--- title: "Double gear on rack, accelerations" subject: "Applied Mechanics (Dynamics)" chapter: 4 key: "semester_3/applied_mechanics_dynamics:double-gear-rack-acceleration" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/double-gear-rack-acceleration site: BCT Engineering --- # Double gear on rack, accelerations Applied Mechanics (Dynamics) · Chapter 4: Kinematics of Rigid Bodies - Asked in: 2075 Asoj, 8 marks; 2073 Shrawan, 8 marks (asked 2 times) - Video: [Dynamics | Double gear systems | Angular Acceleration and acceleration of gear at each point |](https://www.youtube.com/watch?v=xvJnJN1xgPU) — Sudip Dhakal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/double-gear-rack-acceleration - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-4 ## Question The center of the double gear has a velocity and acceleration to the right of 1.2 m/s and 3 m/s$^2$, respectively. The lower rack is stationary. Determine (a) the angular acceleration of the gear and (b) the acceleration of points B, C and D. [Figure: double gear with centre A, radii $r_1 = 150$ mm (to lower rack, contact C) and $r_2 = 100$ mm (to upper rack R, contact B); D on the horizontal line through A to the left; $V_A = 1.2$ m/s and $a_A = 3$ m/s$^2$ to the right.] ## Short answer The gear rolls without slip on the stationary lower rack, so angular acceleration alpha = a_A/r1 = 3/0.150 = 20.0 rad/s^2 clockwise. Using relative acceleration, point B: 8.12 m/s^2 at 52.0° below horizontal, C: 9.60 m/s^2 upward, D: 12.95 m/s^2 at 13.4° above horizontal. ## Answer The lower rack is stationary and the gear rolls on it without slipping, so the contact point $C$ is the instantaneous centre. Take $x$ to the right, $y$ up. $B$ is on the inner radius ($r_2=0.100$ m) above $A$; $C$ is at $r_1 = 0.150$ m below $A$; $D$ is taken on the outer radius, $0.150$ m to the left of $A$. #### (a) Angular acceleration $$ \omega = \frac{v_A}{r_1} = \frac{1.2}{0.150} = 8.00\ \text{rad/s (clockwise)}, \qquad \alpha = \frac{a_A}{r_1} = \frac{3}{0.150} = 20.00\ \text{rad/s}^2\ \text{(clockwise)} $$ #### (b) Accelerations of B, C and D Use $\vec a_P = \vec a_A + \vec\alpha\times\vec r_{P/A} - \omega^2\vec r_{P/A}$ with $\vec\alpha = -20.00\,\hat k$, $\omega^2 = 64.00$ rad²/s², $\vec a_A = 3\,\hat i$. **Point B**, $\vec r_{B/A} = 0.100\,\hat j$: $$ \vec a_B = 3\hat i + (-20.00\hat k)\times(0.1\hat j) - 64.00(0.1\hat j) = (5.00\,\hat i -6.40\,\hat j)\ \text{m/s}^2 $$ $a_B = 8.12$ m/s², directed 52.0° below the horizontal, to the right. **Point C**, $\vec r_{C/A} = -0.150\,\hat j$: $$ \vec a_C = 3\hat i + (-20.00\hat k)\times(-0.15\hat j) - 64.00(-0.15\hat j) = (0.00\,\hat i +9.60\,\hat j)\ \text{m/s}^2 $$ $a_C = 9.60$ m/s², directed upward towards $A$. This is the centripetal acceleration of the contact point; its horizontal part is zero. **Point D**, $\vec r_{D/A} = -0.150\,\hat i$: $$ \vec a_D = 3\hat i + (-20.00\hat k)\times(-0.15\hat i) - 64.00(-0.15\hat i) = (12.60\,\hat i +3.00\,\hat j)\ \text{m/s}^2 $$ $a_D = 12.95$ m/s², directed 13.4° above the horizontal, to the right. **Answer: $\alpha = 20.00$ rad/s² clockwise; $a_B = 8.12$ m/s² (52.0° below horizontal), $a_C = 9.60$ m/s² (upward), $a_D = 12.95$ m/s² (13.4° above horizontal).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the angular velocity be found from w = vA / r1 = 8 rad/s? A. The upper rack is at rest B. The gear rolls without slipping on the stationary lower rack, so the contact point C is the instantaneous centre C. A is the centre of rotation D. The gear is in pure translation Answer: B. The contact point C on the fixed rack has zero velocity, so it is the ICR and vA = w r1. (Key point: Instantaneous centre) 2. What is the angular acceleration of the gear? A. 20 rad/s^2 clockwise (aA / r1) B. 30 rad/s^2 clockwise (aA / r2) C. 8 rad/s^2 clockwise D. 12 rad/s^2 anticlockwise Answer: A. Rolling without slip on the outer radius gives alpha = aA / r1 = 3 / 0.150 = 20 rad/s^2. (Key point: Angular acceleration) 3. Which relation is used to find the acceleration of B, C and D? A. aP = aA + alpha x rP/A - w^2 rP/A B. aP = aA + w x rP/A C. aP = w^2 rP/A only D. aP = alpha x rP/A only Answer: A. General plane motion acceleration = translation of A + tangential (alpha x r) + normal (-w^2 r) parts. (Key point: Acceleration formula) 4. What is the acceleration of the contact point C? A. Zero, since C has zero velocity B. 9.60 m/s^2 directed upward towards A, with zero horizontal part C. 3 m/s^2 to the right D. 9.60 m/s^2 horizontal Answer: B. C has zero velocity but not zero acceleration: aC = w^2 r1 = 64 x 0.15 = 9.6 m/s^2 towards A. 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