Normal and Tangential Kinetics for Planar Problems

Dynamics · FE Reference Handbook section

Dynamics

2 formulas

10 exam-style examples

~49 min

All Dynamics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Normal and Tangential Kinetics for Planar Problems within Dynamics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what normal and tangential kinetics for planar problems describes physically and when it applies.
State every one of the 2 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.

Lecture

Why this section exists. Normal and Tangential Kinetics for Planar Problems is the part of Dynamics that lets you connect a particle or rigid body moving under known forces to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as kinematics first, then a work-energy or impulse-momentum shortcut. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: normal and tangential kinetics for planar problems.
Capstone Studio instructional photograph

Dynamics — Normal and Tangential Kinetics for Planar Problems: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a particle or rigid body moving under known forces. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 2 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

R	Quantity produced by "R=Ft ma=t mdvt /dt" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

When working with normal and tangential directions, the scalar equations may be written as
Fn ma n m `v t2 /t j

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Work–energy: force needed to stop a mass — Normal and Tangential Kinetics for Planar Problems

A 2448 lb vehicle travelling 58 ft/s must stop in 133 ft. What constant resisting force is required?

Given

W = 2448 lb
v₁ = 58 ft/s
d = 133 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 2448/32.2 = 76.02 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(76.02)(58²) = 127,874 ft·lb
Work–energy — F·d = KE
Substituting
$F = 127, 874/133 = 961.5 l b$

Answer: F ≈ 961.5 lb

Why the other options are there

30,959 lb (weight used as mass)
127,874 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

Example 2

Normal and tangential components on a circular path — Normal and Tangential Kinetics for Planar Problems

A car travels a curve of radius 130.0 m at 8 m/s while speeding up at 0.5 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

ρ = 130.0 m
v = 8 m/s
a_t = 0.5 m/s²

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_{n} = v^{2} / ρ$
Substituting
$a_{n} = 8^{2} /130.0 = 0.492 m / s^{2}$
Formula
$a = (a_{t}^{2} + a_{n}^{2})$
Substituting
$a = (0. 5^{2} + 0.49 2^{2}) = 0.702 m / s^{2}$
Formula
$ω = v / ρ$
Substituting
$ω = 8/130.0 = 0.0615 r a d / s$

Answer: a_n = 0.49 m/s², total a = 0.70 m/s², ω = 0.062 rad/s

Why the other options are there

0.99 m/s² (components added directly)
8,320 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

Example 3

Work–energy: force needed to stop a mass — Normal and Tangential Kinetics for Planar Problems (2)

A 4806 lb vehicle travelling 53 ft/s must stop in 162 ft. What constant resisting force is required?

Given

W = 4806 lb
v₁ = 53 ft/s
d = 162 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 4806/32.2 = 149.3 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(149.3)(53²) = 209,628 ft·lb
Work–energy — F·d = KE
Substituting
$F = 209, 628/162 = 1, 294 l b$

Answer: F ≈ 1,294 lb

Why the other options are there

41,667 lb (weight used as mass)
209,628 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

Example 4

Normal and tangential components on a circular path — Normal and Tangential Kinetics for Planar Problems (2)

A car travels a curve of radius 155.0 m at 16 m/s while speeding up at 2.0 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

ρ = 155.0 m
v = 16 m/s
a_t = 2.0 m/s²

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_{n} = v^{2} / ρ$
Substituting
$a_{n} = 1 6^{2} /155.0 = 1.652 m / s^{2}$
Formula
$a = (a_{t}^{2} + a_{n}^{2})$
Substituting
$a = (2. 0^{2} + 1.65 2^{2}) = 2.594 m / s^{2}$
Formula
$ω = v / ρ$
Substituting
$ω = 16/155.0 = 0.1032 r a d / s$

Answer: a_n = 1.65 m/s², total a = 2.59 m/s², ω = 0.103 rad/s

Why the other options are there

3.65 m/s² (components added directly)
39,680 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

Example 5

Work–energy: force needed to stop a mass — Normal and Tangential Kinetics for Planar Problems (3)

A 5872 lb vehicle travelling 60 ft/s must stop in 132 ft. What constant resisting force is required?

Given

W = 5872 lb
v₁ = 60 ft/s
d = 132 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 5872/32.2 = 182.4 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(182.4)(60²) = 328,248 ft·lb
Work–energy — F·d = KE
Substituting
$F = 328, 248/132 = 2, 487 l b$

Answer: F ≈ 2,487 lb

Why the other options are there

80,073 lb (weight used as mass)
328,248 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

Example 6

Normal and tangential components on a circular path — Normal and Tangential Kinetics for Planar Problems (3)

A car travels a curve of radius 95 m at 25 m/s while speeding up at 2.0 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

ρ = 95 m
v = 25 m/s
a_t = 2.0 m/s²

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_{n} = v^{2} / ρ$
Substituting
$a_{n} = 2 5^{2} /95 = 6.579 m / s^{2}$
Formula
$a = (a_{t}^{2} + a_{n}^{2})$
Substituting
$a = (2. 0^{2} + 6.57 9^{2}) = 6.876 m / s^{2}$
Formula
$ω = v / ρ$
Substituting
$ω = 25/95 = 0.2632 r a d / s$

Answer: a_n = 6.58 m/s², total a = 6.88 m/s², ω = 0.263 rad/s

Why the other options are there

8.58 m/s² (components added directly)
59,375 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

Example 7

Work–energy: force needed to stop a mass — Normal and Tangential Kinetics for Planar Problems (4)

A 3227 lb vehicle travelling 70 ft/s must stop in 218 ft. What constant resisting force is required?

Given

W = 3227 lb
v₁ = 70 ft/s
d = 218 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 3227/32.2 = 100.2 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(100.2)(70²) = 245,533 ft·lb
Work–energy — F·d = KE
Substituting
$F = 245, 533/218 = 1, 126 l b$

Answer: F ≈ 1,126 lb

Why the other options are there

36,267 lb (weight used as mass)
245,533 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

Example 8

Normal and tangential components on a circular path — Normal and Tangential Kinetics for Planar Problems (4)

A car travels a curve of radius 130.0 m at 13 m/s while speeding up at 2.5 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

ρ = 130.0 m
v = 13 m/s
a_t = 2.5 m/s²

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_{n} = v^{2} / ρ$
Substituting
$a_{n} = 1 3^{2} /130.0 = 1.300 m / s^{2}$
Formula
$a = (a_{t}^{2} + a_{n}^{2})$
Substituting
$a = (2. 5^{2} + 1.30 0^{2}) = 2.818 m / s^{2}$
Formula
$ω = v / ρ$
Substituting
$ω = 13/130.0 = 0.1000 r a d / s$

Answer: a_n = 1.30 m/s², total a = 2.82 m/s², ω = 0.100 rad/s

Why the other options are there

3.80 m/s² (components added directly)
21,970 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

Example 9

Work–energy: force needed to stop a mass — Normal and Tangential Kinetics for Planar Problems (5)

A 3765 lb vehicle travelling 60 ft/s must stop in 212 ft. What constant resisting force is required?

Given

W = 3765 lb
v₁ = 60 ft/s
d = 212 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 3765/32.2 = 116.9 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(116.9)(60²) = 210,466 ft·lb
Work–energy — F·d = KE
Substituting
$F = 210, 466/212 = 992.8 l b$

Answer: F ≈ 992.8 lb

Why the other options are there

31,967 lb (weight used as mass)
210,466 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

Example 10

Normal and tangential components on a circular path — Normal and Tangential Kinetics for Planar Problems (5)

A car travels a curve of radius 160.0 m at 31 m/s while speeding up at 1.5 m/s². Resolve the acceleration into normal and tangential components, find the total acceleration, and compute the angular velocity of the radius vector.

Given

ρ = 160.0 m
v = 31 m/s
a_t = 1.5 m/s²

Find

a_n, total a, and ω

Start with the thinking

The tangential component changes speed; the normal component changes direction and always points to the centre.
Total acceleration is the vector sum of the two perpendicular components.

Step-by-step solution

Formula
$a_{n} = v^{2} / ρ$
Substituting
$a_{n} = 3 1^{2} /160.0 = 6.006 m / s^{2}$
Formula
$a = (a_{t}^{2} + a_{n}^{2})$
Substituting
$a = (1. 5^{2} + 6.00 6^{2}) = 6.191 m / s^{2}$
Formula
$ω = v / ρ$
Substituting
$ω = 31/160.0 = 0.1938 r a d / s$

Answer: a_n = 6.01 m/s², total a = 6.19 m/s², ω = 0.194 rad/s

Why the other options are there

7.51 m/s² (components added directly)
153,760 m/s² (multiplied by radius)

Reference: FE Reference Handbook — Dynamics → Normal and Tangential Kinetics for Planar Problems

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a particle or rigid body moving under known forces, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Normal and Tangential Kinetics for Planar Problems contains 2 relations; you must be able to find this page in under 15 seconds.
Exam style: kinematics first, then a work-energy or impulse-momentum shortcut.
Unit rule: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.