Center of Mass Cheatsheet

Cheatsheet Content

Understanding the Center of Mass (COM) Imagine trying to balance a weirdly shaped object on your fingertip. The magical spot where it balances perfectly? That's the Center of Mass! It's the average position of all the mass in a system. For us, it's like finding the "balance point" of an object or a collection of particles. COM of a Multiparticle System When you have a bunch of tiny friends (particles) scattered around, their collective balance point is calculated by summing up their individual contributions. Position Vector: $$ \vec{R}_{COM} = \frac{\sum m_i \vec{r}_i}{\sum m_i} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + \dots + m_n\vec{r}_n}{M_{total}} $$ where $M_{total} = \sum m_i$. Coordinates: $$ X_{COM} = \frac{\sum m_i x_i}{\sum m_i} $$ $$ Y_{COM} = \frac{\sum m_i y_i}{\sum m_i} $$ $$ Z_{COM} = \frac{\sum m_i z_i}{\sum m_i} $$ COM of a Two-Particle System The simplest case! If you have two buddies, $m_1$ and $m_2$, their COM always lies on the line connecting them. It's closer to the heavier friend! If $m_1$ is at $x_1$ and $m_2$ at $x_2$: $$ X_{COM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} $$ Important Ratio: The distance of the COM from each mass is inversely proportional to their masses: $m_1 r_1 = m_2 r_2$. COM of Continuous Mass Distribution When mass isn't just in tiny dots but spread out like a delicious cake, we use integration! We chop the cake into infinitesimally small pieces ($dm$) and sum their contributions. Position Vector: $$ \vec{R}_{COM} = \frac{\int \vec{r} \,dm}{\int dm} = \frac{\int \vec{r} \,dm}{M_{total}} $$ Coordinates: $$ X_{COM} = \frac{\int x \,dm}{M_{total}} $$ $$ Y_{COM} = \frac{\int y \,dm}{M_{total}} $$ $$ Z_{COM} = \frac{\int z \,dm}{M_{total}} $$ Mass Element ($dm$): For a 1D rod (length $L$, linear mass density $\lambda = M/L$): $dm = \lambda \,dx$ For a 2D plate (area $A$, surface mass density $\sigma = M/A$): $dm = \sigma \,dA$ For a 3D body (volume $V$, volume mass density $\rho = M/V$): $dm = \rho \,dV$ COM of Standard Bodies (Uniform Mass Density) When the mass is spread evenly, the COM often coincides with the geometric center. Easy peasy! Uniform Rod: At its midpoint. Uniform Ring/Hollow Cylinder: At its geometric center. Uniform Disc/Solid Cylinder: At its geometric center. Uniform Sphere (Solid/Hollow): At its geometric center. Uniform Triangle: At the intersection of its medians (centroid). Uniform Cone (Solid): On its axis, at a height $h/4$ from the base. Uniform Cone (Hollow): On its axis, at a height $h/3$ from the base. Semicircular Arc (Radius R): $(0, \frac{2R}{\pi})$ with origin at center of diameter. Semicircular Disc (Radius R): $(0, \frac{4R}{3\pi})$ with origin at center of diameter. COM of Bodies (Non-uniform Mass Density) Uh oh, the cake isn't uniform! This means $\lambda$, $\sigma$, or $\rho$ aren't constant. You'll need to use the integration method, carefully expressing $dm$ in terms of the varying density function. Example: If $\lambda = kx$ for a rod from $x=0$ to $x=L$, then $dm = kx \,dx$. COM of a Combined Structure If you stick several standard shapes together (like LEGOs!), treat each component as a point mass located at its own COM. Then, use the multiparticle system formula. Strategy: Find the mass ($m_i$) and COM ($\vec{r}_i$) of each individual component. Use the formula: $\vec{R}_{COM} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$. COM in Cavity Problems (2 Methods) Imagine you have a big cookie and you scoop out a smaller cookie from it. How does the COM shift? Think of the scooped-out part as "negative mass" or use subtraction! Method 1: Negative Mass (The "Ghost" Method) Treat the object with a cavity as a complete object (positive mass $M_1$) and a hypothetical object filling the cavity (negative mass $-M_2$). The COM of the remaining body is: $$ \vec{R}_{COM} = \frac{M_1 \vec{r}_1 - M_2 \vec{r}_2}{M_1 - M_2} $$ where $\vec{r}_1$ is COM of the full object, $\vec{r}_2$ is COM of the cavity (if it were filled). This works because removal of mass is like adding negative mass. Method 2: Subtraction This is essentially the same as Method 1 but thinking of it as removing mass. Find the COM of the original body. Find the COM of the removed part. Then use the formula above. It's often clearer to use the negative mass concept. Displacement of COM If particles in a system move, the COM moves too! The displacement of the COM is directly related to the displacement of individual particles. For a multiparticle system: $$ \Delta \vec{R}_{COM} = \frac{\sum m_i \Delta \vec{r}_i}{\sum m_i} $$ If no external force acts on the system, the COM remains at rest or continues to move with constant velocity (i.e., its displacement is zero if it started from rest). Example: A boy walks on a boat. The boy moves one way, the boat moves the other way, but if there's no external friction with water, the COM of the boy+boat system doesn't move horizontally. Velocity of COM The COM isn't just a static point; it can move! Its velocity is the weighted average of the velocities of all the particles. $$ \vec{V}_{COM} = \frac{\sum m_i \vec{v}_i}{\sum m_i} = \frac{d\vec{R}_{COM}}{dt} $$ Total Momentum: The total linear momentum of the system is $P_{total} = M_{total} \vec{V}_{COM}$. This is a big deal! If $\vec{F}_{ext} = 0$, then $\vec{V}_{COM} = \text{constant}$. This means the COM either stays put or moves in a straight line at a constant speed. Acceleration of COM If external forces are acting, the COM will accelerate. Newton's Second Law applies beautifully to the COM! $$ \vec{A}_{COM} = \frac{\sum m_i \vec{a}_i}{\sum m_i} = \frac{d\vec{V}_{COM}}{dt} $$ Newton's Second Law for a System: $$ \vec{F}_{ext} = M_{total} \vec{A}_{COM} $$ where $\vec{F}_{ext}$ is the net external force acting on the system. Internal forces (like particles colliding with each other) do NOT affect the acceleration of the COM. This is super powerful!