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Understanding Gravity and Elasticity: Key Concepts and Applications in Physics 1710

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phoebe
24 diapositives · 368 Views · Oct 21, 2014

This quiz and interactive materials are designed to enhance comprehension of fundamental physics concepts, including gravity, elasticity, and the laws governing these phenomena. Students will explore Newton’s Universal Law of Gravitation, Kepler’s Laws, and the relationships between stress, strain, and the modulus of elasticity. By engaging with peer instruction and practical examples, learners will establish a solid foundation in Physics 1710, preparing them for deeper exploration of these essential topics.

#understanding gravity and elasticity: key concepts and applications in physics 1710 #understanding gravity and elasticity: key concepts and applications in physics 1710 ppt #powerpoint #presentation
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Understanding Gravity and Elasticity: Key Concepts and Applications in Physics 1710
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  1. Physics 1710—Warm-up Quiz Answer Now ! 0 35% 49 of 140 0 I estimate my grade on exam 2 to be: • A • B • C • D • N/A • F

  2. Physics 1710—Chapter 13 Apps: Gravity 0 CM CM Ball Fg = - mg Center of Gravity (comparison) CG Moon Fg = - GmM/r2

  3. Physics 1710—Chapter 13 Apps: Gravity No Talking! Think! Confer! 0 What is the mass of the earth? How do you know? Peer Instruction Time

  4. Physics 1710—Chapter 13 Apps: Gravity 0 What is the mass of the earth? How do you know? g? Fg = - mg Fg = - mg′ Peer Instruction Time

  5. Physics 1710—Chapter 13 Apps: Gravity 0 1′ Lecture The moduli of elasticity (Y, E, B) characterizes the stress-strain relation: stress= modulus • strain; σ = Y ε The force of attraction between two bodies with mass M and m, respectively, is proportional to the product of their masses and inversely proportional to the distance between their centers squared. Fg = - řG Mm/d 2 G = 6.673 x 10 –11 N m2 /kg2 ~ 2/3 x 10 –10 N m2 /kg2

  6. Physics 1710—Chapter 13 Apps: Gravity 0 Elasticity Definitions: Stress σ : the deforming force per unit area. Strain ε : the unit deformation. Stress = modulus x strain σ = F/A = Y ε

  7. Physics 1710—Chapter 13 Apps: Gravity Failure Elastic limit 0 Elasticity Stress σ – Strain ε “Curve” Stress σ (N/m2) σ = Y ε Strain ε = ΔL/L (%)

  8. Physics 1710—Chapter 13 Apps: Gravity 0 Elasticity Stress σ : the deforming force per unit area. Strain ε : the unit deformation. Tensile/Compressive Stress Young’s ModulusE Stress = modulus x strain σ = F/A = E ε = E ΔL/L ΔL L σ = E ε

  9. Physics 1710—Chapter 13 Apps: Gravity 0 Elasticity Stress σ : the deforming force per unit area. Strain ε : the unit deformation. Shear Modulus G Stress = modulus x strain σ = F/A = G ε = G Δx/h L Δx h σ

  10. Physics 1710—Chapter 13 Apps: Gravity 0 Elasticity Stress σ : the deforming force per unit area. Strain ε : the unit deformation. Hydraulic Stress: Bulk Modulus B Stress = modulus x strain σ = F/A = p = B ε = B ΔV/V ΔV p V

  11. Physics 1710—Chapter 13 Apps: Gravity 0 41% 58 of 140 Answer Now ! 0 Which spring should have the larger spring constant, a short spring (0.1 m long) or a longer spring (0.3 m long)? F = -k x • k larger for shorter spring. • k larger for longer spring. • k the same for both springs. • None of the above.

  12. Physics 1710—Chapter 12 Apps: Gravity 0 F = -k x? σ = F/A = E ε = E ΔL/L F = - A E x/ L k = A E/L Solution: What about a thicker (bigger A) wire?

  13. Physics 1710—Chapter 13 Apps: Gravity d apple 0 Isaac Newton’s Universal Law of GravitationF = - G M m/ d 2 d moon

  14. Physics 1710—Chapter 13 Apps: Gravity 0 Kepler’s Laws: The orbits are ellipses. (Contrary to Aristotle and Ptolemy.) A central force: F ∝ 1/ r 2 or r 2 The areal velocity is a constant. Angular momentum is conserved: ½ v r ∆t = constant implies that rmv = L = constant. T 2∝ r 3 implies F ∝ 1/ r 2, only. ⊙ = r3 T 2

  15. Physics 1710—Chapter 12 Apps: Gravity 0 How did Newton figure out UL of G? Fact: a moon circling a planet has an acceleration of a = v 2 /r Fact: a = F/m. Fact: Kepler had found that the square of the period T was proportional to the cube of the radius of the orbit r : T 2 = k r 3 . Thus: v = 2π r / T

  16. Physics 1710—Chapter 12 Apps: Gravity And T 2= (2πr) 2/(F r /m) = k r 3 Thus: F = (2π) 2m/(k r 2) An “inverse square law,” with k = 1/ [(2π) 2G M] F = G Mm/ r 2, But what value is G?

  17. Physics 1710—Chapter 13 Apps: Gravity d apple 0 Isaac Newton’s Universal Law of GravitationF = - G M m/ d 2 d moon F = – g m g =G M ♁ / R♁2

  18. Physics 1710—Chapter 12 Apps: Gravity g =G M / R♁2 G M ⊗=gR ⊗ 2 = (9.80 N/kg)(6.37x10 6 m)2 = 3.99x10 14 N m 2/kg Gravitation Need to know G or M.

  19. Physics 1710—Chapter 13 Apps: Gravity Henry Cavendish And the Cavendish Experiment M m d

  20. Physics 1710—Chapter 13 Apps: Gravity G = 6.673 x 10 -11 N ‧m 2 /kg 2 G ≈ 2/3 x 10 -10 N ‧m 2 /kg 2 (to an accuracy of 0.1%) So, M ⊗ = (3.99x10 14 N m 2/kg)/(6.673 x 10-11 N ‧m 2/kg 2) = 5.98 x10 24 kg

  21. Physics 1710—Chapter 13 Apps: Gravity F = - G M m/ d 2 What is the order of magnitude of the attraction between two people (m~ 100 kg) separated by a distance of ~ 1m? Gravitational Force: M m d

  22. Physics 1710—Chapter 13 Apps: Gravity 10 0% 0 of 1 Answer Now ! What is the order of magnitude of the attraction between two people (m~ 100 kg) separated by a distance of ~ 1m? • 9.8 N. • 980. N • 6.7 X 10 - 7 N • 6.7 X 10 - 9 N • None of the above

  23. Physics 1710—Chapter 13 Apps: Gravity F = - G M m/ d 2 F = - (6.67 x10 –11N ‧m 2 /kg 2)(100 kg)(100kg )/(1 m)2 F = - 6.67 x10 –7N Equivalent weight = F/g = 67 ng Gravitational Force: M m d

  24. Physics 1710—Chapter 13 Apps: Gravity Summary: Kepler’s Laws The orbits of the planets are ellipses. The areal velocity of a planet is constant. The cube of the radius of a planet’s orbit is proportional to the square of the period. F = - G M m/ d 2