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The simple pendulum PowerPoint Presentation
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The simple pendulum

The simple pendulum

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The simple pendulum

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  1. The simple pendulum θ L m

  2. The simple pendulum θ L m mg

  3. The simple pendulum θ L m mg sinθ mg

  4. The simple pendulum θ L x m mg sinθ mg

  5. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L x m mg sinθ mg

  6. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L + x m mg sinθ mg

  7. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ + x m mg sinθ mg

  8. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) + x m mg sinθ mg

  9. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L + x m mg sinθ mg

  10. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma + x m mg sinθ mg

  11. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma ma = - mg x L + x m mg sinθ mg

  12. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma ma = - mg x L + x m mg sinθ mg

  13. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma and a = - g x L + x m mg sinθ mg

  14. Some trig: sin θ = x L For small angles ( < 5 0) θ = x in radians L θ L Restoring force = - mg sinθ ( -ve sign indicates a left displacement and a right restoring force) Restoring force = - mg x L From Newton’s second law F=ma and a = - g x L Compare with SHM equation: a = - (2πf)2 x + x m mg sinθ mg

  15. and a = - g x L Compare with SHM equation: a = - (2πf)2 x θ L + x m mg sinθ mg

  16. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L θ L + x m mg sinθ mg

  17. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L f = 1 g 2π L θ L + x m mg sinθ mg

  18. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L f = 1 g 2π L T = 2π L g θ L + x m mg sinθ mg

  19. and a = - g x L Compare with SHM equation: a = - (2πf)2 x - (2πf)2 = - g L f = 1 g 2π L T = 2π L g θ L + x Discuss: effect of length, mass, gravity, angle of swing. m mg sinθ mg

  20. T = 2πL g

  21. T = 2πL g Put in the form: y = m x + c

  22. T = 2πL g Put in the form: y = m x + c T 2 = 4 π 2L + 0 g

  23. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m

  24. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m Ts Max force on pendulum bob occurs as it passes through the equilibrium: m mg

  25. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m Ts Max force on pendulum bob occurs as it passes through the equilibrium: m mv2 = Ts - mg r mg

  26. T = 2πL g T 2 /s 2 Put in the form: y = m x + c T 2 = 4 π 2L + 0 g L / m Ts Max force on pendulum bob occurs as it passes through the equilibrium: m mv2 = Ts - mg but r = L so mv2 = Ts - mg r L mg