Week 5 — Dynamic Systems and Uniform Circular Motion

Overview

Week 5 ties Newton’s laws together with two practical contexts: multi-body dynamic systems (blocks, springs, inclines, pulleys) and uniform circular motion (objects on strings, banked curves, unbanked curves). The mechanical core is the disciplined use of free body diagrams to write $\sum F=ma$ component-wise, then recognising that for uniform circular motion the net force is always centripetal, $F_c=m{v}^2/r$. The lecture also opens with a study-skills section on agile-style daily check-ins and Pomodoro sprints, intended to manage the rising portfolio/lab workload.

Key concepts

• Free body diagram (FBD) — pictorial representation showing the object as a particle at the origin of a coordinate system with all forces acting on it drawn as labelled vectors. Workflow: Identify → Define → Model → Represent (slide 13).
• Newton’s 1st law — an object persists in rest or uniform straight-line motion unless acted on by a net force; equivalent to $F_{net}=0$ (slide 14).
• Newton’s 2nd law — for constant mass, $F_{net}=ma$ (slide 14).
• Newton’s 3rd law — $F_{A\text{on}B}=-F_{B\text{on}A}$ (slide 14).
• Weight force $F_W=mg$ — vertical, downward; units N. Here $g=9.8{\text{m/s}}^2$.
• Normal force $F_N$ — contact force from a surface, perpendicular to that surface; units N.
• Tension $F_T$ (also $T$) — pulling force transmitted along a string/rope; units N.
• Spring force $F_s=kx$ — restoring force from a stretched/compressed spring; $k$ in N/m, $x$ in m.
• Friction force ${\vec{F}}_{fric}$ — contact force parallel to a surface that opposes (kinetic) or prevents (static) motion. Formula on slide 12: ${\vec{F}}_{fric}=F_N\mu$.
  • Kinetic friction — opposes motion; coefficient ${\mu }_k$; $F_{fric}=F_N{\mu }_k$.
  • Static friction — prevents motion; coefficient ${\mu }_s$; $F_{fric}\le F_N{\mu }_s$.
• Coefficient of friction $\mu$ — dimensionless; always positive (Tutorial Ex. 5 note).
• Uniform circular motion — motion around a circular path at constant speed but with continuously changing velocity direction, hence non-zero acceleration (slide 19).
• Centripetal acceleration $a_c$ — acceleration directed toward the centre of the circular path; magnitude $v^2/r$; units m/s${}^2$.
• Centripetal force $F_c$ — the net (resultant) inward force responsible for $a_c$. It is NOT a separate force you add to the FBD; it is the sum of the real forces (slide 24).
• Banked curve — a curve tilted at angle $\theta$ so the horizontal component of $F_N$ supplies the centripetal force (slide 23).
• Period / radius / speed for circular motion: $r$ (m), $v$ (m/s).

Study-workflow concepts (slides 3–9): Daily Check-in (Morning/Midday), Rule of 3, Pomodoro Sprint (Set Scope 2 min, Sprint 25 min, Reflection 5 min), 4-Pom Rule, Stop the Line, Batch the Admin, high-value vs low-value breaks.

Core formulas

Newton’s laws (slide 14):

$F_{net}=0(\text{1st law})$

${\vec{F}}_{net}=m\vec{a}(\text{2nd law})$

${\vec{F}}_{A\text{on}B}=-{\vec{F}}_{B\text{on}A}(\text{3rd law})$

Symbols: $F$ force (N), $m$ mass (kg), $a$ acceleration (m/s${}^2$).

Component form (per axis):

$\sum F_x=m{a}_x,\sum F_y=m{a}_y$

Friction (slide 12):

${\vec{F}}_{fric}=F_N\mu$

with $\mu ={\mu }_k$ for kinetic and ${\vec{F}}_{fric}\le F_N{\mu }_s$ for static. $F_N$ in N, $\mu$ dimensionless.

Weight:

$F_W=mg$

$g=9.8{\text{m/s}}^2$.

Spring force (Hooke’s law, used in Example 1 and Tutorial 6):

$F_s=kx$

$k$ spring constant (N/m), $x$ extension from equilibrium (m).

Centripetal acceleration (slide 20):

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

$v$ tangential speed (m/s), $r$ radius (m), $a_c$ in m/s${}^2$, directed toward circle centre.

Centripetal (net) force (slide 20):

$\sum F=m{a}_c=\frac{m{v}^2}{r}=F_c$

Horizontal conical pendulum — object on a string (slide 22):

$\sum F_y=T\sin \theta -mg=0\Rightarrow T\sin \theta =mg$

$\sum F_x=T\cos \theta =m{a}_c=\frac{m{v}^2}{r}$

Here $\theta$ is the angle the string makes with the horizontal axis (per slide 22 and Example 4); $T$ is tension (N).

Banked curve, no friction (slide 23):

$\sum F_y=0=F_N\cos \theta -mg\Rightarrow F_N\cos \theta =mg$

$\sum F_x=F_N\sin \theta =m{a}_c=\frac{m{v}^2}{r}$

Dividing gives the design equation $\tan \theta =v^2/(rg)$.

Inclined plane decomposition (Example 2, slide 17):

• Along slope (x, down-slope positive): $\sum F_x=mg\sin \theta -F_{fric}$
• Normal to slope (y): $\sum F_y=F_N-mg\cos \theta =0\Rightarrow F_N=mg\cos \theta$

Combined with kinetic friction:

$a=g\sin \theta -{\mu }_{k}g\cos \theta$

Threshold (no acceleration) condition gives ${\mu }_s=\tan \theta$.

Worked examples

Example 1 — Coupled blocks with a spring (slides 16; handwritten notes pp. 1–2)

A 2.0 kg mass and a 3.0 kg mass on a horizontal frictionless surface are joined by a massless spring with $k=110\text{N/m}$. A 15 N force is applied to the 3 kg mass. Find the spring stretch $x$.

Method 1 — separate FBDs.

FBD of the 2 kg block: horizontal force = spring force $F_s=kx$ pulling it right.

$\sum F_x=m_2{a}_x\Rightarrow 2a_x=kx\Rightarrow a_x=\frac{kx}{2}=\frac{110x}{2}=55x(1)$

FBD of the 3 kg block: applied force 15 N to the right; spring pulls it left with $kx$.

$\sum F_x=m_3{a}_x\Rightarrow 3a_x=15-kx=15-110x(2)$

Sub (1) into (2):

$3(55x)=15-110x$ $165x+110x=15$ $275x=15$ $x=\frac{15}{275}=0.055\text{m}=55\text{mm}$

Method 2 — combined system (alternative, notes p. 2). Treat the two blocks plus spring as a single 5 kg body. $\sum F_x=ma\Rightarrow 5a=15\Rightarrow a=3{\text{m/s}}^2$. Then use the 2 kg FBD: $2(3)=kx\Rightarrow x=6/110=0.0545\text{m}\approx 55\text{mm}$. Same result.

Example 2 — Block on a 30° incline with friction (slide 17; notes pp. 3–4)

Wooden block on a 30° wooden incline; ${\mu }_s=0.4$, ${\mu }_k=0.2$. Does it slide? If so, find $a$.

Choose x along the slope (down-slope positive) and y perpendicular.

$\sum F_y=0:F_N-mg\cos 30°=0\Rightarrow F_N=mg\cos 30°(1)$

$\sum F_x=ma:ma=mg\sin 30°-\mu F_N$

Sub (1):

$ma=mg\sin 30°-\mu mg\cos 30°$ $a=g\sin 30°-\mu g\cos 30°$

Sliding test. For no motion ($a=0$): $0=g\sin 30°-{\mu }_{s}g\cos 30°$, so ${\mu }_s^{\text{required}}=\tan 30°=0.577$. Our ${\mu }_s=0.4<0.577$, so static friction is insufficient — the block slides.

Acceleration (use ${\mu }_k=0.2$):

$a=9.8\sin 30°-0.2(9.8)\cos 30°=4.9-1.697=3.203{\text{m/s}}^2$

Example 3 — Vertical circle: tension at top and bottom (slide 25; notes p. 5)

A 0.5 kg ball on a 1.5 m string, swung in a vertical circle at $v=12\text{m/s}$. Find $T$ at top (A) and bottom (B).

Top of loop. Take inward (downward at top) as positive. Both $T$ and weight point toward the centre.

$\sum F_y=m{a}_c:\frac{m{v}^2}{r}=F_T+mg$

$F_T=\frac{m{v}^2}{r}-mg=\frac{0.5\times 12^2}{1.5}-0.5\times 9.8=48-4.9=43.1\text{N}$

Bottom of loop. Inward (upward at bottom) positive. Tension pulls up (toward centre); weight pulls down.

$\sum F_y=m{a}_c:\frac{m{v}^2}{r}=F_T-mg$

$F_T=\frac{m{v}^2}{r}+mg=\frac{0.5\times 12^2}{1.5}+0.5\times 9.8=48+4.9=52.9\text{N}$

Sanity check: tension is larger at the bottom (string must both support weight and supply centripetal force).

Example 4 — Conical pendulum, horizontal circle (slide 26; notes p. 6)

A ball on a 1.55 m string swings in a horizontal circle; the string makes 12.0° with the horizontal. Find $v$.

Geometry: the horizontal radius is $r=1.55\cos 12°$.

FBD: tension $F_T$ along the string (up and inward), weight $mg$ down. Resolve.

$\sum F_y=0:F_T\sin 12°-mg=0\Rightarrow F_T=\frac{mg}{\sin 12°}(1)$

$\sum F_x=m{a}_c:F_T\cos 12°=\frac{m{v}^2}{r}\Rightarrow F_T=\frac{m{v}^2}{r\cos 12°}=\frac{m{v}^2}{1.55\cos 12°\cos 12°}(2)$

Equate (1) and (2):

$\frac{mg}{\sin 12°}=\frac{m{v}^2}{1.55\cos 12°\cos 12°}$

$v=\sqrt{\frac{1.55\cos 12°\cdot \cos 12°\cdot g}{\sin 12°}}=8.36\text{m/s}$

Mass cancels — speed depends only on geometry and $g$.

Things to practise

Tutorial 5 (Wolfson Ch. 4 & 5). Brief outlines from the solutions PDF.

Exercise 1 — Net force on the middle of three blocks. Three blocks 1, 2, 3 kg on a frictionless table, 12 N applied to the left block. Two methods: (a) write $\sum F_x=ma$ for each block with contact forces $F_{C,12},F_{C,23}$ and add the three equations to get $12=6a\Rightarrow a=2{\text{m/s}}^2$; or (b) treat the 6 kg system to get the same $a$. Then $F_{net,2\text{kg}}=m_{2}a=2(2)=4\text{N}$.

Exercise 2 — Sled on 8.5° slope at constant speed. Constant speed means $\sum F=0$ along the slope, so $mg\sin \theta =\mu mg\cos \theta \Rightarrow \mu =\tan \theta =\tan 8.5°=0.15$.

Exercise 3 — Crate on 30° ramp with rope over a pulley to a hanging weight $w$. Rope is at 20° above the ramp surface; ${\mu }_s=0.25$, ${\mu }_k=0.1$. Find the minimum $w$ to hold the 1520 N crate stationary (friction acts up the ramp at the threshold of sliding down). For the hanging weight: $F_T=w$. For the crate (rotated axes along/perp to ramp):

• $\sum F_y=0$: $F_N=1520\cos 30°-w\sin 20°$
• $\sum F_x=0$: $w\cos 20°-1520\sin 30°+{\mu }_s{F}_N=0$

Solving: $w=\frac{1520\sin 30°+0.25\cdot 1520\cos 30°}{\cos 20°+0.25\sin 20°}=505\text{N}$.

Exercise 4 — 900 g rock on a 1.3 m string in a horizontal circle; breaking strength 120 N.

(a) Minimum angle (with horizontal): $\sum F_y=0\Rightarrow F_T\sin \theta =mg\Rightarrow \theta ={\sin }^{-1}(mg/F_T)={\sin }^{-1}(0.9\times 9.8/120)=4.215°$.

(b) Speed at that angle: $\sum F_x=m{v}^2/r$ with $r=1.3\cos \theta$, giving $v=\sqrt{1.3\cos \theta \cdot F_T\cos \theta /m}\approx 13.13\text{m/s}$.

Exercise 5 — Friction needed for an unbanked turn. Car at 87 km/h = 24.2 m/s on a 130 m radius unbanked curve. $F_N=mg$, and friction supplies $m{v}^2/r$. So $\mu =v^2/(rg)=24.2^2/(130\times 9.8)=0.46$. (The negative sign appearing mid-working just indicates friction points opposite to the assumed axis; $\mu$ itself is always positive.)

Exercise 6 (Challenge) — Mass on a spring undergoing circular motion. $m=2.1\text{kg}$, $k=150\text{N/m}$, natural length $x_0=0.18\text{m}$, $v=1.4\text{m/s}$. The spring force supplies the centripetal force: $k(R-x_0)=m{v}^2/R$. Rearrange to $k{R}^2-k{x}_{0}R-m{v}^2=0\Rightarrow 150R^2-27R-4.116=0\Rightarrow R\approx 0.28\text{m}$.

Common pitfalls

• Centripetal force is not a new force. Don’t add an "$F_c$" arrow to your FBD — it’s the net of the real forces (tension, normal, friction, gravity) projected toward the centre (slide 24).
• Constant speed ≠ constant velocity. Uniform circular motion has constant speed but changing direction, so there is acceleration and therefore a non-zero net force (slide 24).
• Always set the centre of the circle as the positive direction when writing the radial equation (slide 22, 24). It keeps signs honest.
• Sliding check on inclines. Compare required ${\mu }_s$ ($=\tan \theta$ for the no-acceleration threshold) to the actual ${\mu }_s$. If actual is smaller, the block slides and you switch to ${\mu }_k$ for the acceleration calculation (Example 2).
• Friction coefficient is always positive. A negative answer just means the assumed direction of $F_{fric}$ on the FBD was the wrong way (Tutorial Ex. 5).
• For a conical pendulum, $\theta$ is measured from the horizontal in this course (slide 22, Ex. 4). The radius of the circle is $r=L\cos \theta$ where $L$ is the string length, not $L\sin \theta$.
• Unit conversions. Convert km/h → m/s (divide by 3.6) before plugging into $a_c=v^2/r$ (Tutorial Ex. 5).
• Mass cancels in geometry-only problems (banked curve, conical pendulum speed) — a useful sanity check.
• Spring problems with circular motion: the radius $R$ includes the natural length plus the extension, $R=x+x_0$ (Tutorial Ex. 6).
• Pomodoro/study hygiene: if you’re lost after 10 minutes, stop the timer (“Stop the Line”) and find a resource rather than staring; don’t count phone scrolling or “another hard task” as a break (slides 8–9).
• Spend most of your time on Visualise, not Solve — the lecturer’s explicit instruction on the Problem Solving Approach slide (slide 10).

Source citations

• Lecture slides EGD102-Physics/Lecture5_CTP1-1.pdf:
  • Productivity / study skills: slides 3–9
  • Problem-solving approach: slide 10
  • Common forces (friction): slide 12
  • Free body diagram workflow: slide 13
  • Newton’s laws: slide 14
  • Analysis approach: slide 15
  • Example 1 (spring + two blocks): slide 16
  • Example 2 (30° incline with friction): slide 17
  • Uniform circular motion definition: slide 19
  • Centripetal acceleration & force: slides 20–21
  • Conical pendulum (horizontal string): slide 22
  • Banked curves: slide 23
  • Key cautions for circular motion: slide 24
  • Example 3 (vertical circle): slide 25
  • Example 4 (horizontal conical pendulum): slide 26
• Handwritten worked solutions: EGD102-Physics/EGD102 - Lecture5 - Notes.pdf (Examples 1–4, including the combined-system alternative on p. 2 and the static-friction threshold check on p. 4).
• Tutorial problems: EGD102-Physics/Tutorial 5.pdf (Exercises 1–6, Wolfson Ch. 4 & 5).
• Tutorial solutions: EGD102-Physics/Tutorial 5_Solutions.pdf (full step-by-step solutions used above).
• Textbook reference: Wolfson, R. 2020. Essential University Physics, Vol. 1, Global Edition, 4th ed. SI Units, Chapters 4 and 5 (slide 29).