Get Ready: Module 2 — Circular Motion, Gravitation & Torque
Everyone starts somewhere. Before we spin bodies on circles and balance beams, let's lay the floor before the building — one small skill at a time. Go at your own pace; there's no rush and nothing to prove here.
Skills to Build First
Five small building blocks. Master these and Module 2 will feel like the next natural step, not a wall.
Resolving a Vector into Components
What it is & why you need it: A vector at an angle can be split into a horizontal part \(A_x = A\cos\theta\) and a vertical part \(A_y = A\sin\theta\). In Module 2 you'll separate velocities and forces this way — the centripetal force points one direction, gravity another — and you can only add forces along the same axis.
Worked example: split a \(10\) N force at \(30^\circ\) above the horizontal
- Draw the force as the hypotenuse of a right triangle, with the angle measured from the horizontal.
- The horizontal piece uses cosine: \(A_x = A\cos\theta = 10\cos 30^\circ = 10(0.866) = 8.66\) N.
- The vertical piece uses sine: \(A_y = A\sin\theta = 10\sin 30^\circ = 10(0.5) = 5\) N.
- Check: \(\sqrt{A_x^2 + A_y^2} = \sqrt{8.66^2 + 5^2} \approx 10\) N — back to the original magnitude.
Remember: the side next to the angle uses cosine; the side across from it uses sine. "Cosine hugs the angle."
1. A \(20\) N force points \(60^\circ\) above the horizontal. Find its horizontal and vertical components.
2. A velocity of \(8\) m/s points at \(45^\circ\). What are its \(x\)- and \(y\)-components?
Show answer
2. At \(45^\circ\), \(\cos 45^\circ = \sin 45^\circ = 0.707\), so \(v_x = v_y = 8(0.707) = 5.66\) m/s.
Newton's Second Law \(\Sigma F = ma\)
What it is & why you need it: The net force on an object equals its mass times its acceleration: \(\Sigma F = ma\). In Module 2 the acceleration happens to point toward the center of a circle, so the centripetal force is just \(\Sigma F = ma_c\) — the same law, aimed inward.
Worked example: net force on a \(3\) kg cart accelerating at \(4\) m/s²
- Write the law: \(\Sigma F = ma\). Identify \(m = 3\) kg and \(a = 4\) m/s².
- Substitute: \(\Sigma F = (3)(4)\).
- Multiply: \(\Sigma F = 12\) N.
- To go the other way, divide: if a \(12\) N net force acts on the cart, \(a = \dfrac{\Sigma F}{m} = \dfrac{12}{3} = 4\) m/s².
The same triangle works three ways: \(F = ma\), \(a = \dfrac{F}{m}\), and \(m = \dfrac{F}{a}\). Cover the one you want.
1. A \(5\) kg object accelerates at \(2\) m/s². Find the net force.
2. A \(6\) N net force acts on a \(2\) kg block. Find its acceleration.
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2. \(a = \dfrac{\Sigma F}{m} = \dfrac{6}{2} = 3\) m/s².
Arc Length and Radian Measure
What it is & why you need it: An angle in radians is arc length divided by radius: \(\theta = \dfrac{s}{r}\), so the arc swept is \(s = r\theta\). A full circle is \(2\pi\) radians and its arc length (circumference) is \(2\pi r\). In Module 2 this is how a spinning object's angle turns into the distance it travels and its speed.
Worked example: arc length for \(2\) radians on a \(3\) m radius
- Use \(s = r\theta\), with \(r = 3\) m and \(\theta = 2\) rad.
- Substitute: \(s = (3)(2)\).
- Multiply: \(s = 6\) m of arc.
- For a whole lap, \(\theta = 2\pi\), so \(s = r(2\pi) = 2\pi r\) — the circumference, \(2\pi(3) \approx 18.85\) m.
Radians have no units — they're a ratio of two lengths. That's why \(s = r\theta\) comes out in meters whenever \(r\) is in meters.
1. Find the arc length swept by \(\theta = 4\) rad on a wheel of radius \(0.5\) m.
2. Find the full circumference (one lap) of a circle with radius \(5\) m. Leave it in terms of \(\pi\) or round.
Show answer
2. \(C = 2\pi r = 2\pi(5) = 10\pi \approx 31.42\) m.
Inverse-Square Proportional Reasoning
What it is & why you need it: When a quantity depends on \(\dfrac{1}{r^2}\), doubling the distance makes it \(\dfrac{1}{4}\) as strong, tripling makes it \(\dfrac{1}{9}\). Newton's law of gravitation \(F = \dfrac{GMm}{r^2}\) works exactly this way. Being able to reason "the distance tripled, so the force is one-ninth" saves you from re-plugging every number.
Worked example: gravity at double the distance
- Start from \(F = \dfrac{GMm}{r^2}\). Hold the masses fixed and change only \(r\).
- Replace \(r\) with \(2r\): the denominator becomes \((2r)^2 = 4r^2\).
- So the new force is \(\dfrac{GMm}{4r^2} = \dfrac{1}{4}\) of the old force.
- If the original pull was \(80\) N, then at double the distance it is \(\dfrac{80}{4} = 20\) N.
Shortcut: the force scales by \(\left(\dfrac{r_{\text{old}}}{r_{\text{new}}}\right)^2\). Square the ratio of distances — that's the whole trick.
1. A gravitational force is \(90\) N. If the distance triples, what is the new force?
2. If a force of \(12\) N is felt at \(4\) m, what force is felt at \(2\) m (half the distance)?
Show answer
2. Half the distance \(\Rightarrow\) force is \(\left(\dfrac{4}{2}\right)^2 = 4\) times stronger: \(12(4) = 48\) N.
Reading Values from a Data Table
What it is & why you need it: Many physics problems hand you the numbers in a table and expect you to pull the right ones into a formula. In Module 2 you'll read a planet's mass and a satellite's orbital radius from a table, then drop them into \(v = \sqrt{GM/r}\) or \(T = \dfrac{2\pi r}{v}\). Picking the correct row and column — with the right units — is half the battle.
Worked example: pull \(M\) and \(r\) from a table
- Suppose a table lists: Planet A — mass \(M = 6.0\times10^{24}\) kg, orbit radius \(r = 7.0\times10^{6}\) m.
- Match each value to its symbol: \(M\) is the central mass, \(r\) is the orbital radius.
- Check the units before substituting: kilograms for mass, meters for radius — both standard, so no conversion is needed.
- Now they're ready to use, e.g. \(v = \sqrt{\dfrac{GM}{r}}\) with these exact numbers.
Always write the symbol next to the number you copy out: "\(M = 6.0\times10^{24}\) kg." It keeps you from swapping a mass for a radius mid-problem.
1. A table gives a satellite's orbital radius as \(r = 4.0\times10^{7}\) m and period as \(T = 8.6\times10^{4}\) s. Which value would you use as the radius in \(v = \dfrac{2\pi r}{T}\)?
2. The same table lists "mass of central body \(= 5.0\times10^{24}\) kg." Which symbol does that number belong to?
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2. It belongs to \(M\), the central mass — the one inside \(F = \dfrac{GMm}{r^2}\) and \(v = \sqrt{GM/r}\).
Quick Readiness Check
Six short questions across all five skills. Try each one first, then peek at the answer.
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Split a \(10\) N force at \(30^\circ\) above the horizontal into components.
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\(F_x = 10\cos 30^\circ \approx 8.66\) N; \(F_y = 10\sin 30^\circ = 5\) N. -
A \(4\) kg object accelerates at \(3\) m/s². Find the net force.
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\(\Sigma F = ma = 4(3) = 12\) N. -
Find the arc length swept by \(\theta = 5\) rad on a \(2\) m radius.
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\(s = r\theta = 2(5) = 10\) m. -
Find the circumference of a circle of radius \(4\) m.
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\(C = 2\pi r = 2\pi(4) = 8\pi \approx 25.13\) m. -
A gravitational force is \(36\) N. The distance doubles. What is the new force?
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Double the distance \(\Rightarrow\) one-fourth the force: \(\dfrac{36}{4} = 9\) N. -
A table lists \(M = 6.0\times10^{24}\) kg and \(r = 7.0\times10^{6}\) m. Which is the orbital radius?
Show answer
The radius is \(r = 7.0\times10^{6}\) m; \(M\) is the central mass.
If these feel comfortable, you're ready for the module. If a couple felt tricky, that's fine — revisit that skill card above and drill a little with the free links below. No shame, just steps.
Practice More — Free
Want extra reps? These free sites let you drill each skill until it clicks. (Links open in a new tab.)
Resolving vectors
Newton's second law
Arc length & radians
Inverse-square reasoning
Reading data tables
When You're Ready →
No rush. When the skills above feel solid, step into Module 2.
AP Physics is a self-study enrichment track on Dr. Ijezie's STEM Studio — not a TEKS-graded course. There is no syllabus and nothing to submit; these Foundations pages exist only to make the physics easier to pick up.