Introduction to Physics Olympiad

Physics Olympiad

It’s a point

Textbook figure / a block on a smooth incline
\(a=g\sin\theta\) + friction \(a=g(\sin\theta-\mu\cos\theta)\)
Olympiad notes / Part 02 Newton’s laws / Problem 1
A block slides down a fixed, smooth incline at angle θ. Find its acceleration.
Page one of the olympiad book is the same problem as your textbook. The difference is not where you start, but how far you go.
Assumption 01The object is a point
Assumption 02Friction coefficient is constant
Assumption 03There is no air
Assumption 04The object never tips over
Layer 00│Olympiad notes · Newton’s laws · Example 1
Vote

Released together: which reaches the bottom first?

Don’t calculate / go with your gut
Same incline / same start line / released together
ABlock (cannot rotate)
BSolid ball
CSolid cylinder
DHoop (hollow tube)
All four have the same mass and the same radius and are released from the same height. Ignore energy lost to friction.
Ignore frictional losses.
Vote│Answer on the next slide

It’s not a point: it rolls

Rotation takes a share of the energy
Energy equation / one extra term
\[mgh=\tfrac12 mv^{2}+\underbrace{\tfrac12 I\omega^{2}}_{\textstyle\text{rotation}}\]
Write \(\beta=I/mR^{2}\) (how far the mass sits from the axis) and use \(v=\omega R\):
\[\boxed{\;a=\frac{g\sin\theta}{1+\beta}\qquad v=\sqrt{\frac{2gh}{1+\beta}}\;}\]
ObjectβAccelerationPlace
A Block (no rotation)0\(g\sin\theta\)1
B Solid ball2/5\(\tfrac57 g\sin\theta\)2
C Solid cylinder1/2\(\tfrac23 g\sin\theta\)3
D Hoop / hollow tube1\(\tfrac12 g\sin\theta\)4
Same incline / released together / four lanes
Mass and radius don’t change the order. Only the shape does
The result has no m and no R. Lead rolls as fast as wood; a big wheel as fast as a small one.
Layer 01│Demos: raw vs boiled egg · full vs empty can
Insert

Moment of inertia: where the mass is

Cheng Jiafu, Mechanics, pp. 280–281
Same disc / two equal masses / one near, one far
Definition
\[I=\sum_i m_i r_i^{2}\]
Multiply each bit of mass by the square of its distance from the axis, then add them up.
The two point masses are equal; one is near the axis and one is far from it.
Which one makes the disc harder to spin up?
Insert│Cheng Jiafu, Mechanics, pp. 280–281 · definition of moment of inertia

What makes it roll without slipping?

θ = 25° / solid cylinder / critical μ = 0.155
The spin comes from the torque of static friction, and there is enough only if
\(\mu\ \ge\ \dfrac{\beta}{1+\beta}\tan\theta\)
ball 2/7 tanθ  ·  cylinder 1/3 tanθ  ·  hoop 1/2 tanθ
Rollingμ = 0.30 ≥ 0.155
turned 0.00 rev v = ωR ✓
Rolling and slippingμ = 0.08 < 0.155
turned 0.00 rev v > ωR ✗ slipping
Released together · same clock
Which reaches the bottom first?
Layer 02│Cheng Jiafu, Mechanics, p. 290, Fig. 6-9
Core example

Exercise 6-13

Cheng Jiafu, Mechanics, pp. 290–291
Original problem
A uniform solid cylinder of radius R and mass m is placed on an incline at angle θ to the horizontal. The coefficient of friction between cylinder and incline is μ. For θ below a critical angle θc, the cylinder rolls down without slipping.
1) What is the critical angle θc?  2) For θ < θc, what is the acceleration a?

Method 1 · centre of mass + rotation about it5 equations

  • \(mg\sin\theta-f=ma\)
  • \(mg\cos\theta=N\)
  • \(fR=I\alpha=\tfrac12 mR^{2}\alpha\)
  • \(a=R\alpha\) (rolling condition)
  • \(f=\mu N\) (at θ = θc)

Method 2 · about the instantaneous axis1 equation

\[mgR\sin\theta=\Big(\tfrac12 mR^{2}+mR^{2}\Big)\alpha\]
Use the parallel-axis theorem to move the axis to the contact point. Friction then has zero torque, so it never needs to appear.
\(\tan\theta_c=3\mu\) \(a=\tfrac23\,g\sin\theta\)
Olympiad training is not about calculating fast; it is about choosing well.
Layer 02│Cheng Jiafu, Mechanics, Exercise 6-13

In real life, drag is rarely constant

Once the line bends / the three constant-acceleration formulas all fail
Rectangles — m　exact — m
Your textbook says: the area under a v–t graph equals the displacement.
That sentence is an integral.

The textbook only has you compute triangles and rectangles, because only constant acceleration gives a straight line. The olympiad simply bends the line.
Once the line bends, all three fail
\[v=at\quad x=\tfrac12 at^{2}\quad v^{2}=2ax\]
They rest on one assumption: a is constant.
When drag depends on speed
\[m\frac{\mathrm{d}v}{\mathrm{d}t}=mA-kv\ \Longrightarrow\ v_{\mathrm f}=\frac{mA}{k}\]
Layer 05│Terminal velocity · differential equations

A multiple-choice question with drag proportional to speed

2024 Taizhou High School Physics Competition / Q3
Original problem
A ball is thrown straight up from the ground with speed v₀ and lands back with speed v. Air resistance acts throughout, and the drag is proportional to speed. How long is the ball in the air?
\(\mathrm{A}.\;\dfrac{v_0+v}{g}\) \(\mathrm{B}.\;\dfrac{v_0-v}{g}\) \(\mathrm{C}.\;\dfrac{\sqrt{v_0v}}{g}\) \(\mathrm{D}.\;\dfrac{\sqrt{v_0^2+v^2}}{g}\)

Solve the differential equation

Take up as positive, velocity \(u(t)\), and let \(\tau=m/k\).
\[\dot u+\frac{u}{\tau}=-g \ \Rightarrow\ \frac{\mathrm d}{\mathrm dt}(ue^{t/\tau})=-ge^{t/\tau}\]
\[u(0)=v_0\ \Rightarrow\ u(t)=(v_0+g\tau)e^{-t/\tau}-g\tau\]
\[\begin{aligned} x(t)&=\int_0^t u(s)\,\mathrm ds\\ &=\tau(v_0+g\tau)(1-e^{-t/\tau})-g\tau t \end{aligned}\]
On landing, \(x(T)=0\) and \(u(T)=-v\), which give:
\[\begin{aligned} gT&=(v_0+g\tau)(1-e^{-T/\tau})\\ v_0+v&=(v_0+g\tau)(1-e^{-T/\tau}) \end{aligned}\]
\[\therefore\quad T=\frac{v_0+v}{g}\]

Integrate once over the whole flightone-line answer

Integrate \(m\dot u=-mg-ku\) from 0 to T:
\[\begin{aligned} m(-v-v_0)&=-mgT\\ &\quad-k\!\int_0^T u(t)\,\mathrm dt \end{aligned}\]
And \(\int_0^T u(t)\,\mathrm{d}t\) is just the displacement. The ball returns to its start, so it is zero and the drag impulse vanishes.
\(\displaystyle T=\frac{v_0+v}{g}\) Answer A
Here, “area = displacement” wipes out the drag term.
Layer 05│2024 Taizhou (Zhejiang) High School Physics Competition · Q3

It has a size: it can tip over

Don’t give the answer / let students guess
width 0.30 m　height 0.34 m
Test / does the vertical through the centre of mass fall outside the base?
\[\tan\theta \;>\; \frac{w}{h}\ \Longrightarrow\ \text{tips}\]
tan θ = 0.466　w/h = 0.88 stable
Friction is assumed large enough that it won’t slide, so the only question is whether it tips.
Which assumption does this layer remove?
Once it has a width and a height, stability becomes a question of its own.
Layer 06│θ = 25° · critical height h = w/tanθ ≈ 0.64 m
Bonus

As the rod falls, how far does its foot slide?

Smooth floor
Upright → falling → flat / centre of mass on one vertical
Problem
A uniform rod (mass m, length L) stands upright on a smooth floor. A nudge makes it fall from rest.
When it lies flat, how far has its bottom end moved?
The key sentence
Smooth floor → N and gravity both vertical → no horizontal force → the centre of mass falls straight down.
So the answer is pure geometry
Lying flat, the centre of mass is on the same vertical; the end is L/2 away:
\(\Delta x=\dfrac{L}{2}\) at tilt θ　\(\Delta x=\dfrac{L}{2}\sin\theta\)
“No horizontal force” solves the whole problem.
Bonus│Zhao Kaihua, New Concept Physics: Mechanics, Example 21 · Fig. 4-46

Related problems

Stability / no sliding back, no tipping
Cheng Jiafu, Mechanics, Exercise 3-4 · p. 97 / Problem 3-16 · p. 140

A wedge with apex angle θ sits on a horizontal table, and an object rests on its slope. The object and the wedge each have mass m, and every contact surface has friction coefficient μ.
1) If the wedge does not slide back while the object slides down, what relation must μ and θ satisfy?　2) If μ = 1/5, what range of θ satisfies it?

The answer is two inequalities, plus the extra condition tanθ > μ
2015 university admission exams · Topic 02 Forces and equilibrium

Straight tracks of different lengths are made from the same wooden board, keeping the distance from B to the left end A at 0.8 m. The slider is again released from rest at P. To make sure the stand does not tip over while the slider moves along the track, find the allowed range of track lengths.

In olympiad problems, “does not tip over” is a condition you must check, not scenery
Neither problem asks “how much?”; both ask “under what conditions?”
Layer 06│Cheng Jiafu p. 97 · admission exams Topic 02
Extension

Add electric and magnetic fields

Three models / every path is computed from the equations of motion
Magnetic focusing20th CPhO prelim
Fired from one point in many directions, the pitch depends only on the velocity component along the field, so after a whole number of turns they all return to the axis together.
Through two charged shells27th CPhO semifinal Q5
Shells +Q and −Q touch internally, with holes on the line of centres. The field is zero inside a shell, so a force acts only outside the small one: constant speed, then slowing down.
Magnetic mirrorVan Allen belts
The field is strong at both ends and weak in the middle. Heading into the strong field, a particle spirals ever tighter until it bounces back at the mirror point. This is how Earth traps energetic particles.
Same clock · all three paths run together
Extension│20th CPhO prelim · 27th CPhO semifinal Q5 · Van Allen belts

Course plan