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AP Physics · enrichment

Circular Motion, Gravitation & Torque — Visual Lab

Module 2. Spin a body around a central mass and watch the three quantities every circular-motion problem asks for — centripetal acceleration, period, and the inward force — update in real time. Then balance a beam and see why a torque is a force times a lever arm.

Interactive Lab Module 02 · Orbits & Torque a_c = v²/r · F = GMm/r² · τ = rF
Standards & honesty AP Physics · enrichment — not a TEKS-graded course. This module is part of an optional, self-paced enrichment track. It is not aligned to a Texas Essential Knowledge and Skills (TEKS) standard, there is nothing to submit, and nothing here is graded. The physics is real and the numbers in the lab are physically correct — circular motion, Newton's law of gravitation, and rotational equilibrium — but treat it as enrichment, not coursework. If you get stuck, reach out through Student Support.

When something moves in a circle, it is always accelerating — not because it speeds up, but because its direction keeps changing. That acceleration points straight at the center (it's centripetal), and Newton's second law says some real force must supply it. For a planet or a satellite, that force is gravity. The lab below lets you set an orbit's radius and speed and reads back the centripetal acceleration \(a_c = \dfrac{v^2}{r}\), the period \(T = \dfrac{2\pi r}{v}\), and the force the orbit demands — then a second panel balances a beam to show that torque is the rotational cousin of force.

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Orbital Sandbox & Torque-Balance Lab

Set the orbital radius \(r\) and speed \(v\) and watch the body circle a central mass; the readout shows the exact centripetal acceleration, period, and the gravitational force the orbit requires. Toggle Gravity sets the speed to snap to a closed orbit \(v=\sqrt{GM/r}\). The lower panel is a torque-balance beam: \(\Sigma\tau = 0\) solves the force that levels the plank.

Orientation

What you're seeing

  • The inward arrow is the net force. A body on a circle is pulled toward the center; that centripetal acceleration is \(a_c = v^2/r\), and the force behind it is \(F_c = mv^2/r\).
  • The tangent arrow is the velocity. The speed never changes here, but the velocity vector turns constantly — which is exactly why there is an acceleration.
  • Gravity can be that force. \(F = \dfrac{GMm}{r^2}\). Setting it equal to \(\dfrac{mv^2}{r}\) gives the speed that closes a circular orbit: \(v = \sqrt{GM/r}\).
  • Bigger orbit, longer period. The period \(T = \dfrac{2\pi r}{v}\) is the circumference over the speed — the time for one full lap.
  • Torque needs a lever arm. On the beam, the same force does more turning the farther it sits from the pivot: \(\tau = r\,F\). Balance is \(\Sigma\tau = 0\).
Investigation

Try this

  1. Hold the radius and raise the speed. Watch \(a_c = v^2/r\) climb — double the speed and the centripetal acceleration quadruples (it's the speed squared).
  2. Hold the speed and grow the radius. The period \(T = 2\pi r / v\) gets longer and \(a_c\) drops — a wider, lazier circle.
  3. Flip on "Gravity sets the speed." The speed snaps to \(\sqrt{GM/r}\). Now the gravitational pull is exactly the centripetal force the orbit needs — a closed, self-consistent orbit.
  4. On the beam, move the right arm in and out. Pull it closer to the pivot and the balancing force grows; push it out and the force shrinks — same torque, traded between force and lever arm.
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Worked Examples

Three of the moves this module asks for most: a centripetal-force problem, the speed of a circular orbit under gravity, and balancing torques on a beam.

Example 1 — Centripetal force on a car rounding a curve

a_c = v²/r, then F_c = m a_c
  1. Identify the circle. A \(1200\)-kg car takes a flat curve of radius \(r = 50\) m at \(v = 15\) m/s. The turn is circular motion.
  2. Centripetal acceleration. \(a_c = \dfrac{v^2}{r} = \dfrac{15^2}{50} = \dfrac{225}{50} = 4.5\) m/s².
  3. Centripetal force. Newton's second law: \(F_c = m a_c = 1200(4.5) = 5400\) N — supplied here by friction between the tires and the road.
  4. Sanity check. Take the curve faster and \(F_c\) climbs with the speed squared; that's why a too-fast turn skids.
Answer: \(a_c = 4.5\) m/s² and \(F_c = 5400\) N. Check it in the lab by reading the centripetal-acceleration row.

Example 2 — Speed of a satellite in a circular orbit

Gravity supplies the centripetal force

A satellite orbits a planet of mass \(M = 6.0\times10^{24}\) kg at an orbital radius \(r = 7.0\times10^{6}\) m. Find its orbital speed. Use \(G = 6.67\times10^{-11}\).

  1. Set gravity = centripetal force. \(\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}\). The satellite mass \(m\) cancels.
  2. Solve for the speed. \(v = \sqrt{\dfrac{GM}{r}}\) — notice the orbiting body's own mass drops out entirely.
  3. Substitute. \(v = \sqrt{\dfrac{(6.67\times10^{-11})(6.0\times10^{24})}{7.0\times10^{6}}}\).
  4. Evaluate. The numerator inside is \(\approx 5.72\times10^{7}\), so \(v \approx \sqrt{5.72\times10^{7}} \approx 7560\) m/s.
Answer: \(v \approx 7.56\times10^{3}\) m/s. Toggle "Gravity sets the speed" in the lab to see the closing-speed row light up.

Example 3 — Balancing a seesaw with torque

Στ = 0 ⟹ F₁d₁ = F₂d₂
  1. Write the balance condition. A beam balances when the two torques about the pivot are equal: \(F_1 d_1 = F_2 d_2\).
  2. Fill in what you know. A \(60\)-N weight sits \(2\) m left of the pivot, and a second weight sits \(3\) m to the right: \(60(2) = F_2(3)\).
  3. Solve. \(F_2 = \dfrac{60(2)}{3} = \dfrac{120}{3} = 40\) N.
  4. Interpret. The farther weight needs less force because it has a longer lever arm — torque trades force for distance.
Answer: \(F_2 = 40\) N at \(3\) m. Reproduce it in the beam panel with \(F_1 = 60,\ d_1 = 2,\ d_2 = 3\).
Why it matters The same three equations run from a hammer thrower's swing to the Moon. Centripetal force explains why a car skids on a fast turn and why water stays in a bucket spun overhead; gravitation sets the speed of every satellite, the length of a year, and the orbits GPS depends on; and torque is why a long wrench loosens a stubborn bolt, how a door swings, and why a tightrope walker carries a long pole. Master \(a_c = v^2/r\), \(F = GMm/r^2\), and \(\tau = rF\) and you can predict the motion of anything that turns.
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Key Vocabulary

The precise words a physicist uses to describe what the lab is doing.

Centripetal acceleration

The center-pointing acceleration of a body moving on a circle, \(a_c = \dfrac{v^2}{r}\). It changes the direction of the velocity, not its size.

Centripetal force

The net inward force that keeps a body on a circular path, \(F_c = \dfrac{mv^2}{r}\). It is supplied by something real — gravity, tension, friction, or a normal force.

Period

The time for one full revolution, \(T = \dfrac{2\pi r}{v}\) — the circumference divided by the speed.

Universal gravitation

Every pair of masses attracts: \(F = \dfrac{G m_1 m_2}{r^2}\), with \(G = 6.67\times10^{-11}\). The force falls off with the square of the distance.

Torque

The turning effect of a force about a pivot, \(\tau = rF\sin\theta\). It depends on both the force and how far from the pivot it acts (the lever arm).

Rotational equilibrium

A body whose torques cancel, \(\Sigma\tau = 0\), so it does not start to spin. A balanced beam is the simplest case.

What this lab covers

Circular motion, gravitation & rotation

This Visual Lab targets the circular-motion and rotation ideas of an algebra-based first physics course: uniform circular motion and the centripetal relationship, Newton's law of universal gravitation and circular orbits, angular velocity and momentum, and torque and rotational equilibrium. It is enrichment — outside the TEKS math sequence — so explore it for understanding, not for a grade.

Centripetal \(a_c = v^2/r\), \(F_c = mv^2/r\) · Gravitation \(F = GMm/r^2\), \(v = \sqrt{GM/r}\) · Rotation \(\omega = v/r\), \(L = I\omega\), \(\tau = rF\), \(\Sigma\tau = 0\). AP Physics · enrichment, not a graded course.

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AP Physics is a self-study enrichment track on this site — not a TEKS-graded course, with no syllabus and nothing to submit. The physics and the formulas are standard algebra-based first-year content; the worked numbers here are physically correct.