Gears & gear trains: practice problems with step-by-step solutions

Free Gears & gear trains practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: speed ratio, torque and power, compound gear train.

Free practice problemEasy
N1​ = 30N2​ = 48

A driving gear with 30 teeth turns at 1500 rpm and meshes with a gear with 48 teeth. Find the speed n2 of the driven gear.

N1 = 30N2 = 48n1 = 1500 rpm
rpm
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What you practise

  1. Easyspeed ratio
  2. Mediumtorque and power
  3. Hardcompound gear train

Key formulas

Meshing gears
n1·N1 = n2·N2
Torque
T2 = T1·(N2/N1)·η
Power
P = T·ω, ω = 2π·n/60
Compound train
total ratio = product of the stage ratios

Learn Gears & gear trains from scratch

1What gears do

Gears pass rotation from one shaft to another. Their teeth lock into each other, so they cannot slip.

A gearbox does two jobs at once. It changes the speed of rotation, and it changes the torque, the turning force on the shaft. Slower means stronger.

An electric motor spins fast but with little torque. A gear pair slows it down so it can turn a heavy conveyor or a winch. A bicycle in a low gear works the same way.

The gear that is driven by the motor is the driver (often a small gear called the pinion). The gear it turns is the driven gear.

N1​, n1​N2​, n2​

Two gears in mesh turn in opposite directions.

Self-check A small gear drives a large gear. Compared with the small gear, the large gear turns…
  1. at the same speed in the same direction
  2. faster, with more torque
  3. slower, with less torque
  4. slower, with more torque

Answer: slower, with more torque. The large gear needs more of the small gear’s turns for one turn of its own, so it is slower. The power is nearly the same, so the slower shaft carries more torque.

2The speed ratio

At the point where two gears touch, their teeth pass at the same rate. If the pinion turns once with 20 teeth, it pushes 20 teeth of the big gear past. A 60-tooth gear then makes only one third of a turn.

So the speeds n (in rpm) and the numbers of teeth N are linked by:

n1 · N1 = n2 · N2

The gear ratio is i = N2 / N1 = n1 / n2. A ratio of 3 means the output turns three times slower.

In a compound gear train, two gears sit on the same middle shaft, so they turn at the same speed. The total ratio is the product of the stage ratios:

itotal = (NB / NA) · (ND / NC)

An idler gear between two others (each on its own shaft) changes only the direction of rotation. The speed of the last gear follows from the first and last gear alone.

Self-check A 20-tooth gear turning at 900 rpm drives a 60-tooth gear. How fast does the 60-tooth gear turn?
  1. 1.33 rpm
  2. 2700 rpm
  3. 300 rpm
  4. 900 rpm

Answer: 300 rpm. n2 = n1 · N1/N2 = 900 · 20/60 = 300 rpm. Multiplying by 60/20 gives 2700 rpm: the big gear would then turn faster, which is backwards.

3Torque and power

Power is torque times angular speed. The angular speed must be in rad/s, so first convert from rpm:

ω = 2π · n / 60 P = T · ω

A gear pair passes on almost all of the power. Only a few percent are lost to friction, described by the efficiency η (typically 0.95 to 0.98):

P2 = η · P1

Because the speed goes down by the gear ratio while the power stays nearly the same, the torque goes up by that ratio:

T2 = T1 · (N2 / N1) · η
Self-check A 3 kW motor turns at 955 rpm. What torque does it deliver?
  1. 0.03 N·m
  2. 188 N·m
  3. 3.14 N·m
  4. 30 N·m

Answer: 30 N·m. ω = 2π · 955/60 = 100 rad/s, so T = 3000/100 = 30 N·m. Dividing by the rpm directly gives 3.14, leaving the power in kW gives 0.03, and forgetting the 2π gives 188.

4The method step by step

  1. Sketch the gears and write the number of teeth on each. Mark the input shaft and the output shaft.
  2. Find the ratio of each stage, driven teeth / driving teeth, and multiply the stages for a compound train.
  3. Output speed: nout = nin / itotal.
  4. Input torque from the motor: convert rpm to rad/s with ω = 2π·n/60, then T = P/ω.
  5. Output torque: multiply by the ratio and by the efficiency of every gear mesh.

Check the direction of the answer: a reduction makes the output slower and its torque larger.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A 4 kW motor running at 1200 rpm drives a pinion with 20 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 96%. Find the speed of the gear and the output torque on its shaft.

1

Draw the pair

20, 1200 rpm60, n2​ = ?

The pinion is the driver. The gear ratio is i = 60/20 = 3.

2

Output speed

n2 = n1 · N1 / N2 = 1200 · 20 / 60 = 400 rpm
3

Input torque

Convert the motor speed to rad/s first:

ω1 = 2π · 1200 / 60 = 125.7 rad/s
T1 = P / ω1 = 4000 / 125.7 = 31.83 N·m
4

Output torque

T2 = T1 · (N2 / N1) · η = 31.83 · 3 · 0.96 = 91.7 N·m
5

Check the answer

Output power: P2 = 0.96 · 4000 = 3840 W, and ω2 = 2π · 400/60 = 41.89 rad/s.

T2 = 3840 / 41.89 = 91.7 N·m

The same torque, found from the power. The answer is right.

Answer

The gear turns at n2 = 400 rpm with an output torque of T2 ≈ 91.7 N·m.

6Common mistakes

  • Turning the teeth ratio upside down. A bigger driven gear must turn slower.
  • Using rpm in P = T·ω. Convert to rad/s with ω = 2π·n/60.
  • Dividing the torque by the ratio instead of multiplying. A reduction increases the torque.
  • Dividing by the efficiency. Friction makes the output torque smaller, so multiply by η.
  • Counting an idler gear in the ratio, or multiplying the teeth of two gears on the same shaft.
Self-check A pinion with 20 teeth carries 10 N·m and drives a 60-tooth gear. The efficiency is 95%. What is the output torque?
  1. 28.5 N·m
  2. 3.17 N·m
  3. 30 N·m
  4. 31.6 N·m

Answer: 28.5 N·m. T2 = 10 · (60/20) · 0.95 = 28.5 N·m. Dividing by the ratio gives 3.17, ignoring the efficiency gives 30, and dividing by η gives 31.6.

7How exam questions look

  • Easy: the speed ratio. Find the speed of the driven gear, or the number of teeth needed for a given speed. Variations include an idler gear, and the size of gears from the module m: pitch diameter d = m·N, centre distance a = m·(N1 + N2)/2.
  • Medium: torque and power. A motor of given power and speed drives a gear pair: find the input or output torque. Others ask for the tangential force on the teeth, Ft = 2T/d, or involve a worm drive (ratio = wheel teeth / worm starts).
  • Hard: a compound gear train. Find the output speed, the teeth needed on the last gear, or the output torque with an efficiency for every mesh. Some questions use a planetary gear set.

For a compound train such as A (20) → B (60), then C (15) → D (45): i = 3 · 3 = 9, so an input of 900 rpm gives 100 rpm at the output.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
d1​ = 160 mmd2​ = 400 mm1500 rpmn2​ = ?

A motor turning at 1500 rpm drives a 160 mm pulley. A belt connects it to a 400 mm pulley. Assuming no slip, find the speed n2 of the driven pulley.

n1 = 1500 rpmd1 = 160 mmd2 = 400 mm

Find n2 (rpm)

Show hints
  1. The belt moves at one speed, so both rims have the same speed.
  2. π·d₁·n₁ = π·d₂·n₂.
  3. n₂ = 1500·160/400.
Show the answer and the step-by-step solution

Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.

Equal rim speeds

n₁·d₁ = n₂·d₂

Driven speed

n2 = 1500·160/400 = 600 rpm

Answer

n2 = 600 rpm
Example 2Medium
N1​ = 16N2​ = 80

A 7.5 kW motor running at 1450 rpm drives a pinion with 16 teeth, which meshes with a gear with 80 teeth. The gear pair has an efficiency of 98%. Find the output torque T2 on the gear shaft.

P = 7.5 kWn1 = 1450 rpmN1 = 16N2 = 80η = 98%

Find T2 (N·m)

Show hints
  1. Motor torque from power: T = P/ω, with ω = 2π·n/60 in rad/s.
  2. ω₁ = 151.8 rad/s.
  3. The gear multiplies torque by N₂/N₁ (and loses a little to friction).
Show the answer and the step-by-step solution

Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.

Motor torque

ω₁ = 2π·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 7500/151.8 = 49.39 N·m

Output torque

T2 = 49.39·80/16·0.98 = 242 N·m

Answer

T2 = 242 N·m
Example 3Hard
A = 32B = 60C = 50D = ?

In a compound gear train, gear A (32 teeth) on the input shaft turns at 960 rpm and drives gear B (60 teeth). Gear C (50 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 320 rpm. How many teeth ND must gear D have?

nin = 960 rpmA = 32B = 60C = 50nout = 320 rpm

Find ND

Show hints
  1. Find the speed of the intermediate shaft first: nB = nin·A/B.
  2. Then C drives D: ND = C·nB/nout.
  3. nB = 512 rpm.
Show the answer and the step-by-step solution

Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.

Intermediate shaft speed

nB = nC = 960·32/60 = 512 rpm

Teeth on gear D

ND = C·nC/nout = 50·512/320 = 80

Answer

ND = 80

Questions and answers

What is Gears & gear trains?

Gears pass rotation from one shaft to another. Their teeth lock into each other, so they cannot slip.

A gearbox does two jobs at once. It changes the speed of rotation, and it changes the torque, the turning force on the shaft. Slower means stronger.

How do I solve Gears & gear trains problems step by step?

Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.

What are common mistakes in Gears & gear trains?
  • The larger pulley turns more slowly: n₂ = n₁·d₁/d₂.
  • The rim speeds are equal, not the shaft speeds.
  • Torque goes up when the speed goes down: T₂ = T₁·(N₂/N₁)·η.
What do the Gears & gear trains exercises cover?
  • Easy: speed ratio
  • Medium: torque and power
  • Hard: compound gear train

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Calculators

Gear ratio calculatorBelt length calculator (open and crossed belts)

Courses with this topic

DynamicsMachine DesignMechatronic SystemsFE Mechanical exam practice

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Printable worksheet

Gears & gear trains worksheet with answers
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