Systems and Center of Mass
Two people stand at opposite ends of a canoe floating on a calm lake. They walk toward each other and meet in the middle. Does the canoe stay put, or does it slide? To answer, you need one idea: a single point that represents the motion of a whole group of objects. By the end of this page you will be able to say exactly where that point is and why it behaves so simply.
Choosing a system
A system is the object, or group of objects, that you decide to analyze. Everything else is the surroundings. You choose the system. A good choice makes the problem much easier.
Once you pick a system, every force falls into one of two groups:
Look at Emma pushing a box across a floor. If the system is "the box only," then Emma's push, the friction from the floor, and Earth's gravity are all external, because Emma, the floor, and Earth are outside the system. If the system is "box + Emma," her push on the box is now internal (both objects are inside). Friction on her shoes and on the box, and gravity, are still external.
Emma pushes a box across a rough floor. She defines the system as "the box only." Classify each of the following forces as internal or external to this system: Emma's…
Emma pushes a box across a rough floor. She defines the system as "the box only." Classify each of the following forces as internal or external to this system: Emma's push on the box, the friction force from the floor, and the gravitational force from Earth. Then say whether the system would change if she defined it as "box + Emma."
Finding the center of mass
Think of a seesaw with a heavy kid and a light kid. To balance it, the pivot must sit closer to the heavy kid. The balance point is a weighted average of positions, where heavier objects count more. That point is the center of mass.
Why a weighted average? Each position is multiplied by its mass, so a big mass pulls the result toward itself. The result is always between the smallest and largest positions. For a uniform object, such as a meterstick or a canoe, the center of mass is at its geometric center.
Two carts sit on a lab track. A cart of mass $0.50\ \text{kg}$ is at $x = 0$, and a cart of mass $1.50\ \text{kg}$ is at $x = 0.80\ \text{m}$. Calculate the position of…
Two carts sit on a lab track. A cart of mass $0.50\ \text{kg}$ is at $x = 0$, and a cart of mass $1.50\ \text{kg}$ is at $x = 0.80\ \text{m}$. Calculate the position of the center of mass of the two-cart system.
Three point masses lie on the $x$-axis: $2.0\ \text{kg}$ at $x = -1.0\ \text{m}$, $3.0\ \text{kg}$ at $x = 0$, and $1.0\ \text{kg}$ at $x = 4.0\ \text{m}$. Find the…
Three point masses lie on the $x$-axis: $2.0\ \text{kg}$ at $x = -1.0\ \text{m}$, $3.0\ \text{kg}$ at $x = 0$, and $1.0\ \text{kg}$ at $x = 4.0\ \text{m}$. Find the $x$-coordinate of the center of mass.
A $1.0\ \text{kg}$ block is at $(0, 0)$ and a $3.0\ \text{kg}$ block is at $(4.0\ \text{m}, 2.0\ \text{m})$. A constant external force of $8.0\ \text{N}$ in the $+x$…
A $1.0\ \text{kg}$ block is at $(0, 0)$ and a $3.0\ \text{kg}$ block is at $(4.0\ \text{m}, 2.0\ \text{m})$. A constant external force of $8.0\ \text{N}$ in the $+x$ direction acts on the $1.0\ \text{kg}$ block only. The blocks are connected by a light rod. (a) Find the coordinates of the center of mass. (b) Find the magnitude and direction of the acceleration of the center of mass.
Motion of the center of mass
Toss a wrench so it spins in the air. Every part moves in a complicated way, but one point, the center of mass, follows a smooth parabola. That is true for any system.
In a uniform gravitational field, the weight $F_g = Mg$ of the whole system can be treated as acting at the center of mass, so that is the point you use when you calculate the torque of the weight. This works because the mass that resists acceleration (inertial mass) and the mass that gravity pulls on (gravitational mass) are equivalent, so every part falls with the same $g$.
Two important consequences follow.
If the net external force is zero, then $a_{\text{cm}} = 0$: the center of mass stays at rest or moves with constant velocity. The velocity of the center of mass is
$$v_{\text{cm}} = \frac{\sum m_i v_i}{\sum m_i} = \frac{p_{\text{total}}}{M_{\text{total}}}$$so a collision or an explosion (internal forces only) cannot change it.
Another consequence: if a system starts at rest with no net external force, its center of mass does not move, even when parts of it do. That is the canoe puzzle. If Marcus walks along a canoe on still water, the canoe moves the opposite way so that $x_{\text{cm}}$ stays fixed. Set $x_{\text{cm}}$ before equal to $x_{\text{cm}}$ after and solve for the unknown shift.
Jayden (mass $60\ \text{kg}$) and his sister Aaliyah (mass $40\ \text{kg}$) stand at opposite ends of a $4.0\ \text{m}$ long boat of mass $100\ \text{kg}$ floating at…
Jayden (mass $60\ \text{kg}$) and his sister Aaliyah (mass $40\ \text{kg}$) stand at opposite ends of a $4.0\ \text{m}$ long boat of mass $100\ \text{kg}$ floating at rest on a calm lake. Treat the boat's center of mass as its midpoint, and take the origin at Jayden's end. (a) Find the center of mass of the system of boat plus two people. (b) Ignoring water resistance, the two people walk toward each other and meet at the boat's midpoint. Explain what happens to the position of the system's center of mass relative to the shore, and why.
A $2.0\ \text{kg}$ ball moves at $3.0\ \text{m/s}$ in the $+x$ direction and a $3.0\ \text{kg}$ ball moves at $2.0\ \text{m/s}$ in the $-x$ direction. No net external…
A $2.0\ \text{kg}$ ball moves at $3.0\ \text{m/s}$ in the $+x$ direction and a $3.0\ \text{kg}$ ball moves at $2.0\ \text{m/s}$ in the $-x$ direction. No net external force acts on the system. (a) Find the velocity of the center of mass. (b) Describe how this velocity changes when the balls collide with each other, and justify your answer.
Marcus ($70\ \text{kg}$) stands at one end of a uniform $6.0\ \text{m}$ canoe of mass $30\ \text{kg}$ at rest on still water. The end where he stands is $1.0\ \text{m}$…
Marcus ($70\ \text{kg}$) stands at one end of a uniform $6.0\ \text{m}$ canoe of mass $30\ \text{kg}$ at rest on still water. The end where he stands is $1.0\ \text{m}$ from a dock, and the canoe's long axis points away from the dock. Marcus walks to the other end of the canoe. Ignore water resistance. (a) Explain why the center of mass of the Marcus-canoe system does not move relative to the dock. (b) Calculate how far the canoe moves, and in which direction relative to the dock. (c) Calculate Marcus's final distance from the dock.
A firework of total mass $1.2\ \text{kg}$ is launched straight up. At the top of its flight, momentarily at rest, it explodes into two pieces. A $0.40\ \text{kg}$ piece…
A firework of total mass $1.2\ \text{kg}$ is launched straight up. At the top of its flight, momentarily at rest, it explodes into two pieces. A $0.40\ \text{kg}$ piece moves horizontally east at $30\ \text{m/s}$ immediately after the explosion. (a) Determine the velocity (magnitude and direction) of the other piece immediately after the explosion. (b) Describe the motion of the center of mass of the two pieces after the explosion, ignoring air resistance, and explain how it is consistent with your answer to (a). (c) Explain why the explosion's internal forces cannot change the center of mass's motion, while gravity can.
Back to the start: when the two people walk toward each other and meet in the middle of the boat, the boat shifts so that the system's center of mass stays fixed (the boat moves only if the two people have different masses). Water resistance is ignored, and the only forces that move the people are internal.