AP Physics C: Mechanics Unit 2: Force and Translational Dynamics Notes

Introduction

Translational dynamics begins with a declared system and uses the external-force resultant to determine center-of-mass acceleration.

A free-body diagram is an interaction ledger, not a picture of motion. Draw only forces exerted on the chosen system, identify their agents, and project those forces onto useful axes before writing Newton’s law.

The force models in this unit have conditions. Dry friction is piecewise, an ideal spring is linear only over a stated range, linear drag is a low-speed model, and inverse-square gravity relies on source geometry. State the model before solving its differential equation.

System boundary

separates internal interactions from external forces

Newton’s laws

connect interaction vectors to center-of-mass motion

Force models

supply gravitational, contact, spring, and drag terms

Radial dynamics

projects real forces toward a curved path’s center

Current exam guidance: Aligned to the current 2026–27 AP Physics C: Mechanics framework, with calculus-based modeling, derivations, experimental reasoning, and explicit assumptions.

2.1

Systems and Center of Mass

How can a many-particle system have one translational equation?

Short answer

A mass-weighted position defines the center of mass, and internal force pairs cancel when the particle equations are summed, leaving external force to govern its motion.

Know firstVector components · Definite integrals · Newton’s third-law pairs

The boundary decides which interactions survive

For discrete masses, $\vec r_{cm}=M^{-1}\sum_i m_i\vec r_i$ ; for a continuous body, replace the sum with $M^{-1}\int\vec r\,dm$ . The center of mass is a weighted coordinate and need not lie inside material. A system may also change shape, separation, or internal energy while its center-of-mass motion stays simple.

For constant total mass, differentiating twice and summing particle equations gives $M\vec a_{cm}=\sum\vec F_{ext}$ . Internal third-law pairs cancel in this whole-system sum, although they can rearrange constituents and transfer energy internally. Matter or energy crossing the boundary must be entered explicitly; mass flow requires momentum-flux accounting.

Physics model

The center of mass is a weighted position

Change one mass and see how the system center shifts while the coordinate origin stays fixed.

Right mass4 kg

Left mass: 2 kg at −3 m
Right mass: 4 kg at +3 m
Center: 1 m

left massvariable right masscenter of masscoordinate origin

x_{cm}=\frac{\sum_i m_i x_i}{\sum_i m_i}=\frac{1}{M}\int x\,dm

Read this figure without the graphic

Change one mass and see how the system center shifts while the coordinate origin stays fixed.

Current quantity	Value
Left mass	2 kg at −3 m
Right mass	4 kg at +3 m
Center	1 m
Model relation	$x_{cm}=\frac{\sum_i m_i x_i}{\sum_i m_i}=\frac{1}{M}\int x\,dm$

Locate a nonuniform rod’s center of mass

System: A thin rod occupying $0\le x\le L$ .
Frame: One-dimensional laboratory axis along the rod.
Assumptions: Linear density is $\lambda(x)=kx$ with $k>0$ .
Givens: $dm=kx\,dx$ .

Find total mass: $M=\int_0^Lkx\,dx=kL^2/2$ .
Compute the first moment: $\int_0^Lx\,dm=\int_0^Lkx^2\,dx=kL^3/3$ .
Divide first moment by total mass.

Result

x_{cm}=\frac{2L}{3}

Units: The result is a length measured from the low-density end.

Boundary cases

A uniform rod would have $x_{cm}=L/2$ .
Changing the origin changes the coordinate but not the physical point.

Because density increases toward the right end, the center of mass lies at two thirds of the rod length.

Choose a center-of-mass model

System	Expression	Key condition
point masses	$M^{-1}\sum m_i\vec r_i$	all positions use one frame
continuous body	$M^{-1}\int\vec r\,dm$	density supplies $dm$
constant-mass motion	$M\vec a_{cm}=\sum\vec F_{ext}$	no mass crosses boundary

Build the system equation

Draw the boundary and list what crosses it.
Compute total mass and weighted position.
Classify every interaction as internal or external.
Cancel internal pairs only after summing all constituent equations.
Apply the result to center-of-mass motion, not automatically to each part.

Keep these three ideas

Center of mass is a mass-weighted coordinate.
Internal pairs cancel only in a whole-system sum.
External resultant controls constant-mass center-of-mass acceleration.

Force and Translational Dynamics