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0:007 lessons across the whole video1:08:21

You will be able to

State Newton's first, second, and third laws in your own words
Apply F = ma to find the net force or acceleration of an object
Distinguish mass from weight and convert between them
Draw free-body diagrams that show all forces acting on an object
Solve basic problems involving forces on objects at rest and in motion

Good to know first

Basic algebra, including solving linear equations
Definitions of speed, velocity, and acceleration
Basic vector notation and adding forces in one dimension

Syllabus

7 lessons

1
Introduction: uniform circular motion and why it needs a force Free
0:00–9:009 min of video
This lesson introduces the idea that a force changes motion, and shows with uniform circular motion that an object moving in a circle at constant speed is always accelerating toward the centre. We write its position as a vector, differentiate it twice, and find the acceleration points inward with size $ω^{2} r$ .
4 formulas2 exercises3 questions
2
Newton's first law and inertia Beyond Pass
9:00–19:0010 min of video
This lesson explains Newton's first law: an object keeps its state of motion (at rest or moving in a straight line at constant speed) unless a net force acts on it. You will learn what inertia means and how to recognise it in everyday situations.
2 exercises3 questions
3
Newton's second law: F = ma Beyond Pass
19:00–30:0011 min of video
This lesson shows how Newton's second law links the net force on an object to its mass and acceleration, written F = ma. You will learn what each symbol means, how to use the law with units, and how to solve a simple force problem.
2 formulas3 exercises4 questions
4
Mass versus weight and units of force Beyond Pass
30:00–38:008 min of video
This lesson shows that mass measures how much matter an object contains and how hard it is to accelerate, while weight is the force of gravity on that mass. It also introduces the newton as the SI unit of force, using F = ma with kilograms, metres and seconds.
4 formulas3 exercises3 questions
5
Newton's third law: action and reaction Beyond Pass
38:00–47:009 min of video
This lesson explains Newton's third law: forces always come in pairs, and the two forces in a pair are equal in size and opposite in direction. It also shows how the lecture links force, mass and acceleration through F = ma and the idea of cause and effect.
3 formulas4 exercises4 questions
6
Free-body diagrams and worked examples Beyond Pass
47:00–58:0011 min of video
You learn to isolate one object, draw every force acting on it, and write Newton's second law for that object. The lesson then applies this to a two-block push, where the blocks must be treated separately.
4 formulas3 exercises3 questions
7
Applications and closing questions: pulling blocks, springs and elevators Beyond Pass
58:00–1:08:2110 min of video
This lesson shows how Newton's second law solves real problems: two blocks joined by a massless string, a spring hanging under gravity, and a person standing in an elevator. You will learn to draw a force diagram for each object and write one equation per object.
3 exercises3 questions

Free preview · Lesson 1 in full

Introduction: uniform circular motion and why it needs a force

This lesson introduces the idea that a force changes motion, and shows with uniform circular motion that an object moving in a circle at constant speed is always accelerating toward the centre. We write its position as a vector, differentiate it twice, and find the acceleration points inward with size $ω^{2} r$ .

0:00 to 9:00 in the video

Scratchpad

Draw and write next to the video.

The idea

The chapter opens the course on Newton's laws with one question: what do forces do to motion? A force is a push or a pull, and Newton's laws describe exactly how it changes the way an object moves. To describe motion precisely, we first need a clear way to write down where an object is and how it changes.

The lecturer uses a simple, important example: an object moving in a circle at constant speed, like a ball on a string being whirled around, or a point on a spinning wheel. Its position can be written as a vector that has two parts, one along the x direction and one along the y direction. Using $x (t)$ and $y (t)$ for the coordinates, the position is $\vec{r} = \hat{i} x (t) + \hat{j} y (t)$ . For a circle of radius r turning at angular speed $ω$ , this becomes $\vec{r} = \hat{i} r \cos ω t + \hat{j} r \sin ω t$ .

The period T is the time for one full turn. A full turn is an angle of $2 π$ radians, so $ω T = 2 π$ , which gives $ω = 2 π / T$ . Since the frequency f is one over the period, $ω = 2 π f$ . The angular speed tells us how fast the angle is growing.

Next comes the key step. Velocity is the rate of change of position, and acceleration is the rate of change of velocity. Differentiating the position twice with respect to time gives $\vec{a} = \frac{d^{2} \vec{r}}{d t^{2}} = - ω^{2} \vec{r}$ . The acceleration is not zero even though the speed is constant, because the direction of the velocity keeps turning.

The minus sign tells us the direction. The acceleration always points opposite to the position vector, that is, toward the centre of the circle. We call it centripetal acceleration, meaning 'centre-seeking'. Its size is $a = ω^{2} r$ , and writing the unit vector ${\hat{e}}_{r}$ pointing outward from the centre, $\vec{a} = - \frac{v^{2}}{r} {\hat{e}}_{r}$ .

The lesson for Newton's laws: a moving object that goes in a circle at constant speed is accelerating all the time, so something must be pulling it toward the centre. That pull is a force. Forces are what cause changes in motion, and this example shows the link between a changing direction and an inward force that we will study next.

Key points

Position on a circle can be written as a vector with x and y parts that change with time. 4:30
One full turn is 2π radians, so ω = 2π/T = 2πf. 5:11
Acceleration is the second derivative of position: a = d²r/dt² = −ω²r. 7:00
The acceleration points toward the centre, opposite to r, and its size is ω²r = v²/r. 8:30
Constant speed does not mean zero acceleration: a changing direction is still an acceleration. 7:11

Formulas

\vec{r} = \hat{i} r \cos ω t + \hat{j} r \sin ω t

The position on a circle of radius r is a cosine part along x plus a sine part along y, both changing with time.

$\vec{r}$: position vector of the object
$r$: radius of the circle
$ω$: angular speed (how fast the angle grows)
$t$: time
$\hat{i}, \hat{j}$: unit vectors along the x and y directions

ω = \frac{2 π}{T} = 2 π f

Angular speed equals two pi divided by the period, which equals two pi times the frequency.

$T$: period, the time for one full turn
$f$: frequency, turns per second

\vec{a} = \frac{d^{2} \vec{r}}{d t^{2}} = - ω^{2} \vec{r}

Acceleration is the second time derivative of position, and it equals minus omega squared times the position vector.

$\vec{a}$: acceleration vector
$\frac{d^{2} \vec{r}}{d t^{2}}$: second derivative of position with respect to time

a = ω^{2} r = \frac{v^{2}}{r}

The size of the acceleration is omega squared times the radius, which equals speed squared divided by the radius.

$a$: magnitude of the acceleration
$v$: speed of the object

Practice

Calculate

A ball moves in a circle of radius 0.5 m with period T = 2 s. Find its angular speed $ω$ in rad/s.

Hint

Angular speed is two pi divided by the period.

Solution

Use $ω = 2 π / T$ .
$ω = 2 π / 2 = π$ rad/s, which is about 3.14 rad/s.

3.14 rad/s

Calculate

The same ball (radius 0.5 m, period 2 s) moves at constant speed. Find the size of its acceleration $a = ω^{2} r$ in m/s².

Hint

Square omega, then multiply by the radius.

Solution

$ω = π$ rad/s from the previous step.
$ω^{2} = π^{2} \approx 9.87$ .
$a = 9.87 \times 0.5 \approx 4.93$ m/s².

4.93 m/s^2

On your scratchpad

Draw a circle with a dot on it. Draw the velocity arrow tangent to the circle at the dot, then draw the acceleration arrow pointing to the centre. Label which way the force must point.

Quiz

An object moves in a circle at constant speed. What is true of its acceleration?
1. It is zero, because the speed is constant
2. It is not zero and points toward the centre of the circle
3. It points away from the centre
4. It points along the direction of motion
Show the answer
B. It is not zero and points toward the centre of the circle · The direction of the velocity keeps changing, so there is an acceleration. It is centripetal: it points toward the centre.
8:30 watch the moment
For uniform circular motion with angular speed omega, what is the acceleration vector?
1. a = omega r
2. a = minus omega squared times r
3. a = omega squared times r pointing outward
4. a = 2 pi r
Show the answer
B. a = minus omega squared times r · Differentiating the position twice gives a = -omega^2 r. The minus sign means it points opposite to the position vector, toward the centre.
7:00 watch the moment
What is the relation between angular speed and period T?
1. omega = T / 2 pi
2. omega = 2 pi / T
3. omega = 2 pi T
4. omega = T squared
Show the answer
B. omega = 2 pi / T · One full turn is 2 pi radians and takes time T, so omega T = 2 pi. This gives omega = 2 pi / T.
5:11 watch the moment

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Made by Beyond from the video “3. Newton's Laws of Motion” by YaleCourses. The video plays in YouTube's own player. Watch it on YouTube · All lessons

7 lessons

Introduction: uniform circular motion and why it needs a force Free

Newton's first law and inertia Beyond Pass

Newton's second law: F = ma Beyond Pass

Mass versus weight and units of force Beyond Pass

Newton's third law: action and reaction Beyond Pass

Free-body diagrams and worked examples Beyond Pass

Applications and closing questions: pulling blocks, springs and elevators Beyond Pass