CHAPTER 1 POWER TRANSMISSION SYSTEM – GEAR SYSTEM

1.1 Introduction to Gear System 1

1.2 Types of Gear system 1

1.3 Relationship between pitch diameter and pitch circle 7

1.4 Gear Ratio 9

1.5 Gear Train 11

1.6 Gear Efficiency 12

1.7 Power Transmission in a Gear Train System 12

1.8 Equivalent Moment of Inertia 14

1.9 Gear Train Applications 15

1.10 Vehicle Dynamics 22

CHAPTER 1

POWER TRANSMISSION SYSTEM: GEAR SYSTEM

1.1 Gear System

Gears are used for two basic purposes; increase or decrease of rotation speed and increase or decrease of power or torque. Torque is a measure of a force to produce torsion and rotation about an axis. To increase speed and reduce torque a large drive gear is coupled to a smaller driven gear. To reduce speed and increase torque a small Lego gear turning a larger gear is used. They are also used for enhancement for positioning systems.

In gear system, gear that functions as mover mentioned as driver gear, while gear moved name as driven gear.

1.2 Types of Gear System

Gears are used in a wide area of industries including automotive, milling, paper industry etc. Different types of gears are also custom designed and are fabricated by gear manufacturing services as per the specifications. Gears are categorized into several classifications.

1. According to the relative position of the axes of mating gears. The axes of mating gears may be:i. parallel ii. intersectingiii. nonintersecting and nonparallel.

Examples:

Parallel AxesIntersecting

AxesNon-Intersecting

(Non-parallel) AxesRotary to

Translation

Spur Gears Bevel gears Hypoid gears Rack and Pinion

Helical Gears Straight bevel Crossed helical gears -

Herring bone or double helical

gearsZerol bevel Worm gears -

- Spiral bevel - -

Table 2

2. According to the peripheral speed of gears:i. Low velocity gears – velocity < 3 m/sii. Medium velocity gears – 3 m/s < velocity < 15 m/siii. High velocity gears – velocity > 15 m/s

3. According to the type of gearing:i. External gearingii. Internal gearingiii. Rack and pinion

4. According to the type of teeth:i. Straight teethii. Helical teethiii. Curved teethiv. Herringbone teeth

Main classification method: The relative position of the axes of mating gears

a) Parallel Axes Shaft

The shaft axes between driver and driven gear is parallel to each other. Example of this type of gear is Spur Gear.

Figure 1: Spur Gear

Spur gears connect parallel shafts, have involute teeth that are parallel to the shaft and can have internal or external teeth. They cause no external thrust between gears. They are inexpensive to manufacture. They give lower but satisfactory performance. They are used when shaft rotates in the same plane.

The main features of spur gears are dedendum, addendum, flank, and fillet. Dedendum cylinder is a root from where teeth extend, it extends to the tip called the addendum circle. Flank or the face contacts the meshing gear, the most useful feature of the spur gears. The fillet in the root region is kinetically irrelevant.

The speed and change of the force depends on the gear ratio, the ratio of number of teeth on the gears that are to be meshed. One gear among the two is on the input axle; the axle of the motor and the other gear of the pair are on the output axle, the axle of the wheel.

They have higher contact ratio that makes them efficient in operation. They are available for corrosion resistant operation. They are among the most cost-effective type of gearing. They are also used to create large gear reductions.

They are available in plastic, non-metallic, brass, steel and cast iron and are manufactured in a variety of styles. They are made with many different properties. Factors like design life, power transmission requirements, noise and heat generation, and presence of corrosive elements contribute to the optimization of the gear material.

Generally used in simple machines like washing machines, clothes dryer or power winches. They are not used in automobiles because they produce sound when the teeth of both the gears collide with each other. It also increases stress on the gear teeth. They are also used in construction equipment, machine tools, indexing equipment, multi spindle drives, roller feeds, and conveyors.

b) Intersecting Axes Shaft

The shaft axes between driver and driven gear is perpendicular to each other. Example of gear is bevel gears.

Figure 2: bevel gear

Figure 2: Bevel Gear

They connect intersecting axes and come in several types. The pitch surface of bevel gears is a cone. They are useful when the direction of a shaft's rotation needs to be changed. Using gears of differing numbers of teeth can change the speed of rotation. They are usually mounted on shafts that are 90 degrees apart, but can be designed to work at other angles as well.

These gears permit minor adjustment during assembly and allow for some displacement due to deflection under operating loads without concentrating the load on the end of the tooth. For reliable performance, gears must be pinned to shaft with a dowel or taper pin. Bevel gear sets consist of two gears of different pitch diameter that yield ratios greater than 1:1.

The teeth on bevel gears can be straight, spiral or bevel. In straight bevel gears teeth have no helix angles. They either have equal size gears with 90 degrees shaft angle or a shaft angle other than 90 degrees. Straight bevel angle can also be with one gear flat with a pitch angle of 90 degrees. In straight when each tooth engages it impacts the corresponding tooth and simply curving the gear teeth can solve the problem. Spiral bevel gears have spiral angles, which gives performance improvements. The contact between the teeth starts at one end of the gear and then spreads across the whole tooth. In both the bevel types of gears the shaft must be perpendicular to each other and must be in the same plane. The hypoid bevel gears can engage with the axes in different planes. This is used in many car differentials. The ring gear of the differential and the input pinion gear are both hypoid. This allows input pinion to be mounted lower than the axis of the ring gear. Hypoid gears are stronger, operate more quietly and can be used for higher reduction ratios. They also have sliding action along the teeth, potentially reducing efficiency.

A good example of bevel gears is seen as the main mechanism for a hand drill. As the handle of the drill is turned in a vertical direction, the bevel gears change the rotation of the chuck to a horizontal rotation. The bevel gears in a hand drill have the added advantage of increasing the speed of rotation of the chuck and this makes it possible to drill a range of materials.

The bevel gears find its application in locomotives, marine applications, automobiles, printing presses, cooling towers, power plants, steel plants, defenses and also in railway track inspection machine. They are important components on all current rotorcraft drive system.

Spiral bevel gears are important components on all current rotorcraft drive systems. These components are required to operate at high speeds, high loads, and for an extremely large number of load cycles. In this application, spiral bevel gears are used to redirect the shaft from the horizontal gas turbine engine to the vertical rotor.

c) Non-Intersecting (Non-parallel) Axes Shaft

The shaft axes between the driver and driven gears are neither intersecting nor parallel at the same time. A very good example of a non-intersecting-non-parallel gear system is a worm gear as in Figure 3.

Figure 3: Worm Gear

A worm gear is an inclined plane wrapped around a central axle. It is a gear with one or more teeth in the form of screwed threads. Worm gears are made of two parts: the pinion and the worm gear. The pinion has small number of teeth and they wrap around the pitch cylinder. The worm gear has concave faces to fit the curvature of the worm in order to provide line of contact instead of point of contact. They are cut helically for better mating Worm gears can provide a high angular velocity between non-intersecting shafts at right angles.

They are capable of transmitting high tooth loads, the only disadvantage is the high sliding velocities across the teeth. They provide ultimate power ratio. The efficiency of worm gear depends on the lead angle, sliding speed, and lubricant, surface quality and installation conditions. They offer smoothest, quietest form of gearing. They provide high-ratio speed reduction in minimal spaces.

Worm gears are used when large gear reductions are required. Worm gear has a unique property of easily turning the gear. The gear cannot turn the worm because the angle on the worm is shallow and when the gear tries to spin the worm, the friction between the two holds the worm in place.

Worm gears work under difficult conditions, presenting unique lubrication demands. The types of oils most commonly used to lubricate worm gears are compounded mineral oils, EP mineral gear oils and synthetics. Worm gear is always used as the input gear. For the operation of worm gear, torque is applied to the input end of the worm shaft by a driven sprocket or electric motor. The worm and the worm shaft are supported by anti-friction roller bearings. Because of high friction worm gears are very inefficient. There is lot of friction between a worm gear and the gear being driven by the worm gear. When used in high torque applications, the friction causes the wear on the gear teeth and erosion of restraining surface.

There are three types of worm gears. Non throated- a helical gear with a straight worm. Tooth contact is a single moving point on the worm drive. Single throated- has concave helical teeth wrap around the worm. This leads to line contact. Double throated- called a cone or hourglass. It has concave teeth both on the worm and helical gear. Worm gears are widely used in packaging machinery, material handling, machine tools, indexing and food processing. They are used widely in conveyor systems. They are also used in torsen differential, used on some high-performance cars and trucks. They serve as speed reducers in many different industries.

d) Perpendicular Axes Shaft / Rotary to Translation

The shaft axes between driver and driven gear are perpendicular to each other and do not intersect to each other. Example of gear is Rack and Pinion Gear.

Figure 4: Rack and Pinion Gear

Rack and pinion gears are used to convert rotation into linear motion. The speed with which the rack moves as the pinion turns is determined by the diameter of the gear. The flat, toothed part is the rack and the gear is the pinion. A piston coaxial to the rack provides hydraulic assistance force, and an open centered rotary valve controls the assist level. A rack and pinion gears system is composed of two gears. The normal round gear is the pinion gear and the straight or flat gear is the rack. The rack has teeth cut into it and they mesh with the teeth of the pinion gear. Rack and pinion gears provide a less mechanical advantage than other mechanisms, but greater feedback and steering sensation.

Rack and pinion gears are available in three variations: straight teeth metric pitch, straight teeth modular pitch, and helical teeth modular pitch. Rack and pinion gears variations are available in different qualities: 9/10 milled teeth or milled and hardened quality, 7/8 precision cut or precision cut and hardened quality, and 5/6 teeth hardened and ground quality. A rack and pinion gear gives a positive motion especially compared to the friction drive of a wheel in tarmac. In a rack and pinion railway, a central rack between the two rails engages with a pinion on the engine allowing a train to be pulled up very steep slopes. A rack and pinion gear is the differential's critical point of power transfer. A rack and pinion gear set is one of the simplest performance modifications that can be performed on a vehicle. The most common reason to change rack and pinion ratios from the original equipment is to retain power when bigger tires are put on a vehicle. The torque can be increased by a ratio change when there is enhanced pulling or higher take off power from a dead start. A well designed mechanism such as the rack and pinion gears save effort and time.

Rack and pinions gears are commonly used in the steering system of cars to convert the rotary motion of the steering wheel to the side to side motion in the wheels. The steering wheel rotates a gear which engages the rack. As the gear turns, it slides the rack either to the right or left, depending on which way the wheel is turned. Rack and pinion gears are also used in some scales to turn the dial that displays a weight.

1.3 Relationship between Pitch Diameter and Pitch Circle

Figure 5: Gear terminology

Some of the terminology of gear system includes; Pitch surface : The surface of the imaginary rolling cylinder (cone, etc.) that the toothed

gear may be considered to replace.

Pitch circle: A right section of the pitch surface.

Addendum circle: A circle bounding the ends of the teeth, in a right section of the gear.

Root (or dedendum) circle: The circle bounding the spaces between the teeth, in a right

section of the gear.

Addendum: The radial distance between the pitch circle and the addendum circle.

Dedendum: The radial distance between the pitch circle and the root circle.

Clearance: The difference between the dedendum of one gear and the addendum of the

mating gear.

Flank of a tooth: The part of the tooth surface lying inside the pitch surface.

Circular thickness (also called the tooth thickness): The thickness of the tooth

measured on the pitch circle. It is the length of an arc and not the length of a straight line.

Tooth space: The distance between adjacent teeth measured on the pitch circle.

Backlash: The difference between the circle thickness of one gear and the tooth space of

the mating gear.

Circular pitch p: The width of a tooth and a space, measured on the pitch circle.

Diametral pitch P: The number of teeth of a gear per inch of its pitch diameter. A

toothed gear must have an integral number of teeth. The circular pitch, therefore, equals

the pitch circumference divided by the number of teeth. The diametral pitch is, by

definition, the number of teeth divided by the pitch diameter. That is,

and Hence

Where;

p = circular pitch N = number of teeth

P = diametral pitch D = pitch diameter

That is, the product of the diametral pitch and the circular pitch equals .

Module m: Pitch diameter divided by number of teeth. The pitch diameter is usually

specified in inches or millimeters; in the former case the module is the inverse of

diametral pitch.

Pinion: The smaller of any pair of mating gears. The larger of the pair is called simply

the gear.

Velocity ratio: The ratio of the number of revolutions of the driving (or input) gear to the

number of revolutions of the driven (or output) gear, in a unit of time.

Pitch point: The point of tangency of the pitch circles of a pair of mating gears.

Pressure angle : The angle between the common normal at the point of tooth contact

and the common tangent to the pitch circles. It is also the angle between the line of action

and the common tangent.

1.4 Gear Ratio

Gear ratio is defined as ratio of speed of driven gear over the speed of driver gear.

Consider a gear set below;

Figure 6: Gear set

When two gear mate efficiently at point A, the velocity, of both gear are the same.

Thus;

with

Then from will produce

Where = speed of driver gear= speed of driven gear

D1 = pitch diameter of driver gearD2 = pitch diameter of driven gear

Hence, gear ratio, n;

Where N1= number teeth of driver gearN2= number teeth of driven gear

= angular acceleration of driver gear=angular acceleration of driven gear

1.5 Gear Train

Gear trains consist of two or more gears for the purpose of transmitting motion from one axis to another.

Driver gear

Driven gear

Simple Gear Train is the most common of the gear train is the gear pair connecting parallel shafts. The teeth of this type can be spur, helical or herringbone. The angular velocity is simply the reverse of the tooth ratio. The main limitation of a simple gear train is that the maximum speed change ratio is 10:1. For larger ratio, large sizes of gear trains are required; this may result in an imbalance of strength and wear capacities of the end gears.

The sprockets and chain in the bicycle is an example of simple gear train. When the paddle is pushed, the front gear is turned and that meshes with the links in the chain. The chain moves and meshes with the links in the rear gear that is attached to the rear wheel. This enables the bicycle to move.

Compound Gear Train is used for large velocities, compound arrangement is preferred. Two keys are keyed to a single shaft. A double reduction train can be arranged to have its input and output shafts in a line, by choosing equal center distance for gears and pinions.

Gear trains are used in representing the phases of moon on a watch or clock dial. It is also used for driving a conventional two-disk lunar phase display off the day-of-the-week shaft of the calendar.

Figure 7 (a) Simple gear train (b) Compound gear train

When a gear train is complex (consist of many gear sets), it is important for the designer

to identify the rotation of the driver and the final driven gear respectively. However, there is a simple formula to determine the rotation of each successive gear in a gear train.

Figure 8: Network gear short

A gear train may have several drivers and several driven gears. When gear A turns once clockwise, gear B turns 4 times counter-clockwise and gear C turns once clockwise. Hence gear B does not change the speed of C from what it would have been if geared directly to gear A, but it changes its direction from counterclockwise to clockwise.

Figure 9: Odd number of mating gear

“For an ODD number of mating gears, the rotation of Driven gear is the SAME as Driver Gear.”

Figure 10: Even number of mating gear

“For an EVEN number of mating gears, the rotation of Driven gear is REVERSE of Driver Gear.”

Another classification of gear train is called Reverted Gear Train and Epicyclic Gear Train.

1.6 Gear Efficiency

Gear efficiency is defined as the ratio of Output Power from Driven Gear to the Input Power from Driver Gear. Gear efficiency measures how efficient a gear system is to transmit power. High value of gear efficiency reflects a more efficient gear system. Power loss in a gear system may come from sources like friction, slip, backlash and so on. From Power, , then

Gear Efficiency,

Where = Input power from driver gear = Output power from driven gear

= Gear ratio

If the , thus the torque at driver gear is;

If the , thus the torque at driver gear is;

1.7 Power Transmission in a Gear Train System

In a gear train system, power loss normally happen in the bearing and gear due to friction and loading imposed on it and also power loss in overcoming shaft inertia. Consider a gear train consists of two sets of gear reducing arrangement. A motor is attached to the system with is the moment of inertia of motor shaft, is moment of inertia of middle shaft and is the moment of inertia of hoist which acts as the load of the system. Gear ratio and gear efficiency of gear set 1-2 is and , between gear set 3-4 is and respectively. Let;

= Torque of motor = Torque of hoist = Friction torque at bearing X

Draw the free body diagram and using Newton Second Law,

Figure 11: Free body diagram

Assume clockwise direction as positive value.

For (A)……………………………………………….…………… (1)

For (B)…………………………………….………………………. (2)

Since there is gear mating between gear 1 and 2, thus, must include in the analysis its own gear ratio and gear efficiency, and relate it to the inertia of middle shaft, .

Previously, , thus it follows that

………………………………..……………………………….………. (3)

For (C)……………………….………………..……………………….. (4)

also …………………………………………………………………………..(5)

Using power, power transfer to each gear component is;

Hoist

a) Power transfer by the motor

b) Power at gear 1

c) Power at gear 2

d) Power at gear 3

e) Power at gear 4

f) Power at hoist

g) Overall power transfer efficiency,

Thus if friction torque, effect is neglected,

This concludes that

Also;

1.8 Equivalent Moment of Inertia, [ Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg·m2) is a measure of an object's resistance to changes in its rotation rate]

Consider a simple gear system as Figure below. In order for the driver gear A to start rotate, it must have enough torque to overcome its own inertia, first, and then another additional torque to start accelerate the driven gear B. However, to relate torque with the gear parameter, inertia term will be taken into account. For a simple gear system, the solution is straightforward, but when it comes to complex gear train design, it is useful to simplify / group together all inertia term in the system into a single compact inertia expression. The inertia term of each moving gear parts will be referred to a single part in the system, normally at motor side.

Figure 12: Equivalent moment of inertia

1. Torque at B to overcome

Refer to gear A side. Use gear ratio,

Thus,

2. Gear efficiency is related to power and thus torque of the mating gears, thus

3. Therefore, torque at A, to accelerate

4. Therefore total torque at A to accelerate and is

, Or in general form, (referred to motor side)

Thus

The derivation of of this simple gear system can be extended to a double set of gear reducing problem as in section 1.7. By neglecting the friction torque effect, , thus,

Driven gear Driver gear

IB

IA

1.9 Gear Train Applications (Solved Problem)

Example 1(Taken from Roslan, Che’ Abas, Yunus (2001), UTM)

A motor is accelerating a 250 kg load with acceleration of 1.2 m/s2 through a gear system as shown below. The rope that carries the load are encircled on a hoist with diameter 1.2m.Gear for the hoist’s shaft has 200 teeth, gear for motor shaft has 20 teeth. Gear efficiency is 90%. Mass and radius of gyration of each shaft is as below;

Mass (kg) Radius of gyration (mm)Motor shaft 250 100Hoist shaft 1100 500

Calculate the torque of the motor needed to bring up the load with acceleration 1.2 m/s 2. Neglect friction effect.

Figure 13: Gear system attached to hoist

Solution

Total torque at motor to bring up load

Where = Torque to overcome equivalent inertia (refer to motor side).=Torque to accelerate the load through gear system

a) Consider for

From

Thus = Motor shaft inertia

kgm2

= Hoist shaft inertia

kgm2

Dia = 1.2 m

Hoist

Gear ratio,

Put into kgm2

Acceleration of hoist,

Thus rad/s2

From the gear ratio, angular acceleration of motor, rad/s2

Now torque due to equivalent inertia, Nm.

b) Consider for

From Newton 2nd Law,

Then, torque at hoist

Nm

But due to gear efficiency (since the hoist shaft is connected to the gear system), torque to

accelerate the load,

Nm

Then total torque referred to motor side is;

Nm

Figure 14: Free body diagram of hoist and load

Example 2(Taken from Roslan, Che’ Abas, Yunus, (2001), UTM)

250g

F a

TG

250 kg

r

Hoist

Figure 12 below shows a motor used to accelerate a hoist through two sets of gear reducing system. Moment of inertia for the motor shaft is 5 kgm2, middle shaft is 40 kgm2 and hoist shaft is 500 kgm2. Gear ratio for gear set 1 and 2 is 1/3.5 while for gear set 3 and 4 is 1/ 4.5. Gear efficiency for both gear set is 90%. By neglecting the friction effect, find the total torque required by the motor to accelerate the load of 6 tones at acceleration of 0.4 m/s2.

Figure 15: Loading system on gear

Solution

Given that =5 kgm2, kgm2, =500 kgm2, , ,

Neglect friction effect.

Total torque required for the motor is

Where = Torque to overcome equivalent inertia (refer to motor side).=Torque to accelerate the load through gear system

a) Consider for Recall that , but for two set of gear system with friction effect is neglected,

From question, given that, m/s2, thus;

Diameter = 1.2 m

Hoist

rad/s2

From gear ratio,

Thus, rad/s2

Thus, Nm.

It is known that referred to motor side will be denoted as and is related by

Nm

Thus total torque at motor required is

Nm.

Example 3(Taken from Roslan, Che’ Abas, Yunus, (2001), UTM)

4500g

F1a

4500 kg

Hoist

6000g

6000 kg

F2

a

Figure 16: Loading FBD

b) Consider for in Figure 13;

From Newton 2nd Law,

N

N

Resultant torque at hoist

kN

Thus torque at hoist

kN

Figure 17: Gear with inclined loading

Figure 14 above shows a motor accelerating a hoist with diameter 0.9m, through two sets of gear reducing system. Gear ratio for gear 1 and 2 is 1/3.5 while for gear 3 and 4 is 1/ 4.5. Moment of inertia for the motor shaft is 5 kgm2, middle shaft is 20 kgm2 and hoist shaft is 100 kgm2. The rope that is encircled on the hoist must be capable to lift up a load of 5 tones that is sliding on a 1 in 50 slope. Friction on the slope is 1000N and the total torque at motor required to raise the load is 1500N. Use gear efficiency of 90% for both gear set. If there is friction torque effect on the middle shaft, Nm and at hoist shaft is Nm. Calculate the acceleration of the load at the above condition.

Solution

For the overall gear ratio,

Total torque required by motor to raise load

Where = Torque to overcome equivalent inertia (refer to motor side).=Torque to accelerate the load through gear system

Total torque to overcome friction effect.

a) Consider for Previously, For double set of gear reducing system,

kgm2

From , thus,

Dia = 0.9 m

Hoist

Also from gear ratio, , thus,

Thus Nm

b) Consider for as in Figure 15;

c) Consider for

Friction effect can be grouped together to form where;

Nm

From

Thus m/s2

1.10 Vehicle Dynamics

RF5000g

1F

θ

Mg sinθ

r

F1

hoist

Figure 18: inclined loading

From

NThus, torque to accelerate hoist

Nm

Use gear efficiency to relate with

Nm

Figure 19: Vehicle dynamics

For a moving vehicle as in Figure 19, some of the forces acting on it are; Friction due to the vehicle’s body (aerodynamic friction), Forces due to friction from the engine to the wheel such as friction in bearing, shaft,

clutch and gears, . Forces due the acceleration of the vehicle, which is called tractive force, considering

no slip between the wheel and the road surface.We can estimate the speed of the moving vehicle by considering the speed of the wheel itself.

Vehicle speed,

Example 1 (Solved Problem)

Total mass for a two wheeled motorcycle including passenger is 190 kg. The engine produce torque of 25 Nm at speed of 1800 RPM. Moment of inertia for each wheel is 1.4 kgm2 while for other rotating parts in the engine is considered as 0.15 kgm2. The wheel’s effective diameter is 610 mm. If the motorcycle is moving on a road with a speed 23 km/hr at second gear, find

(i) Gear ratio for the second gear(ii) Acceleration at speed 23 km/hr

Assume wind friction is 200 N and gear efficiency is 90%.

Solution

Given that;

=190 kg, =25 Nm at =1800 PM, =1.4 kgm2, =0.15 kgm2, =610 mm, If =23 km/hr at 2nd gear, with =200 N, =90%.

1st Step – Draw Free Body Diagram

D/2

wr

FT

R

v

Wheel

Surface

Figure 20: Free body diagram

2nd StepTotal torque at engine,

Where = Torque due to equivalent inertia of rotating parts in the engine.(referred to engine side).

= Torque to accelerate the wheel.

3rd Step Convert all measurement to SI standard.

Engine speed, rad/s

Wheel speed, m/s

4th Step– Determine In order to find gear ratio for second gear,

(Where )

Thus gear ratio for second gear is

Equivalent moment of inertia is

kgm2

In order to find , use and gear ratio

Engine

Gear system

Wheel

Iint IR

rad/s2

Thus, Nm.

5th Step – Determine as in Figure 18;

Figure 21: dynamic force on the motorcycle

From

Total torque at wheel,

Refer to motor side using gear efficiency

Nm

Thus, total torque at engine

The acceleration at that speed is;By solving the equation for the total torque above, thus

m/s2

PROBLEMS-GEAR SYSTEM

m=190 kg

FT

R

a

1. The axes of two parallel shafts are to be 600mm apart approximately, and have to be

connected by spur gear, having a circular pitch of 30 mm. If gear A rotate at 200 rpm and

gear B rotate at 600 rpm, find the number of teeth on each gear.

2. Figure below showed a motor used to accelerate a hoist through a set of gear system.

Gear for the hoist’s shaft has 200 teeth and gear for motor shaft has 20 teeth. Gear

efficiency is 90 %. Moment of inertia for the motor shaft is 2.5 kgm² and hoist shaft is

275 kgm². The rope that carries a 250 kg load are encircle on hoist with diameter 1.2 m.

By neglecting the friction, find

a) gear ratio,

b) equivalent moment inertia for a gear system,

c) the total torque required by the motor to accelerate the load at acceleration of 1.0

m/s².

Motor

Rope

Set of gear

Load

Hoist

600 mm

3 The diagram above shows a gear train composed of three gears. Gear A revolves at 60

revs/min in a clockwise direction.

GEAR A GEAR B GEAR C

20 TEETH 60 TEETH 10 TEETH

a) What is the output in revolutions per minute at Gear C?

b) In what direction does Gear C revolve ?

[ 120 rpm, clockwise ]