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Study mode: digital notes + auto-check or book-practice

01

Recall and Apply

  • Recall and apply all Phase 2 formulae for circular motion and orbits
02

Identify and Correct

  • Identify and correct common errors in circular motion problems
03

Solve Under Exam Conditions

  • Solve mixed circular motion and orbital problems under exam conditions
Think First

REFLECT — What is the difference between centripetal and centrifugal force? Write your answer before looking at any notes.

Autosaved
Year 12 Physics Module 5: Advanced Mechanics 50 min ★ Consolidation Lesson 13 of 18 IQ2: Circular Motion

📐 Phase 2 Consolidation

No new content. Six formulae. Six common errors. Ten mixed practice questions. Three timed extended responses. This is your Phase 2 exam preparation.

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Phase 2 Formula Summary — All Six Equations
$$a_c = \\frac{v^2}{r} = \\omega^2 r$$
Centripetal acceleration (m/s²). Use for any object moving in a circle at constant speed.
$$F_c = \\frac{mv^2}{r} = m\\omega^2 r$$
Centripetal force (N). The net force required to maintain circular motion. Always directed toward the centre.
$$v = \\frac{2\\pi r}{T} = \\omega r, \\quad \\omega = 2\\pi f$$
Speed and angular frequency. Relates linear speed $v$, period $T$, frequency $f$, and angular velocity $\\omega$.
$$\\text{TOP: } T + mg = \\frac{mv^2}{r} \\qquad \\text{BOTTOM: } T - mg = \\frac{mv^2}{r}$$
Vertical circles. At the top, both tension and gravity point toward the centre. At the bottom, tension points up while gravity points down.
$$v_{\\text{orbital}} = \\sqrt{\\frac{GM}{r}}, \\quad T^2 = \\frac{4\\pi^2}{GM} r^3$$
Orbital mechanics. $v \\propto 1/\\sqrt{r}$ and $T^2 \\propto r^3$ (Kepler's Third Law). Remember $r = R_{\\text{planet}} + h$.
$$KE = \\frac{1}{2}\\frac{GMm}{r}, \\; U = -\\frac{GMm}{r}, \\; E = -\\frac{1}{2}\\frac{GMm}{r}, \\; v_e = \\sqrt{\\frac{2GM}{r}}$$
Energy and escape. PE is always negative (zero at infinity). Total energy is negative for bound orbits. $v_e = \\sqrt{2} \\times v_{\\text{orb}}$.
Formula Sprint
1

Flip Cards — When to Use Each Formula

Phase2 Summary

Phase2 Summary

Click each card to reveal when to use it, common traps, and what it connects to
$$a_c = \\frac{v^2}{r}$$
Click to reveal when to use this
Use when Finding centripetal acceleration of any object in circular motion. Works for cars on curves, satellites, electrons — anything going in a circle.
Trap Using $a = v^2/r$ with $v$ as a velocity vector that is changing. This is magnitude only. Also: forgetting that $v$ may not be constant in vertical circles.
Connects to $F_c = mv^2/r$ (multiply by mass to get force). Also connects to $v = 2\\pi r/T$ when period is given instead of speed.
$$F_c = \\frac{mv^2}{r}$$
Click to reveal when to use this
Use when Finding the net force required to keep an object moving in a circle, or identifying which real force(s) provide that net force.
Trap Inventing "centrifugal force" as a real force. $F_c$ is not a separate force — it is the net result of real forces (tension, gravity, friction, normal).
Connects to Banked curves: $N\\sin\\theta = mv^2/r$. Vertical circles: $T + mg = mv^2/r$ (top). Orbits: $GMm/r^2 = mv^2/r$.
$$T + mg = \\frac{mv^2}{r}$$
Top of vertical loop — click to reveal
Use when Analysing forces at the top of a vertical circle (roller coaster, bucket of water, pendulum). Both tension and gravity point toward the centre.
Trap Using a sign convention that makes both forces positive without thinking. At the top: $T$ and $mg$ both point down (toward centre), so they add. At the bottom: $T$ points up, $mg$ points down, so $T - mg = mv^2/r$.
Connects to Minimum speed at top: set $T = 0$ to find $v_{\\min} = \\sqrt{gr}$. Conservation of energy connects top and bottom speeds.
$$v = \\sqrt{\\frac{GM}{r}}$$
Orbital velocity — click to reveal
Use when Finding the speed of any satellite in stable circular orbit. Derived by equating gravitational force to required centripetal force.
Trap Forgetting $r = R + h$. The orbital radius is centre-to-centre. Also: using $g = 9.8$ m/s² instead of computing $g$ at altitude via $g = GM/r^2$.
Connects to Kepler's Third Law ($T^2 \\propto r^3$), escape velocity ($v_e = \\sqrt{2} \\times v_{orb}$), and gravitational PE ($U = -GMm/r$).
$$U = -\\frac{GMm}{r}$$
Gravitational potential energy — click to reveal
Use when Calculating gravitational potential energy in orbital mechanics. The negative sign indicates a bound orbit. Zero PE is defined at infinity.
Trap Dropping the negative sign. $U$ is always negative for bound orbits. Using $U = mgh$ (which is only valid near Earth's surface) for orbital problems.
Connects to Total energy $E = -\\frac{1}{2}GMm/r = KE + U$. Energy changes when moving between orbits: $\\Delta E = E_2 - E_1$.
$$v_e = \\sqrt{\\frac{2GM}{r}}$$
Escape velocity — click to reveal
Use when Finding the minimum speed needed to escape a planet's gravitational pull from a given distance $r$. At the surface, $r = R_{planet}$.
Trap Forgetting the $\\sqrt{2}$ factor. $v_e = \\sqrt{2} \\times v_{orb} \\approx 1.414 \\times v_{orb}$. Many students use $v_e = \\sqrt{GM/r}$ (orbital velocity) by mistake.
Connects to Total orbital energy $E = -\\frac{1}{2}GMm/r$. Escape requires adding exactly $+\\frac{1}{2}GMm/r$ of energy to bring total to zero.
Error Clinic
2

Six Common Errors — Find the Fix

Each card shows student working with an error. Identify the error and reveal the correction.
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Thinking "centrifugal force" is a real force
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Student working: "In the rotating frame of a spinning ride, the person is pushed outward by centrifugal force, which balances the tension in the chains."
Correction: Centrifugal force is a fictitious force that appears in rotating (non-inertial) reference frames. In an inertial frame, there is no outward force. The only real force is tension (and gravity), which provides the centripetal force toward the centre. The person travels in a circle because the net force points inward. "Centrifugal force" is a felt effect of inertia — the tendency to continue in a straight line — not a real force.
2
Using constant speed equations for vertical circles
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Student working: "A roller coaster loop has radius 5 m. The cart enters at 8 m/s. Find the normal force at the top." Student uses $v = 8$ m/s at the top and calculates $N + mg = mv^2/r$ with $v = 8$.
Correction: In a vertical circle, speed is not constant — gravity does work, converting between kinetic and gravitational potential energy. The entry speed (bottom) is the maximum; speed at the top is lower. Use conservation of energy: $\\frac{1}{2}mv_{bottom}^2 = \\frac{1}{2}mv_{top}^2 + mg(2r)$ to find $v_{top}$, then substitute into $N + mg = mv_{top}^2/r$.
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Forgetting $r = R + h$ for orbital problems
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Student working: "A satellite orbits at altitude 400 km. Find its orbital speed." Student uses $r = 400 \\times 10^3$ m = $4.0 \\times 10^5$ m in $v = \\sqrt{GM/r}$.
Correction: The orbital radius $r$ is always centre-to-centre. You must add the planet's radius: $r = R_{Earth} + h = 6.371 \\times 10^6 + 0.400 \\times 10^6 = 6.771 \\times 10^6$ m. Using just the altitude gives a speed that is too high by a factor of $\\sqrt{R_{Earth}/h} \\approx 4$.
4
Using $g = 9.8$ m/s² for orbital altitude problems
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Student working: "A satellite at 1000 km altitude. Find the gravitational force on a 500 kg satellite." Student calculates $F = mg = 500 \\times 9.8 = 4900$ N.
Correction: $g = 9.8$ m/s² is only valid at Earth's surface. At altitude $h$, gravitational field strength is $g' = GM/(R+h)^2$, which is significantly less. For this satellite: $r = 7.371 \\times 10^6$ m, so $g' = (6.67 \\times 10^{-11} \\times 5.97 \\times 10^{24})/(7.371 \\times 10^6)^2 = 7.33$ m/s². The correct force is $F = 500 \\times 7.33 = 3665$ N. Alternatively, use Newton's law directly: $F = GMm/r^2$.
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Confusing $v_{orbital}$ with $v_{escape}$ — forgetting $\\sqrt{2}$
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Student working: "The orbital speed at Earth's surface is 7.9 km/s. Therefore the escape velocity is also 7.9 km/s."
Correction: Escape velocity is $\\sqrt{2}$ times orbital velocity at the same radius: $v_e = \\sqrt{2GM/r} = \\sqrt{2} \\times \\sqrt{GM/r} = \\sqrt{2} \\times v_{orb} \\approx 1.414 \\times 7.9 = 11.2$ km/s. Orbital velocity gives enough speed to stay in a circular orbit; escape velocity gives enough kinetic energy to reach infinity with zero remaining speed. The factor of 2 inside the square root makes all the difference.
6
Thinking heavier satellites orbit slower
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Student working: "A 1000 kg satellite and a 2000 kg satellite are at the same altitude. The heavier one moves slower because it needs more force to stay in orbit."
Correction: From $v = \\sqrt{GM/r}$, the satellite's own mass $m$ cancels out entirely. Both satellites orbit at exactly the same speed. The heavier satellite does experience twice the gravitational force, but it also has twice the inertia — and these effects exactly cancel. This is a profound result: orbital motion is independent of the orbiting body's mass. A feather and a space station at the same altitude orbit at the same speed.
Quick Reference: All Six Errors at a Glance
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Centrifugal force is real — No. It is a fictitious force in rotating frames. In inertial frames, only real forces exist; centripetal force is the net inward force from real forces.

✗

Constant speed in vertical circles — Speed varies due to gravitational PE changes. Use conservation of energy to relate top and bottom speeds.

✗Using $r = h$ (altitude) — Always use $r = R_{planet} + h$ for centre-to-centre distance.

✗>Using $g = 9.8$ at altitude — Use $g = GM/r^2$ for orbital altitude problems, or use Newton's law of gravitation directly.

✗ v_{orb}$ (forgetting $\\sqrt{2}$) — Escape velocity is $\\sqrt{2}$ times larger: $v_e = \\sqrt{2GM/r}$.

d✗lites orbit slower — Satellite mass cancels: $v = \\sqrt{GM/r}$. All masses orbit at the same speed at the same radius.

Mixed Practice
3

10 Mixed Questions — Band 3 to Band 6

Work through these progressively harder questions. Use the data sheet constants.
Apply Band 3 [2 marks]
A car travels around a curve of radius 30 m at a constant speed of 12 m/s. Calculate the centripetal acceleration.
Apply Band 3 [2 marks]
A 0.40 kg ball on a 0.50 m string is whirled in a horizontal circle at 3.0 revolutions per second. Calculate the centripetal force on the ball.
Understand Band 3 [2 marks]
State Kepler's Third Law and explain what it tells us about satellites orbiting the same central body.
Analyse Band 4-5 [3 marks]
A banked curve has radius 60 m and banking angle $\\theta = 10°$.
(a) Calculate the design speed (no friction needed).
(b) If $\\mu_s = 0.20$, calculate the maximum speed before slipping up the bank.
Analyse Band 4-5 [3 marks]
A roller coaster loop has radius 4.0 m.
(a) Find the minimum speed required at the bottom of the loop to just complete the circle.
(b) If the entry speed at the bottom is 18 m/s, find the normal force at the bottom for a 60 kg rider. ($g = 9.8$ m/s²)
Apply Band 4-5 [3 marks]
A satellite orbits Earth at $r = 7.5 \\times 10^6$ m from Earth's centre. Calculate its orbital speed and period. ($G = 6.67 \\times 10^{-11}$ N m²/kg², $M_E = 5.97 \\times 10^{24}$ kg)
Analyse Band 4-5 [4 marks]
A 0.30 kg bob on a 0.70 m string moves in a vertical circle.
(a) Find the minimum speed at the top of the circle for the string to remain taut.
(b) If the speed at the top is 4.0 m/s, find the tension at the top and at the bottom. ($g = 9.8$ m/s²)
Analyse Band 6 [4 marks]
Derive the relationship between orbital velocity and escape velocity at the same radius. Starting from the energy equations for each case, show clearly that $v_e = \\sqrt{2} \\times v_{orb}$.
Analyse Band 6 [4 marks]
A 500 kg satellite moves from low Earth orbit ($r_1 = 6.8 \\times 10^6$ m) to a higher orbit ($r_2 = 1.2 \\times 10^7$ m). Calculate the energy change required and the new orbital speed. ($G = 6.67 \\times 10^{-11}$ N m²/kg², $M_E = 5.97 \\times 10^{24}$ kg)
Evaluate Band 6 [5 marks]
A 1000 kg car travels over a hill of radius 20 m at 15 m/s. Calculate the normal force at the top of the hill. The car then descends into a valley of radius 15 m at the same speed. Calculate the normal force at the bottom of the valley. Explain the difference in normal forces in terms of centripetal force requirements and the direction of acceleration. ($g = 9.8$ m/s²)
Timed Exam Block
4

Three Extended Response Questions

8 minutes per question. Set a timer and work under exam conditions.
08:00

Q11. Conical Pendulum and Energy Analysis [4 marks]

Suggested time: 8 minutes | Band 5-6

A conical pendulum consists of a 0.50 kg mass on a 1.2 m string, tracing out a horizontal circle of radius 0.40 m with constant speed.

  1. Draw a free-body diagram showing all forces on the mass.
  2. Calculate the tension in the string.
  3. Calculate the speed of the mass.
  4. If the string is shortened to 0.80 m while keeping the same radius, explain how the speed and period change, using energy concepts.

Q12. Banked Curve with Friction [4 marks]

Suggested time: 8 minutes | Band 5-6

A race track has a banked curve of radius 80 m with banking angle 15°. The coefficient of static friction between tyres and track is 0.35.

  1. Calculate the design speed for which no friction is required.
  2. Derive expressions for the maximum and minimum speeds the car can travel without slipping, considering friction acts both up and down the bank.
  3. Calculate the numerical values for $v_{max}$ and $v_{min}$.
  4. Explain what happens if the car travels faster than $v_{max}$.

Q13. Orbital Mechanics — Geostationary Transfer [4 marks]

Suggested time: 8 minutes | Band 5-6

A 2000 kg satellite is to be moved from a parking orbit ($r_1 = 7.0 \\times 10^6$ m) to geostationary orbit ($r_2 = 4.22 \\times 10^7$ m).

  1. Calculate the orbital speed in the parking orbit and in geostationary orbit.
  2. Calculate the total energy change required for this transfer.
  3. Explain why the total energy change is positive even though the satellite slows down in the higher orbit.
  4. A common student error is to use $\\Delta E = \\frac{1}{2}m(v_2^2 - v_1^2)$. Explain why this is incorrect and what the correct approach is.
Answers

Model Answers — All Questions

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Q1. Car on curve [2 marks]

$$a_c = \\frac{v^2}{r} = \\frac{(12 \\text{ m/s})^2}{30 \\text{ m}} = \\frac{144}{30} = 4.8 \\text{ m/s}^2$$

Award 1 mark for correct substitution, 1 mark for answer with units.

Q2. Ball on string [2 marks]

$\\omega = 2\\pi f = 2\\pi \\times 3.0 = 6\\pi$ rad/s

$$F_c = m\\omega^2 r = 0.40 \\times (6\\pi)^2 \\times 0.50 = 0.40 \\times 36\\pi^2 \\times 0.50 = 71 \\text{ N}$$

(Accept $F_c = mv^2/r$ approach: $v = 2\\pi rf = 2\\pi \\times 0.50 \\times 3 = 3\\pi$ m/s, $F_c = 0.40 \\times (3\\pi)^2/0.50 = 71$ N)

Award 1 mark for correct angular velocity or linear speed, 1 mark for final answer.

Q3. Kepler's Third Law [2 marks]

Kepler's Third Law states: $T^2 = \\dfrac{4\\pi^2}{GM} r^3$, or equivalently $\\dfrac{T^2}{r^3} = \\text{constant}$ for all bodies orbiting the same central mass.

This tells us that the square of the orbital period is proportional to the cube of the orbital radius. All satellites orbiting Earth (regardless of their own mass) have the same $T^2/r^3$ ratio, which depends only on Earth's mass.

Award 1 mark for correct statement of the law, 1 mark for explaining what it tells us about satellites.

Q4. Banked curve [3 marks]

(a) Design speed:

$\\tan\\theta = v^2/(gr) \\Rightarrow v = \\sqrt{gr\\tan\\theta}$

$$v = \\sqrt{9.8 \\times 60 \\times \\tan 10°} = \\sqrt{9.8 \\times 60 \\times 0.176} = \\sqrt{103.5} = 10.2 \\text{ m/s}$$

(b) Maximum speed with friction:

At $v_{max}$, friction acts down the slope:

$$v_{max} = \\sqrt{gr \\frac{\\tan\\theta + \\mu_s}{1 - \\mu_s\\tan\\theta}} = \\sqrt{9.8 \\times 60 \\times \\frac{0.176 + 0.20}{1 - 0.20 \\times 0.176}}$$

$$v_{max} = \\sqrt{588 \\times \\frac{0.376}{0.965}} = \\sqrt{229} = 15.1 \\text{ m/s}$$

Award 1 mark for (a), 1 mark for correct friction formula in (b), 1 mark for numerical answer in (b).

Q5. Roller coaster loop [3 marks]

(a) Minimum speed at bottom:

For minimum speed, at the top: $v_{top}^2 = gr$ (set $N = 0$)

$v_{top}^2 = 9.8 \\times 4.0 = 39.2$ m²/s²

By energy conservation: $\\frac{1}{2}mv_{bot}^2 = \\frac{1}{2}mv_{top}^2 + mg(2r)$

$v_{bot}^2 = v_{top}^2 + 4gr = 39.2 + 4 \\times 9.8 \\times 4.0 = 39.2 + 156.8 = 196$

$v_{bot} = \\sqrt{196} = 14.0$ m/s

(b) Normal force at bottom ($v = 18$ m/s):

$$N - mg = \\frac{mv^2}{r} \\Rightarrow N = m\\frac{v^2}{r} + mg = 60 \\times \\frac{18^2}{4.0} + 60 \\times 9.8$$

$$N = 60 \\times 81 + 588 = 4860 + 588 = 5448 \\text{ N}$$

Award 1 mark for (a) with energy method, 1 mark for (b) formula, 1 mark for (b) numerical answer.

Q6. Satellite orbit [3 marks]

$$v = \\sqrt{\\frac{GM}{r}} = \\sqrt{\\frac{6.67 \\times 10^{-11} \\times 5.97 \\times 10^{24}}{7.5 \\times 10^6}} = \\sqrt{5.31 \\times 10^7} = 7.29 \\times 10^3 \\text{ m/s}$$

$v = 7.29$ km/s

$$T = \\frac{2\\pi r}{v} = \\frac{2\\pi \\times 7.5 \\times 10^6}{7.29 \\times 10^3} = 6466 \\text{ s} = 108 \\text{ min}$$

Award 1 mark for correct orbital speed, 1 mark for correct period, 1 mark for unit conversion to minutes.

Q7. Vertical circle bob [4 marks]

(a) Minimum speed at top:

At minimum speed, tension is zero: $mg = mv_{min}^2/r$

$v_{min} = \\sqrt{gr} = \\sqrt{9.8 \\times 0.70} = \\sqrt{6.86} = 2.62$ m/s

(b) Tension at top ($v_{top} = 4.0$ m/s):

$$T_{top} + mg = \\frac{mv_{top}^2}{r} \\Rightarrow T_{top} = \\frac{0.30 \\times 4.0^2}{0.70} - 0.30 \\times 9.8$$

$T_{top} = 6.86 - 2.94 = 3.92$ N

Tension at bottom:

First find $v_{bot}$ using energy: $\\frac{1}{2}v_{bot}^2 = \\frac{1}{2}v_{top}^2 + 2gr$

$v_{bot}^2 = 16 + 4 \\times 9.8 \\times 0.70 = 16 + 27.44 = 43.44$

$$T_{bot} - mg = \\frac{mv_{bot}^2}{r} \\Rightarrow T_{bot} = \\frac{0.30 \\times 43.44}{0.70} + 0.30 \\times 9.8$$

$T_{bot} = 18.62 + 2.94 = 21.6$ N

Award 1 mark for (a), 1 mark for top tension, 1 mark for finding $v_{bot}$, 1 mark for bottom tension.

Q8. Derive $v_e = \\sqrt{2} \\times v_{orb}$ [4 marks]

Step 1 — Orbital velocity:

Equating gravitational force to centripetal force: $\\dfrac{GMm}{r^2} = \\dfrac{mv_{orb}^2}{r}$

$$v_{orb} = \\sqrt{\\frac{GM}{r}}$$

Step 2 — Escape velocity:

For escape, total energy = 0 (object reaches infinity with zero kinetic energy):

$KE + PE = 0 \\Rightarrow \\frac{1}{2}mv_e^2 - \\dfrac{GMm}{r} = 0$

$$v_e = \\sqrt{\\frac{2GM}{r}}$$

Step 3 — Relate the two:

$$v_e = \\sqrt{\\frac{2GM}{r}} = \\sqrt{2} \\times \\sqrt{\\frac{GM}{r}} = \\sqrt{2} \\times v_{orb}$$

Alternative energy argument:

Total orbital energy: $E = -\\dfrac{GMm}{2r}$. To escape, we must add energy $+\\dfrac{GMm}{2r}$ to bring total to zero. This is exactly equal in magnitude to the current kinetic energy $\\frac{1}{2}\\dfrac{GMm}{r}$. Doubling the kinetic energy (hence multiplying speed by $\\sqrt{2}$) provides exactly the energy needed.

Award 1 mark for deriving $v_{orb}$, 1 mark for deriving $v_e$, 1 mark for showing the $\\sqrt{2}$ relationship, 1 mark for clear mathematical reasoning.

Q9. Satellite orbit transfer [4 marks]

Step 1 — Energy in initial orbit:

$$E_1 = -\\frac{GMm}{2r_1} = -\\frac{6.67 \\times 10^{-11} \\times 5.97 \\times 10^{24} \\times 500}{2 \\times 6.8 \\times 10^6} = -1.463 \\times 10^{10} \\text{ J}$$

Step 2 — Energy in final orbit:

$$E_2 = -\\frac{GMm}{2r_2} = -\\frac{6.67 \\times 10^{-11} \\times 5.97 \\times 10^{24} \\times 500}{2 \\times 1.2 \\times 10^7} = -8.29 \\times 10^9 \\text{ J}$$

Step 3 — Energy change:

$$\\Delta E = E_2 - E_1 = -8.29 \\times 10^9 - (-1.463 \\times 10^{10}) = +6.34 \\times 10^9 \\text{ J}$$

Step 4 — New orbital speed:

$$v_2 = \\sqrt{\\frac{GM}{r_2}} = \\sqrt{\\frac{6.67 \\times 10^{-11} \\times 5.97 \\times 10^{24}}{1.2 \\times 10^7}} = \\sqrt{3.32 \\times 10^7} = 5.76 \\times 10^3 \\text{ m/s}$$

$v_2 = 5.76$ km/s

Award 1 mark for each of $E_1$, $E_2$, $\\Delta E$, and $v_2$ with correct reasoning.

Q10. Hill and valley [5 marks]

Normal force at top of hill:

At the top, both normal force and gravity point toward the centre of curvature (downward):

$$mg - N = \\frac{mv^2}{r} \\Rightarrow N = mg - \\frac{mv^2}{r}$$

$$N = 1000 \\times 9.8 - \\frac{1000 \\times 15^2}{20} = 9800 - 11250 = -1450 \\text{ N}$$

$N = 1450$ N upward (or the car would leave the road; this is the minimum speed to stay on the hill at this radius)

Normal force at bottom of valley:

At the bottom, normal force points up (toward centre) while gravity points down:

$$N - mg = \\frac{mv^2}{r} \\Rightarrow N = mg + \\frac{mv^2}{r}$$

$$N = 1000 \\times 9.8 + \\frac{1000 \\times 15^2}{15} = 9800 + 15000 = 24800 \\text{ N}$$

Explanation:

At the top of the hill, the centripetal acceleration is directed downward toward the centre of curvature. The net downward force $(mg - N)$ provides this centripetal acceleration. If $mv^2/r \\geq mg$, the car loses contact with the road. At the bottom of the valley, the centripetal acceleration is directed upward. The normal force must overcome gravity and provide the centripetal force, so $N = mg + mv^2/r$, giving a much larger normal force. The difference arises because at the top, gravity assists the centripetal acceleration, while at the bottom, gravity opposes it.

Award 1 mark for top normal force calculation, 1 mark for bottom normal force calculation, 1 mark for correct force diagrams/directions, 2 marks for clear explanation linking to centripetal force requirements.

Q11. Conical pendulum [4 marks]

(1) Free-body Tension $T$ at angle $\\theta$ to vertical, weight $mg$ downward. Resolve $T$ into vertical ($T\\cos\\theta$) and horizontal ($T\\sin\\theta$) components.

(2) Tension: Vertical equilibrium: $T\\cos\\theta = mg$

$\\cos\\theta = \\sqrt{L^2 - r^2}/L = \\sqrt{1.44 - 0.16}/1.2 = \\sqrt{1.28}/1.2 = 0.943$

$$T = \\frac{mg}{\\cos\\theta} = \\frac{0.50 \\times 9.8}{0.943} = 5.20 \\text{ N}$$

(3) Speed: Horizontal: $T\\sin\\theta = mv^2/r$

$\\sin\\theta = r/L = 0.40/1.2 = 0.333$

$v = \\sqrt{\\frac{rT\\sin\\theta}{m}} = \\sqrt{\\frac{0.40 \\times 5.20 \\times 0.333}{0.50}} = \\sqrt{1.39} = 1.18$ m/s

(4) Effect of shortening string: Shortening the string while keeping the same radius increases the angle $\\theta$. The vertical component must still balance $mg$, so tension increases. The horizontal component $T\\sin\\theta$ increases, requiring greater centripetal force $mv^2/r$, so speed increases. Period $T_{period} = 2\\pi r/v$ decreases because $v$ increases.

Award 1 mark each for FBD, tension, speed, and explanation.

Q12. Banked curve with friction [4 marks]

(1) Design speed: $v_0 = \\sqrt{gr\\tan\\theta} = \\sqrt{9.8 \\times 80 \\times \\tan 15°} = \\sqrt{9.8 \\times 80 \\times 0.268} = \\sqrt{210} = 14.5$ m/s

(2) Expressions:

For $v_{max}$: friction acts down the slope (preventing car from sliding up).

$$v_{max} = \\sqrt{gr \\frac{\\tan\\theta + \\mu_s}{1 - \\mu_s\\tan\\theta}}$$

For $v_{min}$: friction acts up the slope (preventing car from sliding down).

$$v_{min} = \\sqrt{gr \\frac{\\tan\\theta - \\mu_s}{1 + \\mu_s\\tan\\theta}}$$

(3) Numerical values:

$v_{max} = \\sqrt{784 \\times \\frac{0.268 + 0.35}{1 - 0.35 \\times 0.268}} = \\sqrt{784 \\times \\frac{0.618}{0.906}} = \\sqrt{535} = 23.1$ m/s

$v_{min} = \\sqrt{784 \\times \\frac{0.268 - 0.35}{1 + 0.35 \\times 0.268}} = \\sqrt{784 \\times \\frac{-0.082}{1.094}} = \\text{no real solution}$

Since $\\tan\\theta < \\mu_s$, $v_{min}$ is not real — the car will not slide down at any speed (friction alone can hold it at rest).

(4) Above $v_{max}$: The required centripetal force exceeds what gravity and friction can provide. The car slides up the bank, increasing radius until either a barrier stops it or the new radius reduces the required $mv^2/r$ to a sustainable level.

Award 1 mark each for design speed, expressions, numerical values, and explanation.

Q13. Geostationary transfer [4 marks]

(1) Orbital speeds:

$v_1 = \\sqrt{GM/r_1} = \\sqrt{\\frac{6.67 \\times 10^{-11} \\times 5.97 \\times 10^{24}}{7.0 \\times 10^6}} = \\sqrt{5.69 \\times 10^7} = 7.54 \\times 10^3$ m/s = 7.54 km/s

$v_2 = \\sqrt{GM/r_2} = \\sqrt{\\frac{6.67 \\times 10^{-11} \\times 5.97 \\times 10^{24}}{4.22 \\times 10^7}} = \\sqrt{9.44 \\times 10^6} = 3.07 \\times 10^3$ m/s = 3.07 km/s

(2) Energy change:

$E_1 = -\\dfrac{GMm}{2r_1} = -\\dfrac{6.67 \\times 10^{-11} \\times 5.97 \\times 10^{24} \\times 2000}{2 \\times 7.0 \\times 10^6} = -5.69 \\times 10^{10}$ J

$E_2 = -\\dfrac{GMm}{2r_2} = -\\dfrac{6.67 \\times 10^{-11} \\times 5.97 \\times 10^{24} \\times 2000}{2 \\times 4.22 \\times 10^7} = -9.44 \\times 10^9$ J

$\\Delta E = E_2 - E_1 = -9.44 \\times 10^9 + 5.69 \\times 10^{10} = +4.75 \\times 10^{10}$ J $= 47.5$ GJ

(3) Why positive: Although the satellite slows down ($v_2 < v_1$), the increase in gravitational potential energy (less negative $U$) more than compensates for the decrease in kinetic energy. Moving to a higher orbit requires adding energy to climb out of the gravitational well — the satellite is becoming "less bound."

(4) Error in $\\Delta E = \\frac{1}{2}m(v_2^2 - v_1^2)$: This formula only accounts for kinetic energy change but ignores gravitational potential energy change. The correct approach uses total orbital energy: $\\Delta E = E_2 - E_1 = -\\dfrac{GMm}{2r_2} + \\dfrac{GMm}{2r_1}$, which includes both kinetic and potential energy contributions automatically.

Award 1 mark each for orbital speeds, energy change, explanation of positive energy, and critique of the student error.

Copy Into Books

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Circular Motion

  • $a_c = v^2/r = \\omega^2 r$
  • $F_c = mv^2/r = m\\omega^2 r$
  • $v = 2\\pi r/T = \\omega r$
  • Vertical top: $T + mg = mv^2/r$
  • Vertical bottom: $T - mg = mv^2/r$

Orbital Mechanics

  • $v = \\sqrt{GM/r}$
  • $T^2 = (4\\pi^2/GM)r^3$
  • $r = R_{planet} + h$ always
  • $KE = \\frac{1}{2}GMm/r$
  • $U = -GMm/r$ (negative!)
  • $E = -\\frac{1}{2}GMm/r$

Escape Velocity

  • $v_e = \\sqrt{2GM/r}$
  • $v_e = \\sqrt{2} \\times v_{orb}$
  • Escape energy: $+\\frac{1}{2}GMm/r$

Common Errors

  • No "centrifugal force"
  • Speed varies in vertical circles
  • Use $r = R + h$, not just $h$
  • Use $g = GM/r^2$, not 9.8
  • $v_e$ has $\\sqrt{2}$ factor
  • Mass cancels in orbit speed
Revisit Your Answer

Now that you've worked through the consolidation, revisit your initial answer about centripetal vs centrifugal force. Centripetal force is the real net force directed toward the centre of circular motion, provided by real forces such as tension, gravity, or friction. Centrifugal force is a fictitious force that appears to act outward only in a rotating (non-inertial) reference frame — it is not a real force and should never be included in free-body diagrams in inertial frames.

How has your understanding changed? Write a revised explanation:

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Interactive: Phase2 Quiz Interactive
Key Terms — Phase 2
Centripetal acceleration
$a_c = v^2/r$, directed toward the centre of circular motion
Centripetal force
$F_c = mv^2/r$; the net force required for circular motion, not a separate force
Angular velocity
$\\omega = 2\\pi f = 2\\pi/T$; rate of angle swept out in rad/s
Orbital velocity
$v = \\sqrt{GM/r}$; speed for stable circular orbit at radius $r$
Kepler's Third Law
$T^2 \\propto r^3$; relates orbital period to orbital radius
Gravitational PE
$U = -GMm/r$; always negative, zero at infinity
Total orbital energy
$E = -\\frac{1}{2}GMm/r = KE + U$; negative for bound orbits
Escape velocity
$v_e = \\sqrt{2GM/r} = \\sqrt{2} \\times v_{orb}$
Multiple Choice

1. A car moves around a horizontal circular track at constant speed. The net force on the car:

A Is zero because speed is constant
B Acts outward, balancing the centripetal force
C Points toward the centre of the circle
D Points tangential to the circle

2. A satellite in low Earth orbit has speed $v$. What is the escape velocity from the same orbital radius?

A $v/2$
B $v$
C $\\sqrt{2} \\times v$
D $2v$

3. Two satellites orbit Earth at the same altitude. Satellite A has twice the mass of Satellite B. Which statement is correct?

A Satellite A orbits faster
B Both satellites orbit at the same speed
C Satellite A has a longer period
D Satellite A requires more centripetal force but orbits at the same speed

4. At the top of a vertical circular loop, the minimum speed to maintain contact is when:

A The normal force becomes zero
B The speed is zero
C The centripetal force equals the weight
D The kinetic energy equals the potential energy

5. For a satellite moving to a higher orbit, which statement correctly describes the energy change?

A Energy decreases because the satellite slows down
B Energy stays the same by conservation of energy
C Energy increases because the satellite becomes less bound
D Only kinetic energy matters for the energy change
Short Answer

Q1. A conical pendulum has a bob of mass 0.40 kg on a 0.90 m string, tracing a horizontal circle of radius 0.30 m. Calculate the tension in the string and the period of the motion.

[3 marks] Apply Band 4-5

Q2. The Moon orbits Earth with period 27.3 days at a mean distance of $3.84 \\times 10^8$ m. Use this information to calculate the mass of Earth.

[3 marks] Analyse Band 5

Q3. A satellite in circular orbit fires its thrusters briefly in the direction of motion. Evaluate what happens to its orbit, explaining whether it moves to a higher or lower orbit and how its speed and period change.

[4 marks] Evaluate Band 6

Model Answers — Short Answer

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Q1. Conical pendulum [3 marks]

$\\cos\\theta = \\sqrt{L^2 - r^2}/L = \\sqrt{0.81 - 0.09}/0.90 = \\sqrt{0.72}/0.90 = 0.943$

Tension: $T = mg/\\cos\\theta = (0.40 \\times 9.8)/0.943 = 4.16$ N

$\\sin\\theta = r/L = 0.30/0.90 = 0.333$

$T\\sin\\theta = mv^2/r \\Rightarrow v = \\sqrt{rT\\sin\\theta/m} = \\sqrt{0.30 \\times 4.16 \\times 0.333/0.40} = 1.02$ m/s

Period: $T_{period} = 2\\pi r/v = 2\\pi \\times 0.30/1.02 = 1.85$ s

Award 1 mark for tension, 1 mark for speed, 1 mark for period.

Q2. Mass of Earth from Moon [3 marks]

$T = 27.3 \\times 24 \\times 3600 = 2.36 \\times 10^6$ s

From Kepler's Third Law: $T^2 = \\dfrac{4\\pi^2}{GM} r^3$

Rearranging: $M = \\dfrac{4\\pi^2 r^3}{GT^2}$

$$M = \\frac{4\\pi^2 \\times (3.84 \\times 10^8)^3}{6.67 \\times 10^{-11} \\times (2.36 \\times 10^6)^2} = \\frac{4\\pi^2 \\times 5.66 \\times 10^{25}}{3.72 \\times 10^2} = 6.01 \\times 10^{24} \\text{ kg}$$

Award 1 mark for period conversion, 1 mark for rearrangement, 1 mark for answer.

Q3. Satellite thrust evaluation [4 marks]

When thrusters fire in the direction of motion, the satellite's speed increases momentarily. This increases the kinetic energy, making the total energy less negative. The satellite is now moving too fast for its current circular orbit.

The increased speed means the satellite will climb to a higher orbit (apoapsis increases). As it climbs, gravitational potential energy increases (becomes less negative) while kinetic energy decreases. By conservation of angular momentum and energy, the orbit becomes elliptical with the thrust point as perigee.

If the thrust is precisely calculated, the satellite can enter a new circular orbit at a larger radius. In this new stable orbit: speed is lower than before ($v \\propto 1/\\sqrt{r}$), and the period is longer ($T \\propto r^{3/2}$).

The apparent paradox (speeding up to slow down) is resolved because the added energy goes mostly into increasing gravitational potential energy. The satellite ends up moving slower but is less tightly bound to Earth.

Award marks for: explaining initial speed increase (1), describing transfer to higher orbit (1), correctly stating final speed decreases and period increases (1), resolving the paradox with energy reasoning (1).

Boss Battle
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Boss Battle

Phase 2 Consolidation Challenge

Test your mastery of circular motion, vertical circles, banked curves, orbital mechanics, and gravitational energy under timed conditions.

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Phase 2 Consolidation — Circular Motion & Orbits

Physics subject page Module overview Checkpoint 2 Module quiz