AP Physics 1 · Lesson 3

Gravity, weight and circular motion

Know Know it

Every pair of masses attracts: Fg=Gm1m2/r2F_g = Gm_1m_2/r^2 with G=6.67×10−11G = 6.67 \times 10^{-11} N·m²/kg², where r is the distance between the centers. Near a planet’s surface this becomes weight, Fg=mgF_g = mg, where the gravitational field strength is g=GM/R2g = GM/R^2. Mass is the same everywhere; weight depends on g.

Because of the 1/r21/r^2, doubling the distance between centers cuts the force to one quarter, and tripling it cuts the force to one ninth.

An object moving in a circle at constant speed is accelerating toward the center, with ac=v2/ra_c = v^2/r, and v=2πr/Tv = 2\pi r/T for a period T. By Newton’s second law, the forces on it must add to a net inward force of mv2/rmv^2/r.

What supplies the net inward force
SituationForce or forces pointing toward the center
Car turning on a flat roadStatic friction from the road
Ball whirled on a string in a horizontal circleTension in the string
Moon circling EarthEarth’s gravitational pull
Rider at the bottom of a dip in a trackThe normal force up, minus the weight down

Worked example Easy

A 12 kg lander hangs at rest from a spring scale on the surface of a moon, and the scale reads 45 N. What is the gravitational field strength at the moon’s surface?

  1. 0.27 N/kg
  2. 3.8 N/kgAnswer
  3. 9.8 N/kg
  4. 540 N/kg

How to solve it

  1. At rest, the scale reading equals the lander’s weight, 45 N.
  2. Field strength is force per kilogram: g=Fg/mg = F_g/m.
  3. g=45/12=3.75g = 45/12 = 3.75 N/kg, about 3.8 N/kg.

Why each choice is right or wrong

  • A. Incorrect. 0.27 N/kg divides the mass by the weight; field strength is the weight divided by the mass.
  • B. Correct. g=Fg/m=45/12=3.75≈3.8g = F_g/m = 45/12 = 3.75 \approx 3.8 N/kg.
  • C. Incorrect. 9.8 N/kg is Earth’s surface value; the scale reading on this moon gives a different one.
  • D. Incorrect. 540 N/kg multiplies the weight by the mass instead of dividing.

Trap There is no outward force

On a free-body diagram for circular motion, draw only real forces: tension, gravity, the normal force, friction. Never add a separate centripetal force or an outward force. The feeling of being thrown outward in a turning car is your body tending to keep moving in a straight line.

Explain Explain it

Task verbs: calculate, explain

For circular motion, take the direction toward the center as positive. Add the force components along that line, inward minus outward, and set the sum equal to mv2/rmv^2/r.

At the top of a vertical circle, gravity points toward the center and helps; at the bottom it points away from the center, and the normal force or tension must make up for it. That is why, at the same speed on circles of the same radius, a track pushes harder on a rider at the bottom of a dip than at the top of a loop.

Worked example Medium

A 0.25 kg toy car runs on the inside of a vertical loop of track with a radius of 0.40 m. At the top of the loop its speed is 2.5 m/s. What is the size of the normal force from the track on the car at the top?

  1. 1.5 NAnswer
  2. 2.5 N
  3. 3.9 N
  4. 6.4 N

How to solve it

  1. At the top, the center is straight down, so the normal force (down) and gravity (down) both point inward.
  2. FN+mg=mv2/rF_N + mg = mv^2/r, with mv2/r=(0.25)(6.25)/0.40=3.91mv^2/r = (0.25)(6.25)/0.40 = 3.91 N and mg=2.45mg = 2.45 N.
  3. FN=3.91−2.45=1.46F_N = 3.91 - 2.45 = 1.46 N, about 1.5 N.

Why each choice is right or wrong

  • A. Correct. At the top both the normal force and gravity point down, toward the center: FN+mg=mv2/rF_N + mg = mv^2/r. So FN=(0.25)(2.52/0.40)−(0.25)(9.8)=3.91−2.45≈1.5F_N = (0.25)(2.5^2/0.40) - (0.25)(9.8) = 3.91 - 2.45 \approx 1.5 N.
  • B. Incorrect. 2.5 N is about the car’s weight, 2.45 N, which is a different force from the push of the track.
  • C. Incorrect. 3.9 N is the whole net inward force, mv2/rmv^2/r; gravity supplies 2.45 N of it, so the track supplies only the rest.
  • D. Incorrect. 6.4 N adds the weight to mv2/rmv^2/r, which treats gravity as pointing away from the center, as it does at the bottom of the loop.

Explain Your turn

A car drives around a circular roundabout of radius 18 m at a steady 9.0 m/s. Calculate the size of its acceleration. Then explain why the car is accelerating even though its speedometer reading does not change, and say which way the acceleration points.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Gives the size of the acceleration as 4.5 m/s²
  • Explains that the direction of the car’s velocity keeps changing, and any change in velocity is an acceleration
  • Says the acceleration points toward the center of the roundabout

Model answer

ac=v2/r=(9.0)2/18=4.5a_c = v^2/r = (9.0)^2/18 = 4.5 m/s². Acceleration is any change in velocity, and velocity includes direction. Going around the roundabout, the car’s direction of motion changes from moment to moment even though its speed stays 9.0 m/s, so its velocity is changing and the car is accelerating. The acceleration points toward the center of the roundabout, the direction in which the car is being turned.

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