--- title: "Rod AB sliding over wheel C" subject: "Applied Mechanics (Dynamics)" chapter: 4 key: "semester_3/applied_mechanics_dynamics:rod-ab-over-wheel-c-velocity" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/rod-ab-over-wheel-c-velocity site: BCT Engineering --- # Rod AB sliding over wheel C Applied Mechanics (Dynamics) · Chapter 4: Kinematics of Rigid Bodies - Asked in: 2075 Asoj, 8 marks - Video: [(5/88) Linkage Angular Velocity](https://www.youtube.com/watch?v=4YdotfYeRaU) — The Ryder Project - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/rod-ab-over-wheel-c-velocity - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4 ## Question Rod AB moves over a small wheel at C while end A moves to the right with a constant velocity of 635 mm/s. At the instant shown, determine (a) the angular velocity of the rod, (b) the velocity of end B of the rod. [Figure: rod AB inclined, resting on small wheel C at height 178 mm; A on the ground; horizontal distance from A to the foot of C is 254 mm; 508 mm marked along the rod.] ## Short answer The rod slides over the wheel, so the point at C has velocity along the rod. Using v_A sinθ = ω·AC gives ω = 1.1749 rad/s counter-clockwise. Then v_B = v_A + ω×r_B/A yields v_B = 569.6 mm/s at 59.1° above horizontal. ## Answer **Setting up.** The rod slides over the wheel at $C$, so the point of the rod that is at $C$ moves along the rod. Let $A$ be the origin and $\theta$ the angle of the rod with the horizontal, assuming $AB = 508$ mm measured from $A$ along the rod. $$ \tan\theta = \frac{178}{254} \;\Rightarrow\; \theta = 35.02^\circ, \qquad AC = \sqrt{254^2 + 178^2} = 310.2\ \text{mm} $$ ```text B / / C o <- wheel, 178 mm high /| / | A *--+----> vA = 635 mm/s 254 mm ``` #### (a) Angular velocity of the rod Velocity of the rod point at $C$: $\vec v_C = \vec v_A + \vec\omega\times\vec r_{C/A}$. It must be along the rod, so its component perpendicular to the rod is zero. The component of $\vec v_A$ perpendicular to the rod is $v_A\sin\theta$, and the rotation contributes $\omega\,(AC)$ in the opposite sense: $$ v_A\sin\theta = \omega\,(AC) $$ $$ \omega = \frac{v_A\sin\theta}{AC} = \frac{635\times\sin 35.02^\circ}{310.2} = 1.1749\ \text{rad/s (counter-clockwise)} $$ #### (b) Velocity of end B $$ \vec v_B = \vec v_A + \omega\hat k\times\vec r_{B/A}, \qquad \vec r_{B/A} = 508(\cos\theta\,\hat i + \sin\theta\,\hat j) $$ $$ \vec v_B = 635\hat i + 1.1749\,(508)(-\sin\theta\,\hat i + \cos\theta\,\hat j) = (292.5\,\hat i + 488.8\,\hat j)\ \text{mm/s} $$ $$ v_B = 569.6\ \text{mm/s} $$ at 59.1° above the horizontal, to the right. **Answer: $\omega = 1.1749$ rad/s counter-clockwise; $v_B = 569.6$ mm/s at 59.1° above the horizontal.** If the 508 mm is measured from $C$ instead of from $A$, replace $r_{B/A}$ by $AC + 508 = 818.2$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The rod slides over the wheel at C. What condition gives the angular velocity? A. The velocity of the rod point at C is zero B. The velocity of the rod point at C is along the rod, so its component perpendicular to the rod is zero C. The velocity of the rod point at C is perpendicular to the rod D. The velocity of B equals the velocity of A Answer: B. The rod slides along itself over C, so the perpendicular component of the rod point's velocity must vanish. (Key point: Condition at C) 2. Which equation gives omega of the rod? A. v_A cos(theta) = omega (AC) B. v_A = omega (AC) C. v_A sin(theta) = omega (AC) D. v_A tan(theta) = omega (AC) Answer: C. The component of v_A perpendicular to the rod, v_A sin(theta), is cancelled by omega x AC. (Key point: Equation for omega) 3. With theta = 35.02°, AC = 310.2 mm and v_A = 635 mm/s, what is omega? A. 1.17 rad/s counter-clockwise B. 2.05 rad/s counter-clockwise C. 1.17 rad/s clockwise D. 1.68 rad/s counter-clockwise Answer: A. omega = 635 sin 35.02° / 310.2 = 1.1749 rad/s, counter-clockwise. (Key point: Value of omega) 4. How is v_B found once omega is known? A. v_B = omega x (AB) only, perpendicular to the rod B. v_B = v_A + omega x r_B/A, added as vectors C. v_B = v_A, since A and B are on the same rod D. v_B = omega x (CB) along the rod Answer: B. Relative velocity: v_B = v_A + omega k x r_B/A, giving 569.6 mm/s. 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