PHY131H1/Unit 5

Unit 5 · Introduction to Physics I

Friction & Circular Motion

Two applications of Newton's laws that trip people up for opposite reasons. Friction because static and kinetic behave differently, and circular motion because the acceleration points somewhere counterintuitive — inward, even when the speed never changes.

By the end you can

• Distinguish static from kinetic friction and use each correctly.
• Explain why $f_s$ is an inequality and $f_k$ an equation.
• Show that circular motion at constant speed still accelerates.
• Solve banked curves and vertical circles.

Why it matters & where it comes from

Why it exists

Friction is what makes the real world differ from the frictionless problems of Unit 4, and it is why walking, driving and holding anything is possible at all. Circular motion is the first case where acceleration is not along the velocity.

In the real world

Antilock brakes exist because static friction exceeds kinetic: a rolling tyre grips better than a skidding one, so preventing the skid shortens the stop. Banked racetrack curves let cars turn faster than friction alone allows.

Story & open questions

Leonardo da Vinci found the friction laws around 1493, and Coulomb rediscovered them nearly three centuries later. That friction is independent of contact area is still surprising, and the microscopic explanation — real contact happens only at scattered asperities — came only in the twentieth century.

Friction

Definition

Static friction acts on surfaces not sliding. It takes whatever value is needed to prevent sliding, up to a maximum: $f_s \le \mu_s N$.

Kinetic friction acts on surfaces already sliding, and has a fixed value: $f_k = \mu_k N$, directed opposite the motion.

Typically $\mu_s > \mu_k$, so it takes more force to start something moving than to keep it moving.

Static friction is an inequality

Writing $f_s = \mu_s N$ is wrong unless the object is on the verge of slipping. A $10$ N push on a heavy crate that does not move produces $10$ N of static friction, not $\mu_s N$. Static friction is a response, and it only reaches its maximum at the instant sliding begins.

Example Will it move?

A $20$ kg crate on a floor with $\mu_s = 0.5$ and $\mu_k = 0.3$ is pushed horizontally with $80$ N.

$N = mg = 196$ N, so the maximum static friction is $0.5 \times 196 = 98$ N.

The push of $80$ N is less than $98$ N, so it does not move, and the actual friction is $80$ N — exactly balancing the push.

Push with $100$ N instead and it slides. Friction then drops to $f_k = 0.3 \times 196 = 59$ N, so the net force is $41$ N and $a = 2.1$ m/s$^2$. The sudden jump from $98$ to $59$ N is why heavy objects lurch when they finally give.

Example Angle of repose

At what incline angle does a block just begin to slide?

Along the slope: $mg\sin\theta = f_s$. Perpendicular: $N = mg\cos\theta$.

At the point of slipping, $f_s = \mu_s N$, so $mg\sin\theta = \mu_s mg\cos\theta$, giving $\tan\theta = \mu_s$.

The mass cancels. Tilting a surface until an object slides measures $\mu_s$ directly, which is exactly how it is done in the lab.

Circular motion

Rule Constant speed still means accelerating

Velocity is a vector, so changing direction changes velocity even when the speed is fixed. For uniform circular motion the acceleration points to the centre and has magnitude

$a_c = \dfrac{v^2}{r}$

so the net force must also point inward, with $\sum F = \dfrac{mv^2}{r}$.

There is no centrifugal force

Nothing pushes outward. The net force is inward, always. The outward feeling in a turning car is your body continuing straight by the first law while the car turns beneath you — inertia, not a force. Drawing an outward arrow on a free-body diagram is the defining error of this unit.

Rule Centripetal force is a role, not a source

"Centripetal force" names the job of pointing inward, not a new kind of force. Some real force always fills the role:

Car turning — friction between tyres and road.

Ball on a string — tension.

Satellite — gravity.

Bucket swung overhead at the top — gravity plus the normal force from the bucket.

Never add a centripetal force to a diagram alongside the real ones. Set the sum of the real inward forces equal to $mv^2/r$.

Example Flat curve

How fast can a car take a curve of radius $50$ m with $\mu_s = 0.7$?

Friction supplies the inward force: $\mu_s mg = mv^2/r$.

Mass cancels: $v = \sqrt{\mu_s g r} = \sqrt{0.7 \times 9.8 \times 50} = 18.5$ m/s, about $67$ km/h.

Note it is static friction, since the tyres roll rather than skid. Once they skid, $\mu_k$ takes over and the maximum speed drops — the reason skidding into a turn makes it worse.

Example Banked curve

On a frictionless banked curve at angle $\theta$, the normal force tilts and its horizontal component turns the car.

Vertical: $N\cos\theta = mg$. Horizontal: $N\sin\theta = mv^2/r$.

Dividing: $\tan\theta = v^2/(rg)$, so the design speed is $v = \sqrt{rg\tan\theta}$.

Mass cancels again, so the same bank works for a motorcycle and a lorry. That is why banking is used in road design at all.

Example Vertical circle, at the top

What is the minimum speed for water to stay in a bucket swung in a vertical circle?

At the top both gravity and the normal force point down, toward the centre: $mg + N = mv^2/r$.

The slowest case is $N = 0$, where the bucket no longer pushes and gravity alone supplies the turning: $v_{\min} = \sqrt{gr}$.

Slower than that and the required inward force is less than gravity, so the water leaves the circular path — it falls out.

Common pitfalls

• Writing $f_s = \mu_s N$ when the object is not on the verge of slipping.
• Using $\mu_k$ for a rolling tyre. Rolling contact is static.
• Drawing an outward force in circular motion.
• Adding "centripetal force" as an extra arrow instead of recognising which real force does the job.
• Forgetting that $N$ changes on an incline, which changes friction too.

Practice

Work each fully before opening the answer. Original exercises written for this guide in the style of the course's problem sets.

Practice 1 · Does it slide?

A $5$ kg box with $\mu_s = 0.4$ is pushed with $15$ N. Does it move?

Answer

Maximum static friction is $0.4 \times 49 = 19.6$ N, which exceeds $15$ N. It stays put, and friction is $15$ N.

Practice 2 · Angle of repose

A block slips at $25$ degrees. Find $\mu_s$.

Answer

$\mu_s = \tan 25 = 0.47$.

Practice 3 · Direction of acceleration

A car rounds a curve at constant speed. Which way does it accelerate?

Answer

Toward the centre of the curve. Constant speed does not mean zero acceleration when the direction is changing.

Practice 4 · Doubling speed

How does the required centripetal force change if speed doubles at fixed radius?

Answer

Four times, since $F = mv^2/r$. This is why speed limits on curves are strict.

Practice 5 · Conical pendulum

A ball on a string swings in a horizontal circle, the string at angle $\theta$ to the vertical. Find the speed.

Solution

Vertical: $T\cos\theta = mg$. Horizontal: $T\sin\theta = mv^2/r$ with $r = L\sin\theta$. Dividing and substituting gives $v = \sqrt{gL\sin\theta\tan\theta}$.

Practice 6 · Bottom of the loop

At the lowest point of a vertical circle, how does the normal force compare with the weight?

Solution

Larger. Here up is toward the centre, so $N - mg = mv^2/r$ and $N = m(g + v^2/r)$. This is the pressed-into-your-seat feeling at the bottom of a rollercoaster loop.

Practice 7 · Why ABS works

Explain in terms of $\mu_s$ and $\mu_k$.

Answer

A rolling tyre uses static friction, a skidding one kinetic, and $\mu_s > \mu_k$. Preventing the skid keeps the larger coefficient in play, shortening the stop and preserving steering.

Practice 8 · Incline with friction

A block slides down a $30$-degree incline with $\mu_k = 0.2$. Find its acceleration.

Solution

$N = mg\cos 30$, so $f_k = 0.2mg\cos 30$. Along the slope: $mg\sin 30 - 0.2mg\cos 30 = ma$, giving $a = g(0.5 - 0.173) = 3.2$ m/s$^2$. Mass cancels, and the answer is less than the frictionless $4.9$ m/s$^2$ as it must be.

Expanded topic checklist

Course coverage. This unit is part of the complete PHY131H1 topic map. It should cover the following items before you move on:

• Static friction. An inequality; it takes whatever value prevents sliding.
• Kinetic friction. Fixed at $\mu_k N$, opposing motion.
• Centripetal acceleration. $v^2/r$ inward, even at constant speed.
• Applications. Flat and banked curves, vertical circles, and identifying which real force turns the body.

Textbook/source alignment

The calendar lists "friction" explicitly and circular motion follows from the dynamics topics. Check the syllabus on Quercus for assigned problems.

Core idea map

Both halves are Unit 4 applied. Friction adds a force whose size depends on $N$; circular motion fixes the direction of the net force and asks which real force supplies it.

Study routine. In any circular-motion problem, ask "what is pushing or pulling it toward the centre?" before writing anything. Naming that force turns the problem into ordinary Unit 4 work.

Deep practice extensions

Use these after the main practice set. They push on definitions and edge cases rather than routine calculation.

Extension 1 · Static versus kinetic

Why is one an inequality?

Guide

Static friction adjusts to whatever is needed to prevent sliding, up to a limit. Kinetic friction has a single value once sliding starts.

Extension 2 · Direction

Which way does net force point in uniform circular motion?

Guide

Toward the centre, always, with magnitude $mv^2/r$.

Extension 3 · Centrifugal

Is there an outward force?

Guide

No. The outward sensation is inertia — your body continuing straight while the constraint turns.

Extension 4 · Mass cancellation

Why does mass drop out of the banked-curve and angle-of-repose results?

Guide

Because every force involved is proportional to $m$, so it divides out of $\sum F = ma$. That is why one bank angle suits every vehicle.

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