--- title: "Gears A and B, M = 8 N-m, revolutions" subject: "Applied Mechanics (Dynamics)" chapter: 6 key: "semester_3/applied_mechanics_dynamics:gear-system-8nm-540rpm" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/gear-system-8nm-540rpm site: BCT Engineering --- # Gears A and B, M = 8 N-m, revolutions Applied Mechanics (Dynamics) · Chapter 6: Plane motion of rigid bodies: energy and momentum methods - Asked in: 2072 Chaitra, 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/gear-system-8nm-540rpm - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6 ## Question The system is at rest when a moment of $M = 8$ N-m is applied to gear B. Neglecting friction (a) determine the number of revolutions of gear B before its angular velocity reaches 540 rpm and (b) tangential force exerted by gear B on gear A. [Figure: gear A ($r_A = 250$ mm, $m_A = 10$ kg, $\bar k_A = 200$ mm) meshing with gear B ($r_B = 100$ mm, $m_B = 3$ kg, $\bar k_B = 80$ mm).] ## Short answer By work-energy for gears A and B with no slip, the equivalent inertia about B is 0.0832 kg m^2. Starting from rest, gear B reaches 540 rpm after 16.63 rad, or 2.65 revolutions; the tangential force of B on A is 61.54 N. ## Answer **Data.** $r_A=0.250$ m, $m_A = 10$ kg, $\bar k_A = 0.200$ m; $r_B = 0.100$ m, $m_B=3$ kg, $\bar k_B = 0.080$ m; $M = 8$ N m on $B$; the system starts from rest. $$ \bar I_A = 10(0.200)^2 = 0.400\ \text{kg m}^2, \qquad \bar I_B = 3(0.080)^2 = 0.0192\ \text{kg m}^2 $$ The teeth do not slip, so $\omega_A r_A = \omega_B r_B$: $$ \omega_A = \omega_B\frac{r_B}{r_A} = 0.4\,\omega_B $$ #### (a) Revolutions of gear B (work-energy) $$ T = \tfrac12\bar I_B\omega_B^2 + \tfrac12\bar I_A\omega_A^2 = \tfrac12\left[\bar I_B + \bar I_A\left(\frac{r_B}{r_A}\right)^2\right]\omega_B^2 = \tfrac12(0.0832)\,\omega_B^2 $$ $$ \bar I_B + \bar I_A(0.4)^2 = 0.0192 + 0.400(0.16) = 0.0832\ \text{kg m}^2 $$ $$ \omega_B = 540\ \text{rpm} = 56.55\ \text{rad/s} $$ $$ T_1 + M\theta_B = T_2:\quad \theta_B = \frac{\tfrac12(0.0832)(56.55)^2}{8} = 16.63\ \text{rad} = 2.65\ \text{rev} $$ #### (b) Tangential force of B on A Angular acceleration (constant couple): $$ \alpha_B = \frac{M}{I_{eq}} = \frac{8}{0.0832} = 96.15\ \text{rad/s}^2 $$ For gear $B$: $M - F r_B = \bar I_B\alpha_B$ $$ F = \frac{M - \bar I_B\alpha_B}{r_B} = \frac{8 - 0.0192(96.15)}{0.100} = 61.54\ \text{N} $$ (Check with gear $A$: $\alpha_A = 0.4\alpha_B = 38.46$ rad/s², $F = \bar I_A\alpha_A/r_A = 61.54$ N.) **Answer: (a) $\theta_B = 16.63$ rad $= 2.65$ revolutions; (b) $F = 61.54$ N.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the angular velocities of gears A and B related? A. w_A = w_B (r_B / r_A) = 0.4 w_B B. w_A = w_B C. w_A = w_B (r_A / r_B) = 2.5 w_B D. w_A = w_B (r_B / r_A)^2 Answer: A. No slip at the teeth means w_A r_A = w_B r_B, so the larger gear A turns slower, w_A = 0.4 w_B. (Key point: Speed relation) 2. What is the equivalent inertia referred to gear B? A. I_B + I_A = 0.419 kg m^2 B. I_B + I_A (r_B / r_A)^2 = 0.0832 kg m^2 C. I_B + I_A (r_A / r_B)^2 = 2.52 kg m^2 D. I_A only Answer: B. Kinetic energy is (1/2)[I_B + I_A (0.4)^2] w_B^2, so I_eq = 0.0192 + 0.400(0.16) = 0.0832 kg m^2. (Key point: Equivalent inertia) 3. Starting from rest, what gives the angle turned by B to reach 540 rpm (56.55 rad/s)? A. M theta = (1/2) I_eq w^2, so theta = 16.63 rad = 2.65 rev B. theta = w / M C. M theta = I_eq w, so theta = 0.59 rad D. theta = w^2 / 2 pi Answer: A. The work of the constant couple equals the final kinetic energy: theta = 0.5(0.0832)(56.55)^2 / 8 = 16.63 rad. (Key point: Revolutions of B) 4. How is the tooth force F of B on A obtained? A. F = M / r_B B. From gear B alone: M - F r_B = I_B alpha_B, giving F = 61.54 N C. F = I_B alpha_B / r_B D. F = m_B g Answer: B. Gear B has the applied couple M and the opposing tooth force F, so F = (M - I_B alpha_B)/r_B with alpha_B = M / I_eq = 96.15 rad/s^2. 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