AP Physics Notes: Center of Mass & Newton's Laws – Quick Reference

2.1 Center of Mass and Environment

  • System definition: A defined collection of interacting objects, for example, two children, a ball, and a wagon. The environment encompasses all external influences acting on the system, such as Earth (exerting gravity) and the driveway (providing normal force and friction). Air resistance is often assumed negligible in simplified scenarios.

  • Internal vs external forces:

    • Internal forces are forces exerted between components within the defined system (e.g., child pushing the ball, child pulling the wagon). According to Newton's Third Law, these forces occur in equal and opposite pairs that cancel each other within the system. Therefore, internal forces do not change the total momentum of the system and consequently do not affect the velocity of the system's center of mass (COM).

    • External forces originate from outside the system and act upon its components (e.g., gravity from Earth, friction from the driveway, an external push or pull). Only net external forces can cause a change in the system's total momentum or accelerate its COM.

  • External forces and COM acceleration:

    • A net external force on a system will cause its center of mass to accelerate, as described by Newton's Second Law for a system: \sum \vec{F}\{ ext{ext}} = M \{ ext{tot}} \vec{a}\{ ext{COM}}, where \sum \vec{F}\{ ext{ext}} is the vector sum of all external forces, M
      \{ ext{tot}} is the total mass of the system, and \vec{a}\{ ext{COM}} is the acceleration of the center of mass.

    • Examples of external forces that can accelerate the COM include wind pushing the system, friction at the wheels or driveway opposing motion, a slope causing a gravitational component along motion, or an external push from a person.

  • COM during back-and-forth throwing:

    • If the sum of all external forces acting on the system (e.g., children + ball + wagon) is zero, the system's total momentum remains conserved. If the system starts from rest, its COM will remain at rest, meaning its position does not change, even if components within the system move relative to each other.

    • The internal act of throwing and catching a ball causes internal momentum changes within the system components that perfectly cancel out, ensuring the system's overall momentum and COM velocity remain constant (zero, in this case).

  • Ball moves left, wagon moves right (COM balance):

    • When the ball moves one way, the rest of the system (wagon + children) must move in the opposite direction to keep the overall COM fixed, assuming no net external forces. This is a direct consequence of momentum conservation.

    • The lighter component (ball) will undergo a larger displacement for a given momentum change, while the heavier component (wagon + children) will experience a smaller displacement, such that the weighted average position (COM) remains unchanged. Mathematically, \Delta X
      \{ ext{COM}} = \frac{m \{ ext{ball}} \Delta x
      \{ ext{ball}} + m \{ ext{wagon}} \Delta x
      \{ ext{wagon}}}{M \{ ext{tot}}} = 0.

  • Missed ball (ball stops in grass):

    • When the ball exits the system and stops (e.g., due to friction from the grass), that friction becomes an external force acting on the ball, which is now considered part of the environment relative to the original system. More importantly, if the ball was part of the 'children + ball + wagon' system and then leaves it, the system's total mass changes. However, if the question implies the ball stops interacting with the wagon system and encounters friction, then this external friction from the grass on the ball (which if still interacting, is an external force on the system) causes the COM of the original system (if it still includes the ball) to accelerate/change position as a whole, because the friction is an unbalanced external force.

  • Center of mass formula (conceptual):

    • In one dimension (1D), the position of the center of mass (X
      \_{ ext{COM}}) for a system of discrete particles is given by the weighted average of their positions:

    • General formula: \boxed{X
      \{ ext{COM}} = \frac{\sum m \i x
      \i}{\sum m \i} = \frac{\sum m
      \i x \i}{M
      \{ ext{tot}}}} where m \i is the mass of the i-th particle, x
      \i is its position, and M \{ ext{tot}} is the total mass of the system (\sum m
      \_i).

  • Quick takeaway: The motion of a system's center of mass is solely determined by the net external force acting on the system, in accordance with Newton's Second Law (\vec{F}\{ ext{net, ext{ext}}} = M \{ ext{tot}} \vec{a}\_{ ext{COM}}). Internal motions or forces within the system do not cause the COM to accelerate, as they cancel each other out.

2.1 Scenario 2 and 2.3 (brief notes on COM)

  • Scenario 2 (two ice skaters):

    • When two ice skaters push off each other, the forces they exert are internal to the two-skater system. Assuming a frictionless ice surface (no net external horizontal forces), the total momentum of the system is conserved, and thus their center of mass will remain at the same position throughout the interaction. The skaters move apart, but their individual motions compensate to keep the system's COM fixed.

  • Scenario 3 (meterstick + 3M mass):

    • For a system with multiple discrete masses at known positions, the COM is found by taking the weighted average of their positions (or calculating moments).

    • Example calculations:

      • Q9 setup: Consider a system with a mass of 3M at position x = 0 and a mass of M at position x = 4 units. The total mass is 4M.

      • X
        \_{ ext{COM}} = \frac{(0 \times 3M) + (4 \times M)}{4M} = \frac{4M}{4M} = 1 \text{ unit}. The COM is 1 unit from X=0.

      • Q10 added a 2M mass at x = 7: Now the system has three masses: 3M at x = 0, M at x = 4, and 2M at x = 7. The new total mass is 3M + M + 2M = 6M.

      • X
        \_{ ext{COM}} = \frac{(0 \times 3M) + (4 \times M) + (7 \times 2M)}{6M} = \frac{0 + 4M + 14M}{6M} = \frac{18M}{6M} = 3 \text{ units}. The COM is 3 units from X=0.

  • Q11: Penguin of uniform density:

    • For objects with uniform density and a symmetrical shape (like a simplified penguin with a clear axis of symmetry), the center of mass will lie precisely on its axis of symmetry at its geometric center. This is because the mass is distributed evenly around that point.

  • Q12: Heavier monkey vs lighter monkey on a rope:

    • Consider the system of two monkeys. The primary external forces acting on this system are their individual gravitational forces (weights) acting downwards. If they are on a rope and moving, their internal forces (e.g., pulling on the rope against each other) do not affect the system's COM motion. However, the net external force for the system is the sum of their weights ((m
      \{ ext{heavier}} + m \{ ext{lighter}})g) minus any upward tension exerted by the rope from its fixed point. If this net external force is unbalanced (e.g., the rope cannot support their combined weight, or if they are accelerating downwards due to gravity), the COM of the two-monkey system will accelerate downward.

  • Q13: Jupiter–Sun barycenter concept:

    • The barycenter is the center of mass of two or more orbiting bodies. For the Jupiter-Sun system, due to Jupiter's immense mass (it's the most massive planet), the barycenter of the two bodies actually lies outside the Sun's physical radius, although very close to its surface. This means both Jupiter and the Sun orbit this common center of mass. For most other planets in our solar system, their barycenter with the Sun lies within the Sun’s diameter because their masses are much smaller relative to the Sun's.

2.2 Free-Body Diagrams (FBDs): Basics and Common Points

  • FBD essentials:

    • A Free-Body Diagram (FBD) is a visual representation used to analyze the forces acting on a single object or a defined system of objects. To construct an FBD:

      • Isolate the object/system: Draw the object as a single particle or a simplified geometric shape.

      • Identify all forces: Determine all external forces acting on the object (e.g., gravitational force (\vec{F}\{g} = m\vec{g}), normal force (\vec{N}) from contact surfaces, friction (static \vec{f}\{s} or kinetic \vec{f}\{k}), tension (\vec{T}) from ropes/strings, air resistance (\vec{R}), applied push/pull (\vec{F}\{ ext{app}})).

      • Draw vectors: Represent each force as an arrow originating from the object's center (or point of application), pointing in the direction of the force. Label each vector clearly.

      • Do not include internal forces: Internal forces within a multi-object system are not shown on the FBD of the entire system, but would appear if individual objects within the system were diagrammed separately.

  • 7. Action–reaction pairs in a system:

    • According to Newton's Third Law, for every action, there is an equal and opposite reaction. An action-reaction pair involves two forces that:

      • Are equal in magnitude and opposite in direction.

      • Act on different objects.

      • Are of the same type (e.g., gravitational, normal, tension).

    • Examples: The force of a bar on a person is an action, and the force of the person on the bar is the reaction. The gravitational force of the Earth on a person (weight) is an action, and the gravitational force of the person on the Earth is the reaction counterpart (often considered negligible due to Earth's large mass).

  • 8–12 True/False quick checks with explanations:

    • 8: True — A contact force is indeed a macroscopic interaction where two objects are physically touching. This interaction arises from the electromagnetic forces between the atoms at their surfaces.

    • 9: False — At the microscopic scale, the fundamental nature of a contact force is primarily electromagnetic repulsion between the electron clouds of the atoms of the interacting surfaces, not gravitational.

    • 10: False — Objects or systems cannot exert forces on themselves. Forces always involve an interaction between two distinct objects. An object exerts a force on another object, which in turn exerts a reaction force back on the first object.

    • 11: True — To show the components of a force, you typically resolve the force vector into perpendicular components (e.g., horizontal and vertical, or parallel and perpendicular to an inclined plane). These component vectors can be drawn on a separate diagram for clarity or annotated directly on the FBD alongside the original force vector.

    • 12: False — The normal force by definition is a contact force that acts perpendicular (normal) to the surface of contact, pushing outwards from the surface. It is not parallel to the surface; forces parallel to the surface are typically friction forces.

2.3 Center of Mass Calculations (specific problems)

(This section reiterates and expands on calculations introduced briefly in 2.1 Scenario 2 and 2.3)

  • Q9: Calculating COM for two discrete masses:

    • Given a mass of 3M at x = 0 and a mass of M at x = 4 units. The total mass of the system is M
      \_{ ext{tot}} = 3M + M = 4M.

    • Using the COM formula: X
      \{ ext{COM}} = \frac{(m \1 x
      \1) + (m \2 x
      \2)}{M \{ ext{tot}}} = \frac{(3M \times 0) + (M \times 4)}{4M} = \frac{4M}{4M} = 1 \text{ unit}. The center of mass is located at x = 1 unit.

  • Q10: Calculating COM with an added third mass:

    • Adding a third mass of 2M at x = 7 units to the previous system. The new total mass is M
      \_{ ext{tot}} = 3M + M + 2M = 6M.

    • The new COM position: X
      \_{ ext{COM}} = \frac{(3M \times 0) + (M \times 4) + (2M \times 7)}{6M} = \frac{0 + 4M + 14M}{6M} = \frac{18M}{6M} = 3 \text{ units}. The center of mass has shifted to x = 3 units.

  • Q11: Penguin with uniform distribution – COM revisited:

    • For any object with a perfectly uniform mass distribution (uniform density) and a symmetrical shape, its center of mass coincides with its geometric center. This is a powerful shortcut for complex shapes if symmetry and density are known.

  • Q12: Heavier monkey vs lighter on a rope – COM acceleration revisited:

    • When considering the system of two monkeys, the net external force is largely dominated by their combined weights acting downwards. If they are actively scaling a rope, the tension in the rope is an external force upward. However, if the question implies a scenario where their internal interactions lead to a change in their collective state, or if the rope tension itself is insufficient to support their full weight, the system's COM will accelerate downward due to the net gravitational force on the system. The phrase "