---
title: "Gears A, B, C with couple, revolutions and force"
subject: "Applied Mechanics (Dynamics)"
chapter: 6
key: "semester_3/applied_mechanics_dynamics:gears-abc-couple-9-26"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/gears-abc-couple-9-26
site: BCT Engineering
---

# Gears A, B, C with couple, revolutions and force

Applied Mechanics (Dynamics) · Chapter 6: Plane motion of rigid bodies: energy and momentum methods

- Asked in: 2080 Bhadra, 8 marks
- Video: [Simple Gear Ratios, Input and Output Speed, Torque and Power](https://www.youtube.com/watch?v=H9_Llzcz2cY) — Engineers Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/gears-abc-couple-9-26
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6

## Question

Each of the gears A and B has a mass of 2.1 kg and a radius of gyration of 103 mm and gear C has a mass of 10.3 kg and radius of gyration of 187 mm. A couple of M of constant magnitude 9.26 Nm is applied to gear C. Take $r_A = r_B = 107$ mm and $r_C = 265$ mm. Determine: (i) The number of revolutions of gear C required for its angular velocity to increase from 100 to 450 rpm (ii) The corresponding tangential force acting on gear A. [Figure: gear C meshing with gears A and B on either side; couple M on C; $r_A = r_B = 107$ mm, $r_C = 265$ mm.]

## Short answer

Use work-energy for the geared system: equivalent inertia I_eq=0.6335 kg m^2 and speed ratio omega_A/omega_C=r_C/r_A. The required angle is 72.21 rad, so gear C makes 11.49 revolutions; the tangential force on gear A is 7.54 N.

## Answer

**Method.** Work-energy for the whole system: the couple does work $M\theta_C$; friction is neglected, and the weights and bearing reactions do no work. Gears $A$ and $B$ are identical and are driven by $C$, so $\omega_A = \omega_B = \omega_C\,\dfrac{r_C}{r_A}$.

#### Data
$m_A = m_B = 2.1$ kg, $\bar k_A = 0.103$ m, $m_C = 10.3$ kg, $\bar k_C = 0.187$ m, $r_A = r_B = 0.107$ m, $r_C = 0.265$ m, $M = 9.26$ N m

$$
\bar I_A = \bar I_B = 2.1(0.103)^2 = 0.0223\ \text{kg m}^2, \qquad \bar I_C = 10.3(0.187)^2 = 0.3602\ \text{kg m}^2
$$

$$
\frac{r_C}{r_A} = \frac{0.265}{0.107} = 2.4766
$$

```text
        (B)
          \
    M -> ( C ) 
          /
        (A)
```

#### Angular speeds
$$
\omega_1 = 100\ \text{rpm} = 10.47\ \text{rad/s}, \qquad \omega_2 = 450\ \text{rpm} = 47.12\ \text{rad/s}
$$

#### (i) Revolutions of gear C
Kinetic energy in terms of $\omega_C$:

$$
T = \tfrac12\bar I_C\omega_C^2 + 2\left(\tfrac12\bar I_A\omega_A^2\right) = \tfrac12\left[\bar I_C + 2\bar I_A\left(\frac{r_C}{r_A}\right)^2\right]\omega_C^2 = \tfrac12 I_{eq}\,\omega_C^2
$$

$$
I_{eq} = 0.3602 + 2(0.0223)(2.4766)^2 = 0.6335\ \text{kg m}^2
$$

$$
T_1 + M\theta_C = T_2 \;\Rightarrow\; \theta_C = \frac{I_{eq}(\omega_2^2 - \omega_1^2)}{2M} = \frac{0.6335\,(47.12^2 - 10.47^2)}{2(9.26)} = 72.21\ \text{rad}
$$

$$
\text{revolutions} = \frac{\theta_C}{2\pi} = \frac{72.21}{2\pi} = 11.49
$$

#### (ii) Tangential force on gear A
The couple is constant, so the angular acceleration is constant:

$$
\alpha_C = \frac{M}{I_{eq}} = \frac{9.26}{0.6335} = 14.62\ \text{rad/s}^2, \qquad \alpha_A = \alpha_C\frac{r_C}{r_A} = 36.20\ \text{rad/s}^2
$$

For gear $A$ the only moment about its axis is that of the tangential tooth force $F$ (friction neglected):

$$
F\,r_A = \bar I_A\alpha_A \;\Rightarrow\; F = \frac{0.0223\times36.20}{0.107} = 7.54\ \text{N}
$$

(Check on gear $C$: $M - 2Fr_C = \bar I_C\alpha_C$ gives $9.26 - 2(7.54)(0.265) = 5.26$ and $\bar I_C\alpha_C = 5.26$.)

**Answer: (i) $11.49$ revolutions of gear $C$; (ii) $F = 7.54$ N.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which method is used to find the revolutions of gear C?
   A. Work-energy for the whole system, with the couple doing work M theta_C
   B. Impulse-momentum with time
   C. Conservation of angular momentum
   D. Newton's second law for each gear in turn, with time known
   Answer: A. The couple's work M theta_C equals the change in kinetic energy of gears A, B and C; friction is neglected and weights do no work. (Key point: Method)

2. How are the angular speeds of A and C related?
   A. w_A = w_C
   B. w_A = w_C (r_A / r_C)
   C. w_A = w_C (r_C / r_A)
   D. w_A = w_C (r_C / r_A)^2
   Answer: C. Teeth do not slip, so w_A r_A = w_C r_C, giving w_A = w_C r_C / r_A = 2.4766 w_C. (Key point: Speed relation)

3. What is the equivalent moment of inertia referred to gear C?
   A. I_C + 2 I_A (r_C / r_A)^2 = 0.6335 kg m^2
   B. I_C + 2 I_A = 0.405 kg m^2
   C. I_C + I_A (r_C / r_A) = 0.4154 kg m^2
   D. I_C only = 0.3602 kg m^2
   Answer: A. Kinetic energy of the two small gears is 2(1/2) I_A w_A^2 = (1/2)[2 I_A (r_C/r_A)^2] w_C^2, so I_eq = 0.3602 + 2(0.0223)(2.4766)^2 = 0.6335. (Key point: Equivalent inertia)

4. Which formula gives the angle turned by C as it speeds up from 100 to 450 rpm?
   A. theta = I_eq (w2 - w1) / M
   B. theta = I_eq (w2^2 - w1^2) / (2M), which is 11.49 revolutions
   C. theta = M / I_eq
   D. theta = (w2 - w1) / 2 pi
   Answer: B. M theta = (1/2) I_eq (w2^2 - w1^2), so theta = 72.21 rad = 11.49 rev. (Key point: Revolutions)

5. How is the tangential tooth force on gear A found?
   A. F r_A = I_A alpha_A with alpha_A = alpha_C r_C / r_A, giving 7.54 N
   B. F = M / r_C
   C. F = m_A g
   D. F = I_A w_A / r_A
   Answer: A. The only moment on A is F r_A, so F = I_A alpha_A / r_A with alpha_C = M / I_eq = 14.62 rad/s^2 and alpha_A = 36.20 rad/s^2. (Key point: Tangential force on A)

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