--- title: "Two slender rods released, angular velocity of AB" subject: "Applied Mechanics (Dynamics)" chapter: 6 key: "semester_3/applied_mechanics_dynamics:two-rods-released-beta-50" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/two-rods-released-beta-50 site: BCT Engineering --- # Two slender rods released, angular velocity of AB Applied Mechanics (Dynamics) · Chapter 6: Plane motion of rigid bodies: energy and momentum methods - Asked in: 2074 Asoj, 2+6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/two-rods-released-beta-50 - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6 ## Question Differentiate the central and Eccentric impact of the body. Each of the two slender rods as shown in figure below is 0.75 m long and has a mass of 6 kg. If the system is released from rest when $\beta = 50^\circ$, determine (a) the angular velocity of rod "AB" when $\beta = 20^\circ$ (b) the velocity of point 'D' at the same instant. [Figure: two pinned slender rods AB and BD, each 0.75 m, forming an inverted V; A pinned on the ground, D on rollers on the ground; $\beta = 50^\circ$ at A.] ## Short answer Central impact: line of impact passes through both mass centres, causing pure translation. Eccentric impact: line misses at least one mass centre, adding rotation. Using work-energy, gravitational work 18.72 J equals final kinetic energy of both rods; with kinematics ω_BD = ω_AB, solving gives ω_AB = 3.51 rad/s clockwise. Velocity of D = 1.80 m/s horizontally to the right. ## Answer #### Central and eccentric impact | Point | Central impact | Eccentric impact | |---|---|---| | Line of impact | Passes through the mass centres of both bodies | Does not pass through the mass centre of at least one body | | Motion after impact | Pure translation along the line of impact | Translation plus rotation | | Unknowns | Only the final velocities | Final velocities and angular velocities | | Equations used | Linear momentum and restitution | Linear and angular momentum and restitution | | Example | Two smooth spheres colliding head-on | A ball striking the end of a rod | #### Motion of the two rods Rods $AB$ and $BD$ are identical ($L=0.75$ m, $m=6$ kg), joined by a pin at $B$. $A$ is a fixed pin and $D$ slides on a smooth horizontal surface (rollers). Released from rest at $\beta=50^\circ$, the system moves under gravity only, so use the work-energy principle. ```text B / \ / \ / \ A /_______\ D (rollers) beta ``` #### Kinematics Let $\omega_{AB}$ be the angular velocity of $AB$. $\vec v_B \perp AB$ and $\vec v_D$ is horizontal. The ICR $I$ of $BD$ lies on the line $AB$ extended and on the vertical through $D$. By symmetry of the triangle $ABD$, $IB = BD = L$ and $ID = 2L\sin\beta$, so $$ \omega_{BD} = \frac{v_B}{IB} = \frac{\omega_{AB}L}{L} = \omega_{AB} = \omega $$ (the two rods turn in opposite senses). The centre $G_2$ of $BD$ is at $(1.5L\cos\beta,\ 0.5L\sin\beta)$ and $I$ is at $(2L\cos\beta,\ 2L\sin\beta)$, so $$ \left(\frac{IG_2}{L}\right)^2 = 0.25\cos^2\beta + 2.25\sin^2\beta $$ At $\beta = 20^\circ$: $0.25\cos^2 20^\circ + 2.25\sin^2 20^\circ = 0.4840$, so $IG_2 = 0.5218$ m. #### Kinetic energy at $\beta = 20^\circ$ $$ T_{AB} = \tfrac12\left(\tfrac13mL^2\right)\omega^2 = \tfrac12(1.1250)\,\omega^2 $$ $$ T_{BD} = \tfrac12\left[\tfrac1{12}mL^2 + m(IG_2)^2\right]\omega^2 = \tfrac12(1.9146)\,\omega^2 $$ $$ T_2 = \tfrac12(1.1250 + 1.9146)\,\omega^2 = 1.5198\,\omega^2 $$ #### Work of gravity Heights of the mass centres: $\tfrac L2\sin\beta$ for each rod, so the total potential energy is $V = mgL\sin\beta$. $$ U_{1\to2} = mgL(\sin50^\circ - \sin20^\circ) = 6(9.81)(0.75)(0.7660 - 0.3420) = 18.72\ \text{J} $$ #### (a) Angular velocity of AB $$ T_1 + U_{1\to2} = T_2:\quad 0 + 18.72 = 1.5198\,\omega^2 \;\Rightarrow\; \omega_{AB} = 3.51\ \text{rad/s} $$ AB turns clockwise (B moving down and to the right). #### (b) Velocity of D $D$ is $ID = 2L\sin20^\circ$ from the ICR: $$ v_D = \omega_{BD}\,(ID) = 3.51\times2(0.75)\sin20^\circ = 3.51\times0.5130 = 1.80\ \text{m/s}\ (\to) $$ **Answer: $\omega_{AB} = 3.51$ rad/s clockwise; $v_D = 1.80$ m/s horizontally to the right.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What distinguishes central from eccentric impact? A. In central impact the line of impact passes through the mass centres of both bodies, giving pure translation B. Central impact occurs only with spheres of equal mass C. Eccentric impact has e = 0 D. Eccentric impact involves no rotation Answer: A. In eccentric impact the line of impact misses the mass centre of at least one body, so rotation also occurs and angular momentum is needed. (Key point: Central vs eccentric impact) 2. Which equations are used for eccentric impact but not for central impact? A. Only restitution B. Angular momentum equations, together with linear momentum and restitution C. Work-energy only D. Only the parallel-axis theorem Answer: B. Eccentric impact adds rotation, so angular momentum equations must be used besides linear momentum and restitution. (Key point: Equations in eccentric impact) 3. Why is the work-energy principle used for the rods released from rest? A. Only gravity does work (A pin and rollers on the smooth surface do none), and speed at a position is wanted B. The rods collide C. Friction does the work D. Angular momentum is conserved Answer: A. The pin at A and the smooth rollers do no work, so U = change in gravitational potential energy and T1 = 0. (Key point: Method for the two rods) 4. How do the angular velocities of AB and BD compare, using the ICR of BD? A. w_BD = 2 w_AB B. w_BD = w_AB / 2 C. They are equal in magnitude and opposite in sense, since IB = BD = L D. w_BD = 0 Answer: C. v_B = w_AB L and IB = L, so w_BD = v_B / IB = w_AB, with opposite rotation senses. (Key point: Angular velocities of the rods) 5. How is v_D found at beta = 20 degrees? A. v_D = w L B. v_D = w_BD (ID) with ID = 2 L sin beta, giving 1.80 m/s horizontally C. v_D = w_AB (AB) = 2.63 m/s D. v_D = 0, since D is on the ground Answer: B. D is a distance ID = 2L sin 20 = 0.513 m from the ICR, so v_D = 3.51 x 0.513 = 1.80 m/s, directed horizontally along the surface. 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