core34 min studyLesson 116 in Physics

Accelerating coordinates require inertial-force terms

Transform Newton's equation between inertial and accelerating frames, then distinguish translational, centrifugal, Coriolis, and Euler terms.

On this page

What you’ll understand

  • Derive the inertial term for a translating accelerated frame.
  • Identify each term in the rotating-frame acceleration relation.
  • Compare force descriptions of the same event in two frames.

A passenger in an accelerating train sees a loose object move backward even when no ordinary horizontal interaction pulls it backward. The apparent conflict with Newton's law comes from using an accelerating coordinate system as though it were inertial. An inertial-force term repairs that bookkeeping. It predicts the same physical motion as the inertial-frame description and must be derived from the observer's acceleration, rather than added as an unexplained new interaction.

A translating origin shifts every acceleration

Let an inertial observer describe position r, and let the train's origin have position R. If the train's axes stay parallel to the inertial axes, the passenger's relative position is r minus R. Differentiate twice: the passenger measures the physical object's inertial acceleration minus the train origin's acceleration. Substituting this relation into Newton's law leaves the usual interaction-force sum and an additional term equal to minus mass times origin acceleration. This term depends on the coordinate system and acts on every mass in proportion to its inertial mass.

The inertial term does not have a third-law partner on a second object because it is not a new interaction. A real seat or floor force can nevertheless be needed to hold a passenger stationary relative to the accelerating vehicle. In the inertial description that contact force produces the passenger's acceleration. In the vehicle description it balances the inertial term. These are two coordinate descriptions of one contact force and one trajectory. Mixing the vehicle's zero relative acceleration with the inertial-frame force equation would incorrectly conclude that the contact force must vanish.

ma′=∑Finteraction−mAm\mathbf a'=\sum\mathbf F_{\mathrm{interaction}}-m\mathbf A
For nonrotating axes whose origin has inertial acceleration A, a′ is the object's acceleration relative to that origin. The term −mA is a coordinate correction; interaction forces retain their physical sources.↔ Scroll wide equations horizontally.

Rotating axes create three additional structures

A rotating basis changes even a vector whose physical direction is fixed. Its inertial derivative equals its rotating-frame derivative plus the angular velocity crossed into that vector. Applying this relation twice gives terms proportional to relative velocity, position, and the time derivative of angular velocity. Their negatives appear on the force side of the rotating-frame equation. Keep the frame's angular velocity and origin acceleration distinct: one describes orientation change, while the other describes translation. A merry-go-round can rotate about an origin whose translational acceleration is zero.

The centrifugal term points outward from a fixed rotation axis. The Coriolis term depends on velocity measured in the rotating frame and vanishes for an object fixed in that frame. The Euler term appears when the frame's angular velocity changes. For constant rotation it vanishes, but centrifugal and Coriolis terms may remain. These terms are not interchangeable labels for any curved motion. In the inertial description of a body in circular motion, an inward resultant interaction force supplies centripetal acceleration; no centrifugal interaction must be added to cancel it in that inertial equation.

ma′=F−mA−2mΩ×v′−mΩ×(Ω×r′)−mΩ˙×r′m\mathbf a'=\mathbf F-m\mathbf A-2m\boldsymbol\Omega\times\mathbf v'-m\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r')-m\dot{\boldsymbol\Omega}\times\mathbf r'
The terms after F are translational, Coriolis, centrifugal, and Euler corrections respectively. All vectors are expressed in a common instantaneous basis; r′ and v′ refer to the rotating frame, and A is the origin's inertial acceleration.↔ Scroll wide equations horizontally.

Choose one frame and verify the same observation

An effective gravity can combine actual gravitational acceleration with a position-dependent centrifugal term when analyzing objects nearly stationary on a steadily rotating platform. It does not absorb the velocity-dependent Coriolis term for moving objects. Similarly, apparent weight measured by a scale is a contact force, not a direct reading of gravitational attraction alone. An upward-accelerating lift increases the scale reading; a freely falling lift gives zero support force in the ideal point-particle model even though gravitational acceleration remains nonzero in the nearby inertial approximation.

A reliable frame audit names the observer, defines which velocities are relative, and checks a case with no interactions. A free particle moves along a straight line at constant speed in an inertial frame but can trace a curved path in rotating coordinates. Both descriptions must transform into each other. In laboratory work, Earth's rotation can be neglected when its accumulated effects are below the desired uncertainty; it becomes important for long flight times, sensitive pendulums, and large-scale flows. Frame choice is a convenience, while the accuracy requirement decides which transformation terms can safely be omitted.

A parcel in an accelerating train

A 2 kg2\,\mathrm{kg} parcel stays at rest relative to a train accelerating horizontally right at 3 m s−23\,\mathrm{m\,s^{-2}}. Its floor provides static friction, with g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. Find the required friction and minimum static coefficient.

  1. In the ground frame the parcel accelerates right, so its horizontal contact force must be f=ma=2(3)=6 Nf=ma=2(3)=6\,\mathrm N rightward. Vertical balance gives N=20 NN=20\,\mathrm N.

  2. In the nonrotating train frame the parcel has zero relative acceleration. Include a −mA=−6 N-mA=-6\,\mathrm N inertial term leftward; it balances the same 6 N rightward friction.

  3. The no-slip condition requires 6≤μs206\leq\mu_s20, so the minimum coefficient in the ideal model is μs=0.30\mu_s=0.30. A smaller coefficient would invalidate the assumed rest relative to the train.

The result

The two frames agree on the 6 N physical contact force and the friction threshold. Only the train-frame equation includes the translational inertial term.

Explore the model

Interactive lab

Interactive lab · vectors

Resolve the frame acceleration

Set angle to 0° and magnitude to 2 for a normalized frame acceleration A, then change to 90°. Predict the opposite components of the translational inertial acceleration −A before checking the projections. The lab shows vector projections only; it does not transform a moving frame or integrate a passenger trajectory.

Resolve the frame accelerationPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn inertial-force correction follows a specified coordinate acceleration.

Check your understanding

An object is fixed relative to axes rotating at a constant angular velocity about a fixed origin. Which rotating-frame inertial term is necessarily zero?

Try it yourself

A 3 kg3\,\mathrm{kg} mass is fixed 0.5 m0.5\,\mathrm m from the center of a horizontal platform rotating uniformly at 2 rad s−12\,\mathrm{rad\,s^{-1}}. Find the required horizontal interaction force and explain its rotating-frame balance.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Compute inward centripetal acceleration in the inertial frame, then use zero relative velocity on the platform.

Compare with the solution

The inertial acceleration magnitude is Ω2r=4(0.5)=2 m s−2\Omega^2r=4(0.5)=2\,\mathrm{m\,s^{-2}}, so a 6 N horizontal interaction force must act inward. In the rotating frame, the mass is stationary and the 6 N outward centrifugal term balances that same inward interaction force. Coriolis and Euler terms vanish because relative velocity and angular acceleration are zero.

Take it with you

  • Inertial forces arise from accelerating coordinates and are not new pair interactions.
  • Centrifugal, Coriolis, and Euler terms have different position, velocity, and time dependencies.
  • A single event must agree after transforming between consistently used frames.

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