University Physics I · Applications of Newton's Laws · 5.7
Inertial and Accelerating Frames
Newton’s second law holds only where a free object keeps constant velocity. Anywhere else you transform back to such a frame, or add the inertial terms that one demands.
Build the model
Connect the measurement to the mechanism.
Σ F = m a is a statement about a kind of frame, not only about forces. An inertial frame is one in which an object free of interactions keeps constant velocity, and any frame moving at constant velocity relative to it qualifies too. Ride in a frame that accelerates at A and free objects appear to accelerate at −A with nothing touching them, so the law fails as written.
Two repairs work. Transform back to an inertial frame and use only real forces, or stay put and add one inertial term, −mA per object, plus centrifugal and Coriolis terms if the frame rotates. That term names no source object and has no third-law partner, and because it is proportional to mass it imitates gravity exactly.
Every measurable result — a scale reading, a string tension, an angle — comes out the same either way, so the frame is a choice about convenience, never about physics.
- Simple definition
- A non-inertial frame is one in which a free object does not keep constant velocity, so Newton’s second law needs extra inertial terms that name no source object.
- Example
- A bus brakes and a bag slides forward along the seat; from the road the bag simply keeps moving, while in the bus frame you add a forward inertial force to explain it.
Every force in the sum names a source object. This version needs no extra terms.
Unit: N = kg m s⁻²
Velocities and accelerations differ between frames; the list of real forces does not.
A: frame’s acceleration · m s⁻²
Add −mA to every object in the frame, then solve exactly as if the frame were inertial.
Ffict in newtons · one per object
For HORIZONTAL A = 1.2 m s⁻², |geff| = 9.88 m s⁻², tilted 7.0° back from vertical. Add in quadrature only when A is horizontal; for a vertical A the magnitudes simply add or subtract.
θ: plumb-line tilt from vertical
Present for every object in a frame rotating at ω, even one at rest in that frame.
ω in rad s⁻¹ · r: distance from the axis, in m
Acts only on objects moving in the rotating frame, always perpendicular to v′.
φ: angle between ω and v′
The law names a frame before it names a force.
Σ F = m a is not true everywhere. It is true in an inertial frame: a frame in which an object free of interactions keeps constant velocity. That is an experimental test, not a definition you can assert. Once you have found one such frame, every frame moving at constant velocity relative to it passes the same test, so no single inertial frame is privileged. Now let the frame accelerate. Place a ball on the smooth floor of a carriage that accelerates forward at 1.2 m s⁻². Nothing touches the ball horizontally, yet in the carriage frame it accelerates backward at 1.2 m s⁻². Σ F = 0 while a ≠ 0. The law has not been disproved; it has been used outside the domain where it applies.
Differentiate the position twice and the extra term appears.
Let the primed frame have its origin at R(t) in an inertial frame, with no rotation. Any object’s position obeys r = R + r′. Differentiate twice: a = A + a′, where A = d²R/dt² is the frame’s acceleration. The real forces are the same list in both frames — the same hands, cords, and surfaces — so Σ F = m a = m(A + a′). Rearranged, m a′ = Σ F − mA. The primed observer keeps Newton’s form provided one term, −mA, is added to every object. Nothing was assumed about what acts on the object: the term is fixed entirely by the frame’s acceleration and the object’s mass.
An inertial force has no source and no partner.
Test any force by asking which object exerts it. Tension names a cord, the normal force names a surface, weight names Earth. The term −mA names nothing; it comes from the frame’s motion. Three consequences follow. It has no third-law partner, so stop hunting for an equal and opposite force somewhere else. It is exactly proportional to mass, so it gives every object the same acceleration and therefore imitates a uniform gravitational field — inside that carriage a plumb line hangs along an effective gravity of √(9.81² + 1.2²) = 9.88 m s⁻², tilted 7.0° backward. And it vanishes the moment you transform to an inertial frame. None of that makes it optional. Inside the accelerating frame the term is compulsory: omit it and the frame’s own equations give the wrong tension and the wrong angle.
Rotating frames add two terms, not one.
A frame rotating at constant angular velocity ω needs a centrifugal term m ω² r outward from the axis, plus a Coriolis term −2m ω × v′ on anything already moving in the frame. A habitat of radius 100 m reproduces 9.81 m s⁻² with ω = 0.313 rad s⁻¹, one turn every 20.1 s, under 3 rev/min; walk its rim and Coriolis changes your apparent weight depending on which way you go. On Earth ω = 7.29 × 10⁻⁵ rad s⁻¹ gives ω²R = 0.0339 m s⁻² at the equator, 0.35% of g — the larger term, but fixed in size and direction and already folded into the g your balance reads, so bench work never sees it. Coriolis cannot hide, because it depends on velocity: its horizontal part, 2ωv sin(latitude), pushes a 30 m s⁻¹ parcel at latitude 45° sideways at 3.1 × 10⁻³ m s⁻² — 20 km in an hour, turning pressure gradients into weather systems.
Everyday cases, worked both ways.
A 65 kg person in a lift accelerating upward at 1.5 m s⁻². Ground frame: N − mg = ma, so N = m(g + a) = 65 × 11.31 = 735 N. Lift frame: the person is at rest, so N − mg − ma = 0 returns the same 735 N, and a scale calibrated in kilograms reads 735 ÷ 9.81 = 74.9 kg. A 70 kg passenger in a car turning on a 25 m radius at 12 m s⁻¹. Ground frame: seat and door supply a net inward force of m v²/r = 403 N, and the passenger’s body would otherwise travel straight. Car frame: the passenger is at rest, and a 403 N outward centrifugal term is balanced by that same 403 N of inward contact force. The frames never disagree about the reading on the scale or the force from the door. They disagree only about what is doing the pushing.
Declare the frame before you draw the diagram.
Write the frame down as the first line of the solution, then commit to it. If it is inertial, draw real forces only. If it accelerates in a straight line at A, draw the real forces and add −mA to every object; if it rotates, add the centrifugal and Coriolis terms too. Never mix the two: a diagram carrying a centrifugal term alongside a ground-frame acceleration counts the same physics twice. Choose the accelerating frame when it makes objects stationary — a passenger, a crate on a truck bed, water in a towed tank — because a′ = 0 turns dynamics into statics. Choose the inertial frame when the interactions matter more than the geometry. Finish by checking one measurable quantity in both: the same tension, the same reading, the same angle. Disagreement means a wrong sign, or the frame’s acceleration counted twice.
Change one variable at a time
Make the relationship visible.
Raise A and the string tilts further back, so the tilt reads the frame's acceleration. Then change the mass: every force arrow grows in the same proportion, the angle does not move — that is why −mA imitates gravity.
TILT θ17.0 °
EFFECTIVE g10.26 m s⁻²
INERTIAL TERM mA2.40 N
STRING TENSION T8.21 N
Live interpretationTILT θ: 17.0 °. EFFECTIVE g: 10.26 m s⁻². INERTIAL TERM mA: 2.40 N. STRING TENSION T: 8.21 N
Catch the common trap
Explain before calculating.
A carriage accelerates forward at 2.0 m s⁻² along level track. A small steel ball on a light string hangs from the ceiling at a steady angle θ. Take g = 9.81 m s⁻². Which statement is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 58 kg student stands on a bathroom scale in a lift that accelerates downward at 2.4 m s⁻². Work in the lift frame to find the scale reading, then check it from the ground. Take g = 9.81 m s⁻².
- Declare the frame: the lift, accelerating at A = 2.4 m s⁻² downward. The student is at rest in it, so a′ = 0.
- Real forces: weight mg = 58 × 9.81 = 569 N down, and the scale's normal force N up.
- Add the one inertial term. A points down, so −mA points up, with magnitude 58 × 2.4 = 139 N.
- a′ = 0 means the three sum to zero: N + 139.2 − 568.98 = 0, so N = 429.8 N.
- Ground-frame check: N − mg = m(−A), so N = m(g − A) = 58 × 7.41 = 430 N. The same reading, with no inertial term anywhere in it.
AnswerN = 430 N — the scale reads 43.8 kg while the student's mass is still 58 kg.
MediumA 24 kg crate sits on a flat truck bed with μs = 0.35 between crate and bed. Working in the truck's frame, find the largest forward acceleration the truck can have before the crate slides. Take g = 9.81 m s⁻².
- Declare the frame: the truck bed, accelerating forward at A. While the crate does not slide it is at rest in that frame, so a′ = 0.
- Vertical: nothing accelerates vertically in either frame, so N = mg = 24 × 9.81 = 235.4 N.
- Horizontal: the inertial term −mA points backward with magnitude 24A, and static friction is the only real horizontal force, so it must supply f = 24A forward.
- Friction runs out at μs N = 0.35 × 235.4 = 82.4 N, so 24A = 82.4 N gives A = 3.43 m s⁻².
- The mass cancels — A(max) = μs g = 0.35 × 9.81 = 3.43 m s⁻² — because the inertial term and the friction limit both scale with m.
AnswerA(max) = 3.4 m s⁻², whatever the crate's mass.
HardA rotating habitat of radius 90 m is spun so its rim gives 9.81 m s⁻². A 70 kg crew member walks along the rim at 1.5 m s⁻¹ in the direction of the spin. Find the force the floor exerts on her, using the rotating frame, and check it from an inertial frame.
- Spin rate first: the rim must supply g, so ω = √(g/r) = √(9.81/90) = 0.330 rad s⁻¹ — one turn every 2π/ω = 19.0 s.
- Standing still, only the centrifugal term acts: mω²r = 70 × 9.81 = 687 N outward, so the floor pushes 687 N inward. That is the design weight.
- Walking with the spin switches on Coriolis: |−2mω × v′| = 2 × 70 × 0.330 × 1.5 = 69.3 N, and for prograde motion it points outward.
- She is not at rest in this frame either: circling the rim at 1.5 m s⁻¹ needs a′ = v′²/r = 1.5²/90 = 0.025 m s⁻² inward, worth ma′ = 1.75 N.
- Inward positive: ma′ = N − mω²r − 2mωv′, so N = 1.75 + 686.7 + 69.3 = 758 N.
- Inertial check: her true speed is ωr + v′ = 29.71 + 1.50 = 31.21 m s⁻¹, so N = mv²/r = 70 × 31.21² ÷ 90 = 758 N. Same number, no inertial terms used.
AnswerN = 758 N walking with the spin — against 687 N standing still and 619 N walking against it, about ±10% of apparent weight at walking pace.