.80 m/s22.80\,\mathrm{m/s^2} inward. The transverse acceleration is 5.20 m/s25.20\,\mathrm{m/s^2} toward decreasing θ\theta.
Check: Every term has acceleration units. Both components are negative relative to their defined positive directions, matching the stated directions in the answer.

Worked example

Find net force components from the motion

A 1.50 kg1.50\,\mathrm{kg} particle is at r=2.00 mr=2.00\,\mathrm{m} with r˙=0.500 m/s\dot r=0.500\,\mathrm{m/s}, r¨=−0.200 m/s2\ddot r=-0.200\,\mathrm{m/s^2}, θ˙=1.00 rad/s\dot\theta=1.00\,\mathrm{rad/s}, and θ¨=0.600 rad/s2\ddot\theta=0.600\,\mathrm{rad/s^2}. Find the net force components. At the instant shown, the particle is on the positive horizontal ray from the origin.
Net force components
Net force componentspositive radialpositive transverseparticleFr = 3.30 NF_θ = 3.30 N

The net radial force is inward; the net transverse force is positive.

  1. Set the system and signs
    The system is the particle, observed in a stationary frame with the stated fixed origin. At the positive horizontal ray, positive radial points right and positive transverse points up. The requested quantities are net-force components.
  2. Calculate acceleration components
    Use the polar acceleration relations with the rates given at this instant. Keep the negative radial component as a signed value for the force calculation.
    ar=−0.200−(2.00)(1.00)2=−2.20 m/s2,aθ=(2.00)(0.600)+2(0.500)(1.00)=2.20 m/s2a_r=-0.200-(2.00)(1.00)^2=-2.20\,\mathrm{m/s^2},\qquad a_\theta=(2.00)(0.600)+2(0.500)(1.00)=2.20\,\mathrm{m/s^2}
  3. Apply Newton’s second law
    Multiply each acceleration component by the mass. The negative radial force points inward; the positive transverse force points upward at the stated position.
    Fr=(1.50)(−2.20)=−3.30 N,Fθ=(1.50)(2.20)=3.30 NF_r=(1.50)(-2.20)=-3.30\,\mathrm{N},\qquad F_\theta=(1.50)(2.20)=3.30\,\mathrm{N}
Answer: The net force is 3.30 N3.30\,\mathrm{N} inward radially and 3.30 N3.30\,\mathrm{N} in the positive transverse direction.
Check: The units are kg m/s2=N\mathrm{kg\,m/s^2}=\mathrm{N}. Both signs agree with the diagram: radial force is opposite positive radial, while transverse force is along positive transverse.

Common mistakes and how to avoid them

Using r¨\ddot r alone as radial acceleration.
Correction: Include the direction-change contribution: ar=r¨−rθ˙2a_r=\ddot r-r\dot\theta^2.
Forgetting that the radial and transverse directions rotate.
Correction: Use the full transverse component, including 2r˙θ˙2\dot r\dot\theta.
Treating positive radial as inward.
Correction: Positive radial is defined outward; a negative radial component points inward.
Using degrees per second in angular-rate formulas without converting.
Correction: Use radians for angular measures and rates in these kinematic relationships.

Lesson summary

  • Describe planar position using distance rr and angle θ\theta from a fixed origin.
  • Define positive radial outward and positive transverse toward increasing angle.
  • Velocity components are vr=r˙v_r=\dot r and vθ=rθ˙v_\theta=r\dot\theta.
  • Acceleration components are ar=r¨−rθ˙2a_r=\ddot r-r\dot\theta^2 and aθ=rθ¨+2r˙θ˙a_\theta=r\ddot\theta+2\dot r\dot\theta.
  • Apply Newton’s second law separately in the radial and transverse directions when finding net force.

Check your understanding

Question 1

A particle moves on a circle of fixed radius with constant angular speed. Which statement about its acceleration is correct?
  1. Its radial acceleration is inward and its transverse acceleration is zero.
  2. Both acceleration components are zero.
  3. Its radial acceleration is outward and its transverse acceleration is zero.
  4. Its radial acceleration is zero and its transverse acceleration is inward.
Show answer and explanation
Its radial acceleration is inward and its transverse acceleration is zero.
With constant radius and angular speed, ar=−rθ˙2a_r=-r\dot\theta^2 and aθ=0a_\theta=0. The negative radial sign means inward.

Question 2

At an instant, r˙<0\dot r<0 and θ˙>0\dot\theta>0. What is the sign of the contribution 2r˙θ˙2\dot r\dot\theta to transverse acceleration?
  1. Negative
  2. Positive
  3. Zero
  4. It must be inward radial
Show answer and explanation
Negative
A negative value multiplied by a positive value is negative, so this contribution points toward decreasing θ\theta.

Key terms

Radial direction
The direction from the fixed origin outward to the particle.
Transverse direction
The direction perpendicular to the radial direction, positive toward increasing angle.
Angular rate
The time rate of change of the particle’s angle, written θ˙\dot\theta.
Component
A signed part of a vector along a specified direction.

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