---
title: "Gears A, B, C with couple 10 N-m"
subject: "Applied Mechanics (Dynamics)"
chapter: 6
key: "semester_3/applied_mechanics_dynamics:gears-abc-couple-10nm"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/gears-abc-couple-10nm
site: BCT Engineering
---

# Gears A, B, C with couple 10 N-m

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

- Asked in: 2076 Asoj, 8 marks
- Video: [Gear Trains - Epicyclic Gear Problem](https://www.youtube.com/watch?v=31rU8lDHkMQ) — SM Lectures
- 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-10nm
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-6

## Question

Each of gear A and B has a weight of 2.5 Kg and radius of gyration of 100 mm while gear C has a weight of 12.5 kg and radius of gyration of 180 mm. A couple M of magnitude of 10 N-m is applied to gear C. Determine a) number of revolution of gear C required for its angular velocity to increase from 100 to 450 rpm b) the corresponding tangential force on gear A. [Figure: gear C (radius 250 mm) meshing with gears A and B (radius 100 mm each); couple M applied to C.]

## Short answer

Use work-energy with speed ratio omega_A/omega_C=r_C/r_A=2.5 and equivalent inertia 0.7175 kg m^2. Gear C rotates 75.73 rad, i.e. 12.05 revolutions, for the speed increase; the corresponding tangential force on gear A is 8.71 N.

## Answer

(The weights are given in kg, so they are used as masses.)

**Method.** Use 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
$m_A = m_B = 2.5$ kg, $\bar k_A = 0.1$ m, $m_C = 12.5$ kg, $\bar k_C = 0.18$ m, $r_A = r_B = 0.1$ m, $r_C = 0.25$ m, $M = 10$ N m

$$
\bar I_A = \bar I_B = 2.5(0.1)^2 = 0.0250\ \text{kg m}^2, \qquad \bar I_C = 12.5(0.18)^2 = 0.4050\ \text{kg m}^2
$$

$$
\frac{r_C}{r_A} = \frac{0.25}{0.1} = 2.5000
$$

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

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

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

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

$$
I_{eq} = 0.4050 + 2(0.0250)(2.5000)^2 = 0.7175\ \text{kg m}^2
$$

$$
T_1 + M\theta_C = T_2 \;\Rightarrow\; \theta_C = \frac{I_{eq}(\omega_2^2 - \omega_1^2)}{2M} = \frac{0.7175\,(47.12^2 - 10.47^2)}{2(10)} = 75.73\ \text{rad}
$$

$$
\text{revolutions} = \frac{\theta_C}{2\pi} = \frac{75.73}{2\pi} = 12.05
$$

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

$$
\alpha_C = \frac{M}{I_{eq}} = \frac{10}{0.7175} = 13.94\ \text{rad/s}^2, \qquad \alpha_A = \alpha_C\frac{r_C}{r_A} = 34.84\ \text{rad/s}^2
$$

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

$$
F\,r_A = \bar I_A\alpha_A \;\Rightarrow\; F = \frac{0.0250\times34.84}{0.1} = 8.71\ \text{N}
$$

(Check on gear $C$: $M - 2Fr_C = \bar I_C\alpha_C$ gives $10 - 2(8.71)(0.25) = 5.64$, and $\bar I_C\alpha_C = 5.64$.)

**Answer: (a) $12.05$ revolutions of gear $C$; (b) $F = 8.71$ N.**

## Check yourself

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

1. The data give 'weight' in kg. How is this treated?
   A. As force in kg-force multiplied by 9.81 N
   B. As a mass in kg
   C. As moment of inertia
   D. It is ignored
   Answer: B. The values are used directly as masses, so I_A = 2.5(0.1)^2 = 0.025 and I_C = 12.5(0.18)^2 = 0.405 kg m^2. (Key point: Units of weight)

2. With r_C = 0.25 m and r_A = 0.1 m, how fast does gear A turn compared with C?
   A. 0.4 times as fast
   B. Equal speed
   C. 2.5 times as fast
   D. 6.25 times as fast
   Answer: C. Teeth speeds match, so w_A = w_C r_C / r_A = 2.5 w_C. (Key point: Gear ratio)

3. What is I_eq referred to gear C?
   A. 0.405 + 0.025 = 0.430 kg m^2
   B. 0.405 + 2(0.025)(2.5)^2 = 0.7175 kg m^2
   C. 0.405 + 2(0.025)(2.5) = 0.530 kg m^2
   D. 0.405 kg m^2
   Answer: B. A and B each carry kinetic energy (1/2) I_A (2.5 w_C)^2, so their contribution is multiplied by 2.5^2 = 6.25 each. (Key point: Equivalent inertia)

4. From T1 + M theta = T2, how many revolutions does C turn?
   A. About 12.05 revolutions (theta = 75.73 rad)
   B. About 75.73 revolutions
   C. About 6 revolutions
   D. About 2 revolutions
   Answer: A. theta = I_eq (w2^2 - w1^2) / (2M) = 0.7175(47.12^2 - 10.47^2)/20 = 75.73 rad, and dividing by 2 pi gives 12.05 rev. (Key point: Revolutions of C)

5. The tangential force on A is 8.71 N. Which relation gives this and how can it be checked on gear C?
   A. F = M / r_C, checked by M = F r_C
   B. F r_A = I_A alpha_A; check M - 2 F r_C = I_C alpha_C
   C. F = I_C alpha_C, checked by M = 0
   D. F = m_A g, checked by weights
   Answer: B. For gear A only F acts, so F = I_A alpha_A / r_A with alpha_A = 34.84 rad/s^2; gear C feels the couple M and two tooth forces, so M - 2 F r_C = I_C alpha_C. (Key point: Tangential force and check)

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