2. CENTRIPETAL FORCE AND ITS APPLICATIONS Authors: Nirav Shah, Jimit Shah, Aditya Modi, Harshal Parekh

3. Master layout or diagram Make a schematic diagram of the concept Explain the animator about the beginning and ending of the process. Draw image big enough for explaining. In above image, identify and label different components of the process/phenomenon. (These are like characters in a film)‏ Illustrate the basic flow of action by using arrows. Use BOLD lines in the diagram, (minimum 2pts.)‏ In the slide after that, provide the definitions of ALL the labels used in the diagram 5 3 2 4 1 INSTRUCTIONS SLIDE

4. Step1- Master Layout 01 roughness of the surface v (km/hr)‏ 10 vtvt 8

5. Objectives: After interacting with this Learning Object, you will be able to:  identify the variables/factors that affect the car’s movement on a curved road  predict the car’s movement under given conditions of velocity, radius of curve and coefficient of static friction.  calculate the maximum speed with which a car can be safely driven along a curved road.

6. DEFINTIONS Frictional Force (F s ): If we either slide down or attempt to slide a body over a surface, the motion is resisted by a bonding between the body and the surface. This resistance is considered to be a single force, called frictional force. it is the parallel component of contact force exerted by one surface on the in contact with the surface. Normal Reaction Force (F N ): When a body presses against a surface, the surface (even a seemingly rigid one) deforms and pushes on the body with a normal force that is perpendicular to the surface. This force is known as Normal Reaction Force. Centripetal Force: This is the net force exerted on an object moving along a circular path. It is always directed orthogonal to the velocity of the body, toward the instantaneous center of curvature of the path. Centripetal Acceleration (a C ): The acceleration provided by the centripetal force for a body in uniform circular motion is known as centripetal acceleration.

7. DEFINTIONS (contd...) ‏ Coefficient of Static Friction ((μ s ): It is a dimensionless scalar value which describes the ratio of the force of friction between two bodies and the force pressing them together. Another way to define it : ratio of the frictional force exerted on an object (by the surface in contact) to the normal force exerted on the object (by the same surface). Gravitational Force (mg): Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. This force is known as the gravitational force. For an object on Earth, the gravitational force exerted on the object by the Earth points straight down, towards the centre of the Earth.

8. Assumptions: The car is treated as a point object and the forces act on the car’s centre of gravity.

9. Derivation of formula: The frictional force is given by- F s =μ s F n ….(1) where μ s is the coefficient of static friction. F s = ma c = mv 2 /r …..(2) ‏ μ s F N = mv 2 /r …….. (3)………from (1) and (2) ‏ The normal reaction force F n balances the weight mg. F N = mg ………..(4) ‏ μ s mg = mv 2 /r …… from (3) and (4) ‏

10. Derivation of formula: (contd..) ‏ μ s g = v 2 /r v 2 = μ s g r Hence, v = This is the maximum speed with which a car can be safely driven along a curved road.

11. Diagrams for reference: F n –normal force exerted on the car by the road F s – frictional force exerted on the car by the road F g – gravitational force exerted on the car by earth

12. F n – normal force exerted by the road F s – frictional force exerted by the road F g – gravitational force exerted by earth The above figure shows the forces exerted on the car.

13. Explain the process 1 5 3 2 4 In this step, use an example to explain the concept. It can be an analogy, a scenario, or an action which explains this concept/process/topic Try to use examples from day-to-day life to make it more clear You have to describe what steps the animator should take to make your concept come alive as a series of moving images. Keep the examples simple to understand, and also to illustrate/animate.

14. Analogy / Scenario / Action 1 5 3 2 4 The concept is of a car which moves at a certain velocity around the corner to take a turn. So depending on the radius of the curve and the velocity at which the car takes the turn there will be a possibility that a car will overturn or not.

15. Animation design Please see the design template provided in the next slide. This is a sample template, and you are free to change as per your design requirements. Try and recreate the sections/subsections as shown in the template. 1 5 2 4 3

16. Want to know more… (Further Reading) ‏ Definitions Formula with derivation Diagram (for reference) ‏ Animation Area Test your understanding (questionnaire) ‏ Lets Learn! Assumptions Lets Sum up (summary) ‏ Interactivity options Sliders(IO1) ‏ / Input Boxes(IO2) ‏ /Drop down(IO3) ‏ (if any) ‏ Play/pauseRestart What will you learn (objectives) ‏ Credits

17. Interactivity and Boundary limits In this section, you will add the ‘Interactivity’ options to the animation. Use the template in the next slide to give the details. Insert the image of the step/s (explained earlier in the Section 3) in the box, and provide the details in the table below. The details of Interactivity could be: Types: Drop down, Slider bar, Data inputs etc. Options: Select one, Multiple selections etc Boundary Limits: Values of the parameters, which won’t show results after a particular point Results: Explain the effect of the interaction in this column Add more slides if necessary 1 2 5 3 4

18. Step 3 Interactivity and Boundary limits

19. Interactivity option:Step No:1 v 1 2 5 3 4 Interact- ivity type Instructions to the learner Boundary limits Instructions to the animator Results (audio narration)‏ Slider bar. Change the value of velocity or (mu)µ, one at a time and click the play button. Observe what happens. The value of ‘g’ is fixed as 9.8m/s 2. Either the user will change “v” or “µ. 1) Speed of the car is within the calculated speed limit, hence the car moves safely along the curved road. The value of ‘r’ is fixed as 0.3 km = 300 m Once the user chooses value for “v” or µ, calculate v t using the formula: v t = √µ x g x r and then highlight v t with red colour as shown above. 1) If v<v t or if v=v t the car will move along the curved road. 2) If v>v t then car will skid. The value of µ ranges from 0 to 1 (values are 0.1, 0.2,0.3, 0.4, …., 0.9,1) ‏ Default value for µ is 0.5 2) Speed of the car is more than the calculated speed limit, hence the car skids. For “v” values are 10, 11, 12,… ….. 98, 99, 100 km/hr and so on. Default value for v is 50 Refer to master layout for the image

20. Interactivity option:Step No:2 v 1 2 5 3 4 Interact- ivity type Instructions to the learner Boundary limits Instructions to the animator Results (audio narration)‏ Slider bar. Now that you have chosen one value keep it fixed and change the value of the other variable. Click on the play button. Observe what happens. The value of ‘g’ is fixed as 9.8m/s 2. Either the user will change “v” or “µ. 1) Speed of the car is within the calculated speed limit, hence the car moves safely along the curved road. The value of ‘r’ is fixed as 0.3 km = 300 m Once the user chooses value for “v” or µ, calculate v t using the formula: v t = √µ x g x r and then highlight v t with red colour as shown above. 1) If v<v t or if v=v t the car will move along the curved road. 2) If v>v t then car will skid. The value of µ ranges from 0 to 1 (values as 0.1, 0.2,0.3, 0.4, …., 0.9,1) Default value for µ is 0.5 2) Speed of the car is more than the calculated speed limit, hence the car skids. For “v” values as 10, 11, 12,… ….. 98, 99, 100 km/hr and so on. Default value for v is 50 Refer to master layout for the image

21. Test your understanding Step 1: Questionnaire Step 2: Summative question

22. QUESTIONNAIRE: Q.1. In an uniform circular motion, the angle between the velocity vector and centripetal acceleration is a) 0 b) 90c) 180d) 45 Q.2. When a car takes a turn on a horizontal road, the centripetal force is provided by the a) weight of the car b) normal reaction of the road c) frictional force between the surface of the road and the tires of the car d) centrifugal force Q.3. A car takes a turn on a slippery road at a safe speed of 9.8m/s. If the coefficient of friction is 0.2, the minimum radius if the arc in which the car takes a turn is a) 20 mb) 49 mc) 24.5 md) 80 m Q.4. If the coefficient of friction between a car and a dry road is 0.7, what would be the maximum safe speed on an exit circular ramp with a radius of 56 m? a) 1190 mb) 1960 mc) 1160 md) 1690 m Answers: 1) c 2) c 3) b 4) b Feedback: If user clicks correct answer then display “Correct!” If user clicks incorrect answer then display “Try again!”

23. Summative question: step 1 v 1 2 5 3 4 Interact- ivity type Options/instr uctions to the user Boundary limits DescriptionResults Slider bar Change the value of velocity or (mu) µ one at a time The value of ‘g’ is fixed as 9.8m/s 2. Once the user chooses value for “v” or µ, calculate v t using the formula: v t = √µ x g x r The value of ‘r’ is fixed as 0.3 km = 300 m The value of µ ranges from 0 to 1 (values as 0.1, 0.2,0.3, 0.4, …., 0.9,1) ‏ Default value for µ is 0.5 For “v” values as 10, 11, 12,… ….. 98, 99, 100 km/ hr and so on. Default value for v is 50 Refer to master layout for the image

24. Summative question: step 2 v 1 2 5 3 4 Interact-ivity type Options/instructi ons to the user Boundary limits Description Click Will the car smoothly take the turn? Click on the appropriate answer. The value of ‘g’ is fixed as 9.8m/s 2. The question and options will appear below the animation area or as a pop-up The value of ‘r’ is fixed as 0.3 km = 300 m The value of µ ranges from 0 to 1 (values as 0.1, 0.2,0.3, 0.4, …., 0.9,1) ‏ Default value for µ is 0.5 For “v” values as 10, 11, 12,… ….. 98, 99, 100 km/hr and so on. Default value for v is 50 Refer to master layout for the image YesNo

25. Summative question: step 3 v 1 2 5 3 4 Description for the animatorFeedback to the learner Refer to the calculated v t in step 1 (slide 23) ‏ 1) If v<v t or if v=v t the car will move along the curved road. 2) If v>v t then car will skid. If user’s answer is correct, display – “Your answer is correct. Click on play button to confirm your answer.” If user’s answer is incorrect, play the animation and go back to show step 2 If user’s answer is incorrect, display – “Click on the play button to view the animation and answer the question again.” Refer to master layout for the image

26. Further reading http://en.wikipedia.org/wiki/Centrifugal_force http://en.wikipedia.org/wiki/Centripetal_force http://dev.physicslab.org/Document.aspx?doct ype=3&filename=CircularMotion_CentripetalF orceExamples.xml

27. Summary  An object traveling in a circular path with a constant speed experiences an acceleration that is constant in magnitude and always directed radially, or towards the center of the circle.  An object’s movement on a circular path which moves at a certain velocity around the corner to take a turn the possibility that a car will overturn or not depends on o the radius of the curve o the velocity at which the car takes the turn o the coefficient of static friction  The maximum speed with which a car can be safely driven along a curved road is given by – v = √µgr, where µ - coefficient of static friction g – acceleration due to gravity r – radius of the curve