--- title: "Equations of motion for plane motion" subject: "Engineering Mechanics" chapter: 6 key: "semester_3/engineering_mechanics:equations-plane-motion-rigid-body" source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/equations-plane-motion-rigid-body site: BCT Engineering --- # Equations of motion for plane motion Engineering Mechanics · Chapter 6: Plane Kinetics of Rigid Bodies: Force, Mass and Acceleration - Asked in: year not given, 6 marks - Video: [Rigid Bodies Equations of Motion General Plane Motion (Learn to solve any question)](https://www.youtube.com/watch?v=UNuRhIHthhY) — Question Solutions - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/equations-plane-motion-rigid-body - Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-6 ## Question State the equations of motion of a rigid body in plane motion. Explain how they are applied to (a) translation, (b) rotation about a fixed axis and (c) general plane motion, including the use of the moment equation about a fixed point and about the centre of mass. ## Short answer The equations of motion are ΣF_x = m a_Gx, ΣF_y = m a_Gy, and ΣM_G = I_G α. For translation α=0 and ΣM_G=0. For rotation about a fixed axis O, ΣM_O = I_O α with I_O = I_G + m r̄². For general plane motion, use ΣM_P = I_G α + moment of m a_G about P and add kinematic constraints. ## Answer For a rigid body of mass $m$ in plane motion with mass centre G, acceleration $\mathbf{a}_G$ and angular acceleration $\alpha$, acted on by external forces and couples in the plane, the equations of motion (Newton-Euler equations) are: $$ \Sigma F_x = m\,a_{Gx} \qquad \Sigma F_y = m\,a_{Gy} \qquad \Sigma M_G = I_G\,\alpha $$ The first two give the motion of the mass centre (it moves as if all the mass and forces were at G); the third gives the rotation about G. $I_G$ is the mass moment of inertia about the axis through G perpendicular to the plane. Alternatively, the **inertia force** $-m\mathbf{a}_G$ (through G) and the **inertia couple** $-I_G\alpha$ can be added and the body treated as being in static equilibrium (D'Alembert's principle). #### (a) Translation $\alpha = 0$, so $\Sigma M_G = 0$ and the resultant force passes through G: $\Sigma F = m\mathbf{a}_G$. Example: a crate sliding on a surface (check tipping with $\Sigma M_G = 0$). #### (b) Rotation about a fixed axis O G moves in a circle, with $a_{Gt} = \alpha\bar r$ and $a_{Gn} = \omega^2\bar r$. Taking moments about O: $$ \Sigma M_O = I_O\,\alpha \qquad I_O = I_G + m\bar r^2 $$ The reactions at O follow from $\Sigma F_t = m\bar r\alpha$ and $\Sigma F_n = m\bar r\omega^2$. #### (c) General plane motion Use the three equations with $a_G$ and $\alpha$ unknown, plus kinematic relations (for example, rolling without slipping, $a_G = \alpha r$). Moments may be taken about any point P if the moment of $m\mathbf{a}_G$ is included: $$ \Sigma M_P = I_G\alpha + (\text{moment of } m\mathbf{a}_G \text{ about P}) $$ For a body rolling without slipping on a fixed surface, taking P as the contact point C gives $\Sigma M_C = I_C\alpha$ only when G moves parallel to the surface (such as a homogeneous round body on a plane). #### Procedure 1. Draw the free-body diagram with all forces and the unknown reactions. 2. Show $m\mathbf{a}_G$ and $I_G\alpha$ in a kinetic diagram. 3. Write the three equations and add the kinematic constraint equations. 4. Solve for the unknowns. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which set is the equations of motion of a rigid body in plane motion? A. Sum F = 0 and Sum M_G = 0 B. Sum Fx = m a_Gx, Sum Fy = m a_Gy, Sum M_G = I_G alpha C. Sum F = m v and Sum M = I omega D. Sum Fx = m a_Gx, Sum Fy = m a_Gy, Sum M_G = 0 always Answer: B. Force components give the motion of G and the moment about G gives the angular acceleration. (Key point: Equations of plane motion) 2. For pure translation, what is true? A. alpha = 0, so Sum M_G = 0 and the resultant force passes through G B. The body must be in equilibrium C. The resultant force passes through the contact point D. Sum F = 0 Answer: A. With no rotation, I_G alpha = 0, so Sum F = m a_G acts through G. (Key point: Translation) 3. For rotation about a fixed axis O, taking moments about O gives: A. Sum M_O = I_G alpha B. Sum M_O = I_O alpha, with I_O = I_G + m r-bar^2 C. Sum M_O = m r-bar alpha D. Sum M_O = 0 Answer: B. About the fixed pivot the moment equals I_O alpha, with I_O from the parallel axis theorem. (Key point: Fixed-axis rotation) 4. When can Sum M_C = I_C alpha be used for a body rolling without slipping? A. Always, for any rolling body B. Only when G moves parallel to the surface, such as a homogeneous round body on a plane C. Only when there is no friction D. Only when the body is at rest Answer: B. Otherwise the moment of m a_G about C must be included. 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