Introduction
The center of mass (CM) is the weighted average position of mass in a system. It moves as if all mass were concentrated there and all external forces acted there. Understanding CM simplifies analysis of complex systems, allowing us to separate internal motions from overall translation.
Definition of Center of Mass
For discrete masses: R_CM = (Σ m_i r_i)/(Σ m_i) = (1/M)Σ m_i r_i. For continuous distribution: R_CM = (1/M)∫ r dm. In components: X_CM = (1/M)Σ m_i x_i, Y_CM = (1/M)Σ m_i y_i, Z_CM = (1/M)Σ m_i z_i. The CM position depends on mass distribution, not on coordinate system choice.
Velocity and Acceleration of CM
CM velocity: V_CM = dR_CM/dt = (1/M)Σ m_i v_i = P_total/M. CM acceleration: A_CM = dV_CM/dt = (1/M)Σ m_i a_i. Key result: M A_CM = F_ext (total external force). The CM moves according to Newton's Second Law as if it were a single particle of mass M subject to net external force.
\nIntroduction
The center of mass (CM) is the weighted average position of mass in a system. It moves as if all mass were concentrated there and all external forces acted there. Understanding CM simplifies analysis of complex systems, allowing us to separate internal motions from overall translation.
Definition of Center of Mass
For discrete masses: R_CM = (Σ m_i r_i)/(Σ m_i) = (1/M)Σ m_i r_i. For continuous distribution: R_CM = (1/M)∫ r dm. In components: X_CM = (1/M)Σ m_i x_i, Y_CM = (1/M)Σ m_i y_i, Z_CM = (1/M)Σ m_i z_i. The CM position depends on mass distribution, not on coordinate system choice.
Velocity and Acceleration of CM
CM velocity: V_CM = dR_CM/dt = (1/M)Σ m_i v_i = P_total/M. CM acceleration: A_CM = dV_CM/dt = (1/M)Σ m_i a_i. Key result: M A_CM = F_ext (total external force). The CM moves according to Newton's Second Law as if it were a single particle of mass M subject to net external force.
\nImportance in Problem Solving
CM motion separates from internal motion. For isolated systems (F_ext = 0), CM moves with constant velocity regardless of complex internal motions. Exploding projectile fragments have CM continuing original trajectory. Walking person moves CM to maintain balance. Jumping - CM follows parabola even as body parts move. This separation simplifies many problems.
Calculating CM for Various Objects
Uniform rod: at midpoint. Uniform disk: at center. Uniform triangle: at centroid (intersection of medians). Semicircle: at 4R/(3π) from center along symmetry axis. System of multiple objects: treat each as point mass at its own CM. For composite objects with holes, treat hole as negative mass. Symmetry often determines CM location without calculation.
\nApplications
Applications: Balancing objects (support must be under CM for stable equilibrium); Figure skating (pulling arms in moves mass closer to rotation axis); High jumping (Fosbury flop - pass body over bar while CM goes under); Stability analysis (lower CM is more stable); Vehicle design (CM height affects rollover risk); Robotic locomotion (controlling CM position).
Solved Example: Three-Mass System CM
Three masses are placed: m1 = 2 kg at (0, 0), m2 = 3 kg at (4, 0) m, m3 = 5 kg at (2, 3) m. Find CM position. Solution: X_CM = (m1x1 + m2x2 + m3x3)/(m1+m2+m3) = (2×0 + 3×4 + 5×2)/10 = (0 + 12 + 10)/10 = 22/10 = 2.2 m. Y_CM = (2×0 + 3×0 + 5×3)/10 = (0 + 0 + 15)/10 = 1.5 m. CM is at (2.2, 1.5) m. If system rotates about this point, it will be balanced (no net torque from gravity). If system is in uniform gravitational field, this is where you could support entire system with single force.
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