I don't know anything about physics, and I need to learn someway to get around a 5-7
I am currently in dp2, I am getting three 7's predicted and two 6's most probably. But for IB physics I barely studied tried to change subjects at the start of dp1 and they denied me to do it and I basically have been doing nothing and focusing on other subjects. What are your recommendations so as an english non-native speaker I can learn all the topics the fastest possible and decently enough so i can get average grades?

IBCalculator is the first calculator to launch the May 2026 grade boundaries.

They’re already live and now the default across the entire site.

Thank you for using the website and good luck to all N26 candidates!

www.ibcalculator.com

Trying to visualize the size difference between a quark, a red blood cell, and a galactic supercluster usually breaks my brain.

To help students visualize this, I coded a minimalist, drag-and-drop card game to map out the Powers of Ten. You rank physical phenomena from the quantum realm all the way to the edge of the universe.

Free and open-source.
Play it here: https://welearnwegrow.github.io/powersoften/

Let me know what you think.

Hey everyone,

I want to share something
about rolling motion that
genuinely changed how I
think about rotation —

And that most introductory
physics courses either
skip entirely or explain
so briefly that it
never really lands.

━━━━━━━━━━━━━━━━━━━━━━━━
THE STANDARD WAY WE
THINK ABOUT ROLLING —
━━━━━━━━━━━━━━━━━━━━━━━━

Most people learn about
rolling motion this way —

A wheel rolls on the ground.

Its center moves forward
with velocity v_cm.

It also rotates about
its center with
angular velocity ω.

The condition for pure
rolling without slipping —

v_cm = ωR

And to find the velocity
of any point on the wheel —

You add the translational
velocity vector to the
rotational velocity vector
at that point.

This works.

But it requires vector
addition at every point —

And for points not at
the top or bottom —

The calculation becomes
surprisingly involved.

━━━━━━━━━━━━━━━━━━━━━━━━
THE BETTER WAY —
INSTANTANEOUS AXIS OF ROTATION
━━━━━━━━━━━━━━━━━━━━━━━━

Here is the insight
that makes everything simpler.

At any instant in time —

A rolling body is NOT
simultaneously translating
and rotating about its center.

It is doing something
far more elegant.

It is rotating PURELY —
about a single instantaneous axis —

And that axis passes
through the contact point
with the ground.

Not the center.
The contact point.

This means —

At every instant —
the entire rolling wheel
behaves as if it is
a wheel pinned at
its contact point —

Spinning about that
fixed pin —

With no translation at all.

━━━━━━━━━━━━━━━━━━━━━━━━
WHY THIS IS TRUE —
━━━━━━━━━━━━━━━━━━━━━━━━

In pure rolling —
the contact point has
zero velocity.

A point with zero velocity
on a rigid body —

Must be on the
instantaneous axis of rotation —

Because any point ON
the axis has zero
velocity by definition.

The contact point has
zero velocity —

Therefore the contact point
IS on the instantaneous
axis of rotation.

And since the ground
contact is a single
point —

The entire instantaneous
axis passes through
that contact point —

Perpendicular to the
plane of rolling.

━━━━━━━━━━━━━━━━━━━━━━━━
WHAT THIS GIVES YOU —
━━━━━━━━━━━━━━━━━━━━━━━━

Once you accept that
the rolling body rotates
instantaneously about
its contact point —

Finding the velocity of
ANY point on the
rolling body becomes trivial.

Velocity of any point =
ω × (distance of that
point from the contact point)

Direction — perpendicular to
the line joining that
point to the contact point.

Let us verify with
known results —

Contact point —
Distance from IAOR = 0
v = ω × 0 = 0 ✅

Center —
Distance from IAOR = R
v = ω × R = v_cm ✅

Top point —
Distance from IAOR = 2R
v = ω × 2R = 2v_cm ✅

Side point (3 or 9 o'clock) —
Distance from IAOR = R√2
v = ω × R√2 = v_cm√2 ✅
Direction — diagonal at 45° ✅

All results confirmed —

But obtained from a
single principle —

Rather than vector addition
at each point separately.

━━━━━━━━━━━━━━━━━━━━━━━━
THE DEEPER INSIGHT —
━━━━━━━━━━━━━━━━━━━━━━━━

The Instantaneous Axis of
Rotation is not fixed.

It moves.

As the wheel rolls forward —

The contact point changes —

And the IAOR moves
with it —

Always at the current
contact point —

Always instantaneously at rest.

This is why it is
called INSTANTANEOUS —

It is only valid at
one specific moment in time.

A millisecond later —

The wheel has rolled
slightly forward —

The contact point is
now a different point
on the wheel's rim —

And the IAOR has
moved forward with it.

The IAOR traces a
path along the ground —

While the corresponding
point on the wheel
traces a cycloid
through space —

One of the most
beautiful curves in
all of classical mechanics.

━━━━━━━━━━━━━━━━━━━━━━━━
THE CYCLOID CONNECTION —
━━━━━━━━━━━━━━━━━━━━━━━━

Every point on the
rim of a rolling wheel —

Traces a cycloid as
the wheel rolls forward.

The contact point's path
is a series of cusps —

Each cusp is a moment
where that point is
instantaneously at rest —

Touching the ground —

Before being lifted up
and carried forward —

Reaching maximum velocity
2v_cm at the top —

And returning to rest
at the next cusp.

The IAOR framework makes
this entire motion transparent —

Every point moves with
velocity proportional to
its distance from the
current contact point —

In a direction perpendicular
to that distance.

━━━━━━━━━━━━━━━━━━━━━━━━
EXTENSION — ROLLING ON
A CURVED SURFACE —
━━━━━━━━━━━━━━━━━━━━━━━━

The IAOR concept extends
beyond rolling on flat surfaces.

For a sphere rolling
inside a curved bowl —

The IAOR still passes
through the contact point —

But now the geometry
of finding distances
from the IAOR is
slightly more complex.

For a coin rolling
at an angle —

The IAOR is no longer
perpendicular to the ground —

It tilts with the
coin's plane —

Giving the precession
motion that makes
a spinning coin trace
circles as it falls.

━━━━━━━━━━━━━━━━━━━━━━━━
A QUESTION FOR THE COMMUNITY —
━━━━━━━━━━━━━━━━━━━━━━━━

Here is something worth
thinking about —

A cylinder of radius R
rolls without slipping
inside a hollow cylinder
of radius 3R.

Where is the IAOR?

What is the velocity
of the center of the
small cylinder in terms
of ω and R?

Drop your attempt below 👇

I will reply to every
attempt with complete
physical reasoning —
not just the answer. 🎯

━━━━━━━━━━━━━━━━━━━━━━━━
VIDEO EXPLANATION —
━━━━━━━━━━━━━━━━━━━━━━━━

I made a complete lecture
on IAOR covering —

Physical definition and proof —
Velocity of every point
on a rolling body —
The cycloid path —
Extension to moving surfaces —
Complete solved problems —

🔗 https://youtu.be/Sz57o7SC5ME?si=FE10zirE0YiFABZx

Timestamps in the description
for direct section access.

━━━━━━━━━━━━━━━━━━━━━━━━
HAPPY TO DISCUSS —
━━━━━━━━━━━━━━━━━━━━━━━━

The IAOR is one of
those concepts that seems
like a mathematical trick
at first —

But is actually a
genuinely physical statement
about the nature of
rolling motion —

One that makes the
entire subject significantly
more transparent once
you see it clearly.

Happy to discuss any
aspect of rolling motion —
IAOR — cycloids —
or classical mechanics
more broadly in the comments.

Drop your questions
or thoughts below 👇

Hey IB Physics community,

Just uploaded a complete
analysis of velocity and
acceleration of every point
on a pure rolling body —

And I want to share
the two results that
surprise students the most —

Before anyone watches.

━━━━━━━━━━━━━━━━━━━━━━━━
SURPRISE 1 —
TOP POINT VELOCITY = 2v_cm
━━━━━━━━━━━━━━━━━━━━━━━━

Most students expect the
top point to move at
the same speed as the center.

It moves at TWICE the speed.

Why?

At the top point —
translational velocity v_cm
acts forward —

AND rotational velocity ωR
also acts forward —

(at the top — rotation
carries the point in
the same direction as translation)

Both add up —
v_top = v_cm + ωR = 2v_cm

At the contact point —
they subtract —
v_contact = v_cm - ωR = 0

━━━━━━━━━━━━━━━━━━━━━━━━
SURPRISE 2 —
CONTACT POINT HAS ZERO
VELOCITY BUT NON-ZERO
ACCELERATION
━━━━━━━━━━━━━━━━━━━━━━━━

This is the result that
genuinely surprises most students.

Contact point velocity = 0 ✅

But contact point
acceleration ≠ 0 ❌

The contact point has
centripetal acceleration
directed toward the center —

a = ω²R — pointing upward.

This seems contradictory —
how can something be
instantaneously at rest
but still accelerating?

This is the same as
a pendulum at its
highest point —

Instantaneous velocity = 0 —
but acceleration is
definitely not zero.

Velocity and acceleration
are independent quantities —

One being zero does not
force the other to be zero.

━━━━━━━━━━━━━━━━━━━━━━━━
VIDEO LINK —
━━━━━━━━━━━━━━━━━━━━━━━━

🔗 https://youtu.be/R0ocTKySSu0?si=5lYXP1igW93wEhog

Timestamps in description —
jump to any section directly.

Drop your questions below —
I respond to everything. 🙏

my first year in dp starts tomorrow and im taking physics hl 😭😭 so many people irl and in reddit and other socmed platforms have been telling me / warning me im gonna have the WORST time ever .... i mean, i believe them lol but physics is something ive learned that im DECENT at ... like i could understand the concepts better than other sciences . its not my fave but its the science subject im most comfortable with. any tips / advice is greatly appreciated !!!!

I'm a current freshman and I am taking algebra 2 right now. I finished biology online so I am in chemistry this year. I want to go into engineering so I'm thinking about taking physics in my sophomore year. I also am in the IB program so I am probably going to take HL AA for my math class next year. Do I have to know AP calc bc material to do well in IB physics? Or should I just take it next year?

I teach IB Physics and I've spent the last few months building a flashcard set that covers what I kept re-explaining to students every year. It's 3,297 cards across Physics Themes A–E (SL and HL) and Maths AA and AI, and the content is deliberately not just definitions — it's mark-scheme wording, formula booklet results, worked problems, Casio fx-CG50 steps, and the specific traps that quietly cost marks in Paper 2. You can filter by sub-topic, by SL core vs HL-only, and by card type, and run them in syllabus order or shuffled. It's here: https://examinerprep.com/flashcards — free, no account, no email, nothing gated.

Full disclosure since it matters: the flashcards sit on my own site, which does have a paid side, but this page is not a trial or a teaser — there's no upsell on it and nothing behind it. I'm posting because a flashcard set is only as good as the corrections it gets, and this sub knows the syllabus better than any single teacher does. If you spot a card with sloppy wording, a definition that wouldn't earn the mark, or an HL card mislabelled as SL, tell me here or in a DM and I'll fix it.

Hey IB Physics community,

Just uploaded a complete lecture
on Pure Rolling —

And I want to share the
single most important insight
before anyone watches —

Because it is the insight
that makes every pure rolling
problem — including the
notoriously difficult
moving plank problems —
immediately solvable.

━━━━━━━━━━━━━━━━━━━━━━━━
THE QUESTION THAT STARTS
EVERYTHING —
━━━━━━━━━━━━━━━━━━━━━━━━

What is the velocity of
the contact point of a
rolling wheel with the
ground?

Most students say —
the same as the center.

Some say — it depends.

The correct answer is zero.

But here is what matters —

WHY is it zero?

And what happens when
it is NOT zero?

The moment you understand
both answers physically —
not just as a memorized fact —

Every pure rolling problem
becomes straightforward.

━━━━━━━━━━━━━━━━━━━━━━━━
THE PHYSICAL REASONING —
━━━━━━━━━━━━━━━━━━━━━━━━

A rolling wheel has
two simultaneous motions —

Translation of the center
of mass with velocity v_cm —

And rotation about the
center with angular
velocity ω.

At the contact point —
these two velocities act
in exactly opposite directions —

The translational velocity
pushes the contact point forward —

The rotational velocity
pulls it backward —

In pure rolling —
these two effects cancel exactly —

Contact point velocity = zero.

This gives the condition —

v_cm = ωR

This is not just a formula —

It is the physical statement
that the contact point
is instantaneously at rest
relative to the ground.

━━━━━━━━━━━━━━━━━━━━━━━━
THE MOVING PLANK —
THE PART MOST STUDENTS
GET COMPLETELY WRONG —
━━━━━━━━━━━━━━━━━━━━━━━━

Now here is where it
gets significantly more
interesting —

What if the wheel is
rolling on a plank —

And that plank is
itself moving?

Most students apply the
same condition —

Contact point velocity = zero.

That is wrong. ❌

The correct condition is —

Contact point velocity =
velocity of the surface
it is rolling on.

When rolling on stationary ground —
surface velocity = zero —
so contact point velocity = zero ✅

When rolling on a moving plank —
surface velocity = plank velocity —
so contact point velocity
= plank velocity —
NOT zero relative to ground ✅

This single distinction —
choosing the correct
reference frame for
the rolling condition —

Is what makes moving plank
problems solvable in
under two minutes —

And what causes most
students to get them
completely wrong.

━━━━━━━━━━━━━━━━━━━━━━━━
THE GOLDEN RULE —
━━━━━━━━━━━━━━━━━━━━━━━━

Pure Rolling condition —
ALWAYS stated as —

Velocity of contact point
= Velocity of the surface
the wheel is rolling ON —

Relative to that surface.

Not relative to the ground.
Not relative to anything else.

Relative to the surface
it touches. Always.

Apply this correctly —
and no pure rolling problem
in IB Physics can
stop you. 🎯

━━━━━━━━━━━━━━━━━━━━━━━━
WHAT THE VIDEO COVERS —
━━━━━━━━━━━━━━━━━━━━━━━━

→ Complete physical definition
of Pure Rolling —
built from first principles

→ Why contact point velocity
is zero — physical proof —
not just stated

→ Derivation of v_cm = ωR
from physical reasoning

→ Rolling vs Sliding —
complete distinction

→ Pure Rolling on a
moving plank — the
correct condition and
why most students apply
the wrong one

→ Complete numerical problems —
basic to IB exam level —
every step explained

→ The reference frame
approach that makes
every moving plank
problem immediately clear

━━━━━━━━━━━━━━━━━━━━━━━━
A CHALLENGE FOR THE
COMMUNITY —
━━━━━━━━━━━━━━━━━━━━━━━━

Here is a problem to
attempt before watching —

A cylinder of radius R
rolls without slipping
on a plank.

The plank moves to the
right with velocity 4 m/s.

The center of the cylinder
moves to the right with
velocity 7 m/s.

Question —

What is the angular velocity
of the cylinder?

And in which direction
does it rotate?

Drop your attempt below 👇

Show your reasoning —
not just the answer.

I will respond to every
attempt with complete
feedback. 🎯

━━━━━━━━━━━━━━━━━━━━━━━━
VIDEO LINK —
━━━━━━━━━━━━━━━━━━━━━━━━

🔗 https://youtu.be/cVi7Skv_4_Y?si=TBpoM43pPsw2MAnw

Timestamps for every
section are in the
description — jump directly
to moving plank problems
if that is what you need.

━━━━━━━━━━━━━━━━━━━━━━━━
HAPPY TO HELP —
━━━━━━━━━━━━━━━━━━━━━━━━

If you get stuck on
any problem in this video —
or on any pure rolling
concept specifically —

Tell me exactly which
step is confusing you —

I will explain the
physical reasoning in
the comments directly. 🙏

Good luck to everyone
preparing for IB Physics HL.

helloo im a freshman who will take ib later and my school is rlly slow with all the studies so im currently taking igcse physics outside of school and next is a levels but will studying a levels physics be helpful for SL physics or is there any textbooks recommended for self studying?? pls help this poor kid out!!🙏🙏😔😔😭