Teacher Mechanics Note to the teacher On this page, students will learn about the relationship between gear ratio, gear rotational speed, wheel radius, diameter, circumference, revolutions and distance. Students will determine gear ratio by dividing the number of teeth on the driven gear by the number of teeth on the driving gear. They will determine the rotational speed of the driving gear by multiplying the gear ratio by the rotational speed of the driven axle. They will determine distance by multiplying this rotational speed by the wheel circumference. They will also have to know how to determine circumference from diameter and radius. Students will have to use both fractions and decimals to make these calculations. While the worksheet is designed to help students learn how to determine gear ratio and derive rotational speed from it, and to derive distance from wheel circumference and rotational speed, and may be successfully completed by students with little background in these areas, the existing ability to multiply and simplify fractions, the ability to use decimals, and the ability to determine circumference from diameter or radius will be necessary to successfully complete the worksheet. Teachers may wish to review any or all of these skills depending on their students background. Note that this exercise is more challenging than the exercises in Gears 1-4. To successfully complete the worksheet, students must have a working understanding of most of the concepts and equations introduced in both the Gears and Wheels worksheets. Students will also have to perform more calculations than in any of the previous Gears and Wheels worksheets. Note that there are no instructions regarding rounding. The answers assume rounding to 2 digits beyond the decimal place, except for known fractions. Teachers may wish to supply additional instructions. If they do not, students answers may vary slightly according to what rounding conventions they use. The formulas are: Wheel Circumference Revolutions Distance 1 (# Teeth on driven gear) (# Teeth on driving gear) = Gear Ratio x = 3 9.82 1 9.82 2 (Gear Ratio) X (Speed of driven gear) = (Speed of driving gear) 4 (Speed of driving gear) (Gear Ratio) = (Speed of driven gear) The necessary formulas (above) are included with the instructions to reinforce the concepts. 5.61
Mechanics Teacher A. B. C. 1. Assuming the circumference of the wheel and the RPM of the motor are exactly the same for all experiments, which gear ratio would create the most speed, A, B or C? Why did you choose that answer? The gear ratio B is lowest, meaning it would create the highest speed for the driven axle. 2. If the driving gear is moving at 100 RPM, how fast will the driven gear move for pictures A, B and C? A. 58.3 RPM B. 300 RPM C. 100 RPM Students are expected to use the following procedure: Identify gear sizes by counting teeth Determine the gear ratio of each pair of gears Identify the information required by the question Reconstruct the equations provided as an example Enter the data provided into these equations Manipulate the equations if necessary Solve the equations Note that the first question tests comprehension of the concept of gear ratios. Some students may have more difficulty with this question than the others, which only require entering numbers in equations and solving them. Approximate classroom time: 15-25 minutes depending on students background 5.62
Teacher Mechanics Students successfully completing the worksheet will be able to: 1. Define gear ratio 2. Understand the concept of gear ratio, in particular that low gear ratios will create more wheel speed than high gear ratios when motor speed is held constant 3. Identify gear size by counting teeth 4. Simplify fractions 5. Describe the geometry of a circle 6. Describe the relationship between radius, diameter, circumference, revolutions and distance for a wheel 7. Calculate diameter from radius 8. Calculate circumference from diameter 9. Derive driving gear RPM from driven gear RPM if given the gear ratio 10. Calculate distance from wheel circumference and revolutions 11. Multiply decimals and fractions 12. Reconstruct equations relating diameter, circumference, revolutions and distance 13. Identify data provided in word problems 14. Identify information required in the word problems 15. Manipulate these equations to solve for the required information Standards addressed: Math Standards Numbers and Operations Algebra Geometry Measurement Problem Solving Connections Technology Standards The Nature of Technology Standard 1 Design Standards 8, 9 Abilities for a Technological World Standard 12 Using Technology to Design the Future Standards 16, 18, 19 Science Standards Content Standard B Content Standard E Note: Workbook answers begin on the next page. 5.63
Mechanics Teacher Instructions Use the formulas and pictures below to answer the following questions Note: The driving gear is always on the right. Possible gear sizes are 40, 24, 14 and 8 tooth gears. The formulas are: Wheel Circumference Revolutions Distance 1 (# Teeth on driven gear) (# Teeth on driving gear) = Gear Ratio x = 3 9.82 1 9.82 2 (Gear Ratio) X (Speed of driven gear) = (Speed of driving gear) 4 (Speed of driving gear) (Gear Ratio) = (Speed of driven gear) A. B. C. 1. Assuming the circumference of the wheel and the RPM of the motor are exactly the same for all experiments, which gear ratio would create the most speed, A, B or C? Why did you choose that answer? We know that the lower the gear ratio, the faster the driven gear will revolve for every revolution of the driving gear. By counting teeth, we know that the lowest gear ratio is the one in picture B. 2. If the driving gear is moving at 100 RPM, how fast will the driven gear move for pictures A, B and C? To find the answer to this question, we can use Formula 4. For picture A, we can see that the gear ratio is 24/14, which simplifies to 12/7, and we know from the question that the driving gear is moving at 100 RPM. So 100 RPM/(12/7) = 100 RPM x 7/12 = 58.3 RPM. For picture B, we can see that the gear ratio is 8/24, which simplifies to 1/3, and we know from the question that the driving gear is moving at 100 RPM. So 100 RPM/(1/3) = 100 RPM x 3 = 300 RPM. For picture C, we can see that the gear ratio is 24/24, which simplifies to 1/1, and we know from the question that the driving gear is moving at 100 RPM. So 100 RPM/(1) = 100 RPM. 5.64
Teacher Mechanics 3. If the driving gear moves at 100 RPM and the circumference of the wheel is 4.4 cm, how far will the robot move in 2 minutes for pictures A, B and C? To find the answer to this question, we can use Formula 4 and then Formula 3. For picture A, we can see that the gear ratio is 24/14, which simplifies to 12/7, and we know from the question that the driving gear will move 100 revolutions/minute x 2 minutes = 200 revolutions. So 200 revolutions/(12/7) = 200 revolutions x 7/12 = 116.6 revolutions. Now we can use formula 3, above. So 116.67 x 4.4 cm = 513.35 cm. For picture B, we can see that the gear ratio is 8/24, which simplifies to 1/3, and we know from the question that the driving gear will move 100 revolutions/minute x 2 minutes = 200 revolutions. So 200 revolutions/(1/3) = 200 revolutions x 3 = 600 revolutions. Now we can use formula 3, above. So 600 x 4.4 cm = 2640 cm. For picture C, we can see that the gear ratio is 1/1, and we know from the question that the driving gear will move 100 revolutions/minute x 2 minutes = 200 revolutions. So 200 revolutions/1 = 200 revolutions Now we can use formula 3, above. So 200 x 4.4 cm = 880 cm. 4. If the driving gear moves at 200 RPM and the diameter of the wheel is 1.9 cm, how far will the robot move in 3 minutes for pictures A, B and C? To find the answer to this question, we can use Formula 4 and then Formula 3. For picture A, we can see that the gear ratio is 24/14, which simplifies to 12/7, and we know from the question that the driving gear will move 200 revolutions/minute x 3 minutes = 600 revolutions. So 600 revolutions/(12/7) = 600 revolutions x 7/12 = 350 revolutions. The wheel circumference is π x diameter, or π x 1.9 = 5.97 cm. Now we can use formula 3, above. So 350 x 5.97 cm = 2089.5 cm. For picture B, we can see that the gear ratio is 8/24, which simplifies to 1/3, and we know from the question that the driving gear will move 200 revolutions/minute x 3 minutes = 600 revolutions. So 600 revolutions/(1/3) = 600 revolutions x 3 = 1800 revolutions. The wheel circumference is π x diameter, or π x 1.9 = 5.97 cm. Now we can use formula 3, above. So 1800 x 5.97 cm = 10746 cm. For picture C, we can see that the gear ratio is 1/1 and we know from the question that the driving gear will move 200 revolutions/minute x 3 minutes = 600 revolutions. So 600 revolutions/(1) = 600 revolutions. The wheel circumference is π x diameter, or π x 1.9 = 5.97 cm. Now we can use formula 3, above. So 600 x 5.97 cm = 3582 cm. 5.65
Mechanics Teacher 5. If the driving gear moves at 200 RPM and the radius of the wheel is 2.5 cm, how far will the robot move in 3 minutes for pictures A, B and C? To find the answer to this question, we can use Formula 4 and then Formula 3. For picture A, we can see that the gear ratio is 24/14, which simplifies to 12/7, and we know from the question that the driving gear will move 200 revolutions/minute x 3 minutes = 600 revolutions. So 600 revolutions/(12/7) = 600 revolutions x 7/12 = 350 revolutions. The wheel circumference is 2 x π x r = 2 x π x 2.5 cm = 15.7 cm. Now we can use formula 3, above. So 350 x 15.7 cm = 5495 cm. For picture B, we can see that the gear ratio is 8/24, which simplifies to 1/3, and we know from the question that the driving gear will move 200 revolutions/minute x 3 minutes = 600 revolutions. So 600 revolutions/(1/3) = 600 revolutions x 3 = 1800 revolutions. The wheel circumference is 2 x π x r = 2 x π x 2.5 cm = 15.7 cm. Now we can use formula 3, above. So 1800 x 15.7 cm = 28260 cm. For picture C, we can see that the gear ratio is 1/1, and we know from the question that the driving gear will move 200 revolutions/minute x 3 minutes = 600 revolutions. So 600 revolutions/(1) = 600 revolutions. The wheel circumference is 2 x π x r = 2 x π x 2.5 cm = 15.7 cm. Now we can use formula 3, above. So 600 x 15.7 cm = 9420 cm. 5.66