[{"content":"Welcome to gdpark.blog — notes on physics, finance, CFA, programming, and AI.\nMath rendering test:\n$$ \\vec{r}_{cm} = \\frac{\\sum \\vec{r}_i m_i}{\\sum m_i} $$Inline math works too: $E = mc^2$.\n","permalink":"https://gdpark.blog/posts/hello-world/","summary":"\u003cp\u003eWelcome to \u003cstrong\u003egdpark.blog\u003c/strong\u003e — notes on physics, finance, CFA, programming, and AI.\u003c/p\u003e\n\u003cp\u003eMath rendering test:\u003c/p\u003e\n$$\n\\vec{r}_{cm} = \\frac{\\sum \\vec{r}_i m_i}{\\sum m_i}\n$$\u003cp\u003eInline math works too: $E = mc^2$.\u003c/p\u003e","title":"Hello, world"},{"content":"Dynamics of a particle system So suppose the system we\u0026rsquo;re dealing with…\n…is made of many particles, each at its own position.\nHandling all of them individually is painfully complicated — so we bring in the concept of the center of mass (CM).\nRough idea: lump all the particles together and pretend the total mass sits at a single point.\nLet\u0026rsquo;s start with the formula (it\u0026rsquo;s a definition, not an axiom):\n$$ \\vec{r}_{\\mathrm{cm}} = \\frac{\\sum \\vec{r}_i \\, m_i}{\\sum m_i} $$Stare at it for a moment — a vector divided by a scalar, so the result is still a vector. Good.\nNow look at the numerator.\nThe position of the $i$-th particle,\n$$\\vec{r}_i$$is multiplied by that particle\u0026rsquo;s mass,\n$$m_i$$and we sum them all.\nSo:\nA particle with larger mass gets its position vector amplified in the sum. A particle with mass less than one (in whatever units) gets its position vector shrunk. Then all those weighted vectors are added together — that\u0026rsquo;s the numerator.\nImagine every $m_i$ is the same. Then\n$$\\sum \\vec{r}_i m_i$$is a uniformly-weighted sum of position vectors — which will \u0026ldquo;lean toward\u0026rdquo; wherever the particles are clustered. Makes sense.\nSo we\u0026rsquo;re taking each position vector $\\vec{r}_i$, scaling it by the scalar weight $m_i$, and summing. Dividing by the total mass gives us a weighted average.\nIn other words, the CM vector is just the mass-weighted average position of the particles.\nNow let the particles move.\nBy the same logic, the CM velocity should be the mass-weighted average of each particle\u0026rsquo;s velocity, divided by the total mass. Let\u0026rsquo;s see what that actually equals:\n$$ \\vec{v}_{\\mathrm{cm}} = \\frac{\\sum m_i \\dot{\\vec{r}}_i}{\\sum m_i} = \\frac{\\sum m_i \\frac{d}{dt}\\vec{r}_i}{\\sum m_i} = \\frac{\\frac{d}{dt}\\sum m_i \\vec{r}_i}{\\sum m_i} = \\frac{d}{dt}\\vec{r}_{\\mathrm{cm}} = \\dot{\\vec{r}}_{\\mathrm{cm}} $$The time derivative of the CM position vector is exactly the CM velocity vector. Obvious? Yeah, obvious. But it\u0026rsquo;s nice to see it fall out cleanly.\nCM acceleration follows the same pattern — I\u0026rsquo;ll skip writing that one out.\nOriginally written in Korean on my Naver blog (Nov 2014). Translated to English for gdpark.blog.\n","permalink":"https://gdpark.blog/posts/center-of-mass/","summary":"Treat a system of many particles as if all the mass were concentrated at a single point. That\u0026rsquo;s what the center of mass is for.","title":"Center of Mass (CM) — Classical Mechanics Notes #13"}]