3.4 Apply force–acceleration equations in radial and transverse coordinates | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Particle Kinetics: Force and Acceleration

Applying Newton’s second law to a particle described by its distance and angle

When a particle moves along a curved path, its acceleration may be easier to describe relative to a point than relative to fixed horizontal and vertical axes. Radial and transverse coordinates do this using the particle’s distance from a chosen origin and its angle around that origin. In this lesson, the system is one particle, and the observer uses a fixed planar reference frame. The radial direction points outward from the origin; the transverse direction is perpendicular to it in the direction of increasing angle. Choose those directions before resolving forces. The method is Newton’s second law, applied to the two components of acceleration.

What you will learn

Describe a particle’s position using radial and transverse unit directions.
Use radial and transverse acceleration components, including the terms caused by changing direction.
Apply Newton’s second law separately in the radial and transverse directions.
Check component signs, units, and physical meaning.

1. Set up the coordinates and directions

Choose a fixed origin in the plane. The particle’s position is described by its radial distance

r

from the origin and its angular position

\theta

, measured from a chosen fixed reference line. The radial unit vector

\mathbf e_r

points outward from the origin. The transverse unit vector

\mathbf e_\theta

is perpendicular to

\mathbf e_r

and points in the direction that

\theta

increases.

These directions rotate as the particle moves. That is the key difference from fixed horizontal and vertical axes: even if the particle’s radial distance is constant, its radial and transverse directions can still change. Use radians when differentiating angular position with respect to time.

The particle’s position vector is

\mathbf r=r\mathbf e_r

. Differentiating it gives velocity and acceleration. Since the unit vectors rotate, their derivatives are

\dot{\mathbf e}_r=\dot\theta\mathbf e_\theta

and

\dot{\mathbf e}_\theta=-\dot\theta\mathbf e_r

.

\mathbf v=\dot r\mathbf e_r+r\dot\theta\mathbf e_\theta

State the origin, reference line, and direction of increasing angle.
Positive radial direction is outward; positive transverse direction follows increasing $\theta$ .
A component can be negative: that means it points opposite its chosen positive direction.

2. Acceleration and force equations

Differentiating velocity accounts for both changes in the component values and changes in the unit-vector directions. The resulting radial acceleration is

a_r=\ddot r-r\dot\theta^2

. The term

-r\dot\theta^2

points inward; it is present whenever the particle has angular motion, even when its angular speed is constant.

The transverse acceleration is

a_\theta=r\ddot\theta+2\dot r\dot\theta

. Its first term is associated with changing angular speed. Its second term matters when both radial motion and angular motion occur. Neither acceleration component should be omitted just because one coordinate seems momentarily steady.

Draw a free-body diagram of the particle and resolve each force along the local radial and transverse directions. Let

\sum F_r

and

\sum F_\theta

mean the signed sums of force components along the chosen positive directions. Newton’s second law then applies independently in each direction. These equations use the net force, not a single force unless it is the only force in that direction.

\sum F_r=m(\ddot r-r\dot\theta^2),\qquad \sum F_\theta=m(r\ddot\theta+2\dot r\dot\theta)

Radial acceleration includes the inward term from turning.
Transverse acceleration includes a term when both $\dot r$ and $\dot\theta$ are nonzero.
Use a free-body diagram to identify all forces before summing components.

3. A reliable solution sequence

First define the system as the particle and identify the fixed observer’s frame and the chosen origin. State the known position, velocity components, accelerations, and forces at the instant of interest. Distinguish the path length travelled from displacement: neither is automatically the same as the coordinate

r

.

Next draw the path if it is stated and useful, or draw a particle free-body diagram. Mark

\mathbf e_r

and

\mathbf e_\theta

at the particle’s current position. Resolve forces into those directions, keeping their signs consistent with the marked axes.

Choose force–acceleration equations when forces or acceleration components are required at a specified instant. Substitute the kinematic quantities at that instant, solve the two component equations, and combine components only if a resultant is requested. Do not use a constant-acceleration shortcut unless the acceleration is actually constant over the interval.

Check that each side of Newton’s second law has units of newtons. A negative component indicates a direction opposite the positive axis, not an error by itself. For circular motion with constant radius and constant angular speed, the equations reduce to inward radial acceleration and zero transverse acceleration.

1\,\mathrm{N}=1\,\mathrm{kg}\cdot\mathrm{m/s^2}

Evaluate all quantities at the same instant.
Resolve forces before applying the component equations.
Check signs, units, and simple limiting cases.

Worked example

Uniform circular motion

A

2.0\,\mathrm{kg}

particle moves counterclockwise at constant speed

6.0\,\mathrm{m/s}

on a circle of radius

4.0\,\mathrm{m}

. At the instant shown at the rightmost point, determine its acceleration and the net force. The system is the particle, observed from a fixed planar frame with the circle’s centre as origin.

Particle on a circle

At the rightmost point, inward is left and the counterclockwise velocity is upward.

Choose positive directions
At the rightmost point, outward radial is right and positive transverse is upward because the motion is counterclockwise. The radius is constant, so the radial speed and radial acceleration are zero. The angular speed is constant.
$r=4.0\,\mathrm{m},\quad \dot r=0,\quad \ddot r=0,\quad \dot\theta=\frac{v}{r}=1.50\,\mathrm{rad/s},\quad \ddot\theta=0$
Find acceleration components
The radial equation gives an inward acceleration, while the transverse equation gives zero. The negative radial sign means the acceleration is opposite outward positive.
$a_r=0-(4.0)(1.50)^2=-9.0\,\mathrm{m/s^2},\quad a_\theta=0$
Apply Newton’s second law
Multiply the acceleration components by the particle’s mass to obtain the net force components. The inward net force is the required centripetal resultant; it is not an additional force to add to the free-body diagram.
$\sum F_r=(2.0)(-9.0)=-18\,\mathrm{N},\quad \sum F_\theta=0$

Answer: The acceleration is

9.0\,\mathrm{m/s^2}

inward, and the net force is

18\,\mathrm{N}

inward.

Check: The result points toward the circle’s centre, as expected. The units are

\mathrm{kg}\cdot\mathrm{m/s^2}=\mathrm{N}

.

Worked example

Radial motion with angular motion

A

1.5\,\mathrm{kg}

particle is at

r=3.0\,\mathrm{m}

with

\dot r=1.0\,\mathrm{m/s}

outward,

\ddot r=0.50\,\mathrm{m/s^2}

,

\dot\theta=2.0\,\mathrm{rad/s}

, and

\ddot\theta=0

. Find the net force components. Use the origin as the fixed reference and positive directions outward and toward increasing angle.

Instantaneous force components

The arrows show the signs of the calculated net-force components.

Calculate the radial acceleration
Use the instantaneous radial coordinate and angular speed. The turning contribution is inward, so the result may be negative even though the radial acceleration itself is outward positive.
$a_r=0.50-(3.0)(2.0)^2=-11.5\,\mathrm{m/s^2}$
Calculate the transverse acceleration
Angular acceleration is zero, but the changing radius still contributes because the particle has both radial and angular velocity.
$a_\theta=(3.0)(0)+2(1.0)(2.0)=4.0\,\mathrm{m/s^2}$
Find the net force components
Apply Newton’s second law separately. A negative radial component points inward; a positive transverse component points toward increasing angle.
$\sum F_r=(1.5)(-11.5)=-17.25\,\mathrm{N},\quad \sum F_\theta=(1.5)(4.0)=6.0\,\mathrm{N}$

Answer: The net force components are

17.25\,\mathrm{N}

inward and

6.0\,\mathrm{N}

in the positive transverse direction.

Check: Each component has force units. The positive transverse acceleration comes from outward radial motion combined with positive angular speed.

Worked example

Force components from given motion

At one instant, a

2.0\,\mathrm{kg}

particle has

r=1.5\,\mathrm{m}

,

\dot r=-0.40\,\mathrm{m/s}

,

\ddot r=0.60\,\mathrm{m/s^2}

,

\dot\theta=2.0\,\mathrm{rad/s}

, and

\ddot\theta=1.2\,\mathrm{rad/s^2}

. Determine the net force components using outward radial and increasing-angle transverse directions.

Net force at an instant

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

Evaluate radial acceleration
The particle is moving inward radially, but the radial acceleration equation depends on radial acceleration and the turning term, not on the sign of radial velocity alone.
$a_r=0.60-(1.5)(2.0)^2=-5.4\,\mathrm{m/s^2}$
Evaluate transverse acceleration
The angular acceleration contributes positively. Since radial velocity is negative, the mixed term reduces the transverse result.
$a_\theta=(1.5)(1.2)+2(-0.40)(2.0)=0.20\,\mathrm{m/s^2}$
Convert acceleration to net force
Multiply both signed components by the same mass. The signs give the directions relative to the selected axes.
$\sum F_r=(2.0)(-5.4)=-10.8\,\mathrm{N},\quad \sum F_\theta=(2.0)(0.20)=0.40\,\mathrm{N}$

Answer: The net force is

10.8\,\mathrm{N}

inward radially and

0.40\,\mathrm{N}

in the positive transverse direction.

Check: Both components have units of newtons. The force directions agree with the signs of the calculated accelerations.

Common mistakes and how to avoid them

Writing radial acceleration as only

\ddot r

.

Correction: Include the inward turning term

-r\dot\theta^2

whenever angular speed is nonzero.

Dropping the transverse term

2\dot r\dot\theta

because angular acceleration is zero.

Correction: That term can remain nonzero when both radial and angular velocities are nonzero.

Treating a negative component as an invalid answer.

Correction: A negative sign means the component points opposite the selected positive direction.

Using the same radial and transverse directions at every point on the path.

Correction: Redraw the local directions at the particle’s current position; they rotate as the particle moves.

Lesson summary

Describe position with radial distance and angle from a fixed origin.
Use $a_r=\ddot r-r\dot\theta^2$ and $a_\theta=r\ddot\theta+2\dot r\dot\theta$ .
Resolve the net force along the local directions and apply Newton’s second law component by component.
Check that signs match the chosen directions and that force units are newtons.

Check your understanding

Question 1

A particle moves on a circle of fixed radius at constant angular speed. Which acceleration components are nonzero?

Only the radial component, directed inward
Only the transverse component, directed forward
Both components, each directed outward
Neither component

Show answer and explanation

Only the radial component, directed inward

With constant radius and angular speed,

a_r=-r\dot\theta^2

and

a_\theta=0

.

Question 2

At an instant,

\dot r=0

,

\dot\theta=0

,

\ddot r=2.0\,\mathrm{m/s^2}

, and

\ddot\theta=1.0\,\mathrm{rad/s^2}

. What are the acceleration components?

$a_r=2.0\,\mathrm{m/s^2}$ and $a_\theta=r(1.0\,\mathrm{rad/s^2})$
$a_r=0$ and $a_\theta=0$
$a_r=-r(1.0\,\mathrm{rad/s^2})^2$ and $a_\theta=2.0\,\mathrm{m/s^2}$
$a_r=2.0\,\mathrm{m/s^2}$ and $a_\theta=0$

Show answer and explanation

a_r=2.0\,\mathrm{m/s^2}

and

a_\theta=r(1.0\,\mathrm{rad/s^2})

At this instant the angular speed terms vanish, leaving

a_r=\ddot r

and

a_\theta=r\ddot\theta

.

Question 3

If the computed radial force component is negative while outward is positive, what does the sign mean?

The force points inward.
The force is zero.
The particle must be moving inward.
The chosen coordinate system is invalid.

Show answer and explanation

The force points inward.

The sign reports direction relative to the chosen positive radial direction; it does not, by itself, determine the particle’s velocity.

Key terms

Radial direction: The direction from the chosen origin outward to the particle.
Transverse direction: The direction perpendicular to radial, positive in the direction of increasing angle.
Angular speed: The rate of change of angular position with time, represented by $\dot\theta$ .
Net force: The vector sum of all forces acting on the particle.

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 3.4. 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

3.4 · Apply force–acceleration equations in radial and transverse coordinates

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Applying Newton’s second law to a particle described by its distance and angle

What you will learn

1. Set up the coordinates and directions

2. Acceleration and force equations

3. A reliable solution sequence

Uniform circular motion

Radial motion with angular motion

Force components from given motion

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review