3.3 Apply force–acceleration equations in normal and tangential coordinates | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Particle Kinetics: Force and Acceleration

Resolve particle motion along its path and toward the path’s centre of curvature

Newton’s second law is often easiest to apply in axes that move with a particle’s path. The tangential axis follows the direction of motion, while the normal axis points toward the centre of curvature. These directions separate two effects: changing speed and changing direction. In this lesson, the system is one particle, and the observer uses a fixed, inertial reference frame. The path-based axes are attached to the particle only to make force resolution convenient; they do not change the underlying law of motion.

What you will learn

Define tangential and normal directions for a particle moving along a curved path.
Use the force–acceleration equations in tangential and normal coordinates.
Relate tangential acceleration to speed change and normal acceleration to path curvature.
Solve particle problems with consistent signs, directions, and SI units.

1. Define the path-based coordinates

At a chosen instant, let the particle have position measured from a fixed origin and velocity tangent to its path. The positive tangential direction,

t

, is the direction of velocity. The positive normal direction,

n

, points inward toward the centre of curvature. For a straight path, curvature is zero and the radius of curvature is effectively infinite.

Speed is the magnitude of velocity; it is not the same as position, path length, or displacement. Tangential acceleration changes speed. Normal acceleration changes the direction of velocity, even if the speed stays constant. With

v

denoting speed and

\rho

the local radius of curvature, the acceleration components are as shown below.

Use these directions consistently when resolving forces. A force component opposite the positive tangent is negative. A force component outward, away from the centre of curvature, is negative in the normal direction. The normal direction is not automatically upward or downward: it depends on the path at the instant being studied.

\mathbf a=\dot v\,\mathbf e_t+\frac{v^2}{\rho}\,\mathbf e_n

System: the particle being analysed; frame: a fixed inertial frame.
Positive $t$ : along velocity. Positive $n$ : toward the centre of curvature.
Tangential acceleration changes speed; normal acceleration changes direction.

2. Apply Newton’s second law along each axis

Draw a free-body diagram of the particle and include every force acting on it. Then identify the local tangent and inward normal before resolving the forces. The sum of the force components along each direction equals mass times the corresponding acceleration component. These are scalar equations in the selected directions, not separate laws.

The tangential equation determines how quickly speed changes. The normal equation determines the inward acceleration needed for the path’s curvature. Normal acceleration is nonnegative with the chosen inward direction; the signed normal force sum must therefore equal this inward requirement. If a calculated force component is negative, its actual direction is opposite the assumed positive direction.

Knowns and unknowns vary by problem. You may know speed and curvature and seek a force, or know forces and seek acceleration. Choose the coordinates because they match the path geometry, not because they make every force point along an axis. Forces that do not align with an axis must be resolved with trigonometry.

\sum F_t=m\dot v,\qquad \sum F_n=m\frac{v^2}{\rho}

Resolve forces into the local $t$ and inward $n$ directions.
Use the tangential equation for speed change and the normal equation for curvature.
Keep force signs consistent with the chosen positive directions.

3. A reliable solution and checking process

First state the particle, observer, frame, and instant being analysed. Sketch the stated path when curvature or direction matters, and make a free-body diagram when forces are involved. Record given quantities and the requested result, including initial conditions if the question gives them. Then choose positive tangent and inward normal directions.

Write the force–acceleration equations before substituting numbers. This keeps the geometry and signs visible and makes it easier to solve symbolically. Use SI units: mass in kilograms, speed in metres per second, radius in metres, and force in newtons. Acceleration components must have units of metres per second squared.

Check direction as well as magnitude. If speed is constant, the tangential acceleration is zero, but the normal acceleration may not be. If the path becomes straight, the normal acceleration tends to zero. If the speed is zero at an instant on a smooth path, the normal acceleration is zero at that instant. These checks help expose sign, unit, and setup errors.

[v^2/\rho]=\mathrm{m/s^2}

Write symbolic component equations before inserting numerical values.
The units of $v^2/\rho$ are acceleration units.
Constant speed does not imply zero acceleration on a curved path.

Worked example

1. Given force components, find acceleration

A particle of mass

5.0\,\mathrm{kg}

moves on a circular path of radius

12\,\mathrm{m}

. At one instant its speed is

6.0\,\mathrm{m/s}

. The net force has a tangential component of

10\,\mathrm{N}

in the direction of motion and an inward normal component of

15\,\mathrm{N}

. Find both acceleration components and the acceleration magnitude.

Local force components

The local axes are tangent and inward normal at the instant shown.

Set the frame and directions
Treat the particle as the system and use a fixed inertial observer. At the instant considered, positive tangent is along its velocity and positive normal is inward. The stated force components are already resolved in these coordinates.
Apply the component equations
The tangential force controls the rate of change of speed. The normal force supplies the inward acceleration needed for the curved path.
$a_t=\frac{10}{5.0},\qquad a_n=\frac{15}{5.0}$
Calculate the magnitude
The two acceleration components are perpendicular, so their magnitudes combine by the Pythagorean theorem.
$a=\sqrt{a_t^2+a_n^2}=\sqrt{2.0^2+3.0^2}=3.61\,\mathrm{m/s^2}$

Answer: The tangential acceleration is

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

along the motion, the inward normal acceleration is

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

, and the acceleration magnitude is

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

.

Check: The normal result also matches

v^2/\rho=(6.0)^2/12=3.0\,\mathrm{m/s^2}

. Both force-to-mass calculations have acceleration units.

Worked example

2. Normal force at the top of a circular curve

A

1200\,\mathrm{kg}

car travels over the top of a circular road crest of radius

50\,\mathrm{m}

at

15\,\mathrm{m/s}

. Find the road’s normal force on the car at the top, assuming contact is maintained.

Forces at the crest

At the crest the inward normal direction is downward; weight is inward and road force is outward.

Choose the inward direction
The system is the car, viewed from a fixed inertial frame. At the crest, the centre of curvature is below the car, so choose downward as positive normal. The car’s speed is constant at this instant for the stated calculation; no tangential force information is needed to find the normal force.
Resolve forces inward
Weight acts downward and the road’s normal force acts upward. Thus weight is positive and the road force is negative in the inward direction.
$mg-N=m\frac{v^2}{\rho}$
Solve and substitute
Rearrange for the road force, then use $g=9.81\,\mathrm{m/s^2}$ . The mass cancels from the speed-curvature term only after the equation is set up.
$N=1200\left(9.81-\frac{15^2}{50}\right)=6372\,\mathrm{N}$

Answer: The road’s normal force is approximately

6.37\times10^3\,\mathrm{N}

upward.

Check: The result is positive, so the assumed upward contact force is possible. It is less than the car’s weight, as expected when gravity supplies part of the required downward acceleration.

Worked example

3. Tension and tangential acceleration of a swinging particle

A

0.50\,\mathrm{kg}

particle is attached to a light string of length

2.0\,\mathrm{m}

and moves in a vertical circular arc. At an instant when the string is

30^\circ

from the downward vertical, the particle’s speed is

4.0\,\mathrm{m/s}

. Find the string tension and the tangential acceleration. Assume the string remains taut.

Particle on a circular arc

The string pulls toward the centre; gravity has inward and tangential components.

Define local directions
The particle is the system, and the observer uses a fixed inertial frame. Choose inward along the string as positive normal. Choose the positive tangent toward the bottom of the swing. At the stated position, the inward component of weight is $mg\cos 30^\circ$ and its tangential component toward the bottom is $mg\sin 30^\circ$ .
Find the normal force balance
Tension acts inward. The inward component of weight also acts inward, so both contribute to the required normal acceleration. Solve for tension using the given instantaneous speed and radius.
$T+mg\cos 30^\circ=m\frac{v^2}{L}$
Find tension and tangential acceleration
Substitute the values for tension. Along the tangent, only gravity has a component, so the tangential acceleration follows from the tangential force equation.
$T=0.50\left(\frac{4.0^2}{2.0}-9.81\cos30^\circ\right)= -0.248\,\mathrm{N}$

Answer: The stated data are inconsistent with a taut string: the normal equation gives a negative tension, which a string cannot provide. The tangential acceleration is

9.81\sin30^\circ=4.91\,\mathrm{m/s^2}

toward the bottom of the swing.

Check: A negative calculated tension signals that the assumed constrained circular motion cannot be maintained at that instant. This is a useful physical check, not a negative physical tension.

Common mistakes and how to avoid them

Taking the normal direction as upward in every problem.

Correction: Define positive normal toward the centre of curvature at the instant being analysed.

Setting acceleration to zero whenever speed is constant.

Correction: Constant speed makes tangential acceleration zero, but curved motion still has inward normal acceleration.

Using

v^2/\rho

as the tangential acceleration.

Correction: The normal acceleration is

v^2/\rho

; tangential acceleration is the rate of change of speed.

Accepting a negative string tension as a real force.

Correction: A negative result means the assumed taut-string motion is not physically possible under those conditions.

Lesson summary

Choose positive tangent along the velocity and positive normal inward.
Use $\sum F_t=m\dot v$ and $\sum F_n=mv^2/\rho$ .
Resolve each force according to the local path directions and check signs, units, and physical feasibility.

Check your understanding

Question 1

A particle moves at constant speed around a circle. Which statement is correct?

Both tangential and normal acceleration are zero.
Tangential acceleration is zero, but normal acceleration is inward.
Tangential acceleration is inward, but normal acceleration is zero.
Both acceleration components point along the motion.

Show answer and explanation

Tangential acceleration is zero, but normal acceleration is inward.

Constant speed means no change in speed, so tangential acceleration is zero. The velocity direction changes, requiring inward acceleration

v^2/\rho

.

Question 2

At a point on a curved path, which direction is positive normal under the convention in this lesson?

Always vertically upward.
Always opposite the velocity.
Toward the centre of curvature.
Along the tangential direction.

Show answer and explanation

Toward the centre of curvature.

Positive normal is defined inward toward the local centre of curvature; its orientation depends on the path.

Key terms

Tangential direction: The direction along the path, chosen positive in the direction of the particle’s velocity.
Normal direction: The direction perpendicular to the tangent and chosen inward toward the centre of curvature.
Radius of curvature: The radius of the circle that locally matches the curve at the particle’s position.
Tangential acceleration: The acceleration component that changes the particle’s speed.
Normal acceleration: The inward acceleration component that changes the direction of velocity.

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3.3 · Apply force–acceleration equations in normal and tangential coordinates

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Resolve particle motion along its path and toward the path’s centre of curvature

What you will learn

1. Define the path-based coordinates

2. Apply Newton’s second law along each axis

3. A reliable solution and checking process

1. Given force components, find acceleration

2. Normal force at the top of a circular curve

3. Tension and tangential acceleration of a swinging particle

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through EN PH 131

About this lesson and its review