1 / 24

Flashback

Flashback. An object experiencing circular motion has an angular speed of 10 rad /s and is traveling 5 m away from its axis of rotation. Find the object’s centripetal acceleration. Hint: a c = r ω 2. Announcements. Units & Symbols Quiz on Wednesday Circular Motion Quiz on Thursday

jude
Télécharger la présentation

Flashback

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Flashback An object experiencing circular motion has an angular speed of 10 rad/s and is traveling 5 m away from its axis of rotation. Find the object’s centripetal acceleration. Hint: ac = rω2

  2. Announcements • Units & Symbols Quiz on Wednesday • Circular Motion Quiz on Thursday • My contact information: • Email: david.mccallister@thomasmore.edu • Cell: 859 466 6402

  3. Centripetal Force & Newton’s Universal Law of Gravitation Circular Motion Physics Mr. McCallister

  4. Recall: ac = vt2/r ac = rω2 Newton’s Second Law: F = ma

  5. Centripetal Force • Centripetal Force is the force responsible for circular motion • Symbol: Fc • Unit: N • Recall: F = ma • Fc = mvt2/r • Fc = mrω2

  6. Centripetal Force • Centripetal force is not a new kind of force. • It is the name given to any kind of force responsible for keeping an object in circular motion. • Ex: Car turning = friction is Fc David’s sling = tension is Fc

  7. Practice 7H p 261 #2-4 • #1 • A girl sits in a tire that is attached to an overhanging tree limb by a rope 2.10 m in length. The girl’s father pushes her to a tangential speed of 2.50 m/s. If the magnitude of the force that maintains her circular motion is 88.0 N, what is the girl’s mass? • Formula: Fc = mvt2/r • Substitute: 88 = m(2.50)2 / 2.10 • Solve: m = 29.6 kg

  8. Fundamental forces • All forces, at their most basic fundamental level, are field forces! • Ex: Atoms in hand repelling atoms in wall • All mass in the universe attracts all other mass in the universe with a gravitational force. The size of the force depends on: • the amount of mass the objects have • how far apart the objects are.

  9. Gravitational Force • Gravitational force: an attractive force between all mass in the universe. The size of the force depends on: • the amount of mass the objects have • how far apart the objects are. • Symbol: Fg • Unit: N

  10. Newton’s Universal Law of Gravitation • Fg = G m1 m2 / r2 • Where • m1 & m2 are masses in kg • r is the distance the masses are separated in m. r is measured from the center of mass of large objects (ex: planet’s core) • G is the constant of universal gravitation • 6.673 x 10-11 N m2 / kg2

  11. Stop to Think… Compare the gravitational force on the moon from the earth to the gravitational force on the earth from the moon. They are equal! (Newton’s Third Law)

  12. Calculating g. • Let m1 = person’s mass, m2 = Earth’s mass • Recall weight = mass x g, where g = 9.81 m/s2 • So Fg = mg and Fg = G m1 m2 / r2 • mg = G m mE / rE2 • The person’s mass cancels… • g = G mE / rE2

  13. Calculating g. • g = G mE / rE2 • mE = 5.98 x 1024 kg • rE = 6.37 x 106 m • G = 6.673 x 10-11 N m2/kg2 • Substitute and Solve: • g = 6.673 x 10-11 x 5.98 x 1024 (6.37 x 106)2 • g = 9.83 m/s2

  14. Practice 7I pg 265 #2-4 • #1 • If the mass of each ball in Sample Problem 7I is 0.800 kg, at what distance between the balls will the gravitational force between the balls have the same magnitude as that in Sample Problem 7I? How does the change in mass affect the magnitude of the gravitational force?

  15. Practice 7I pg 265 #2-4 • #1 • Formula: • Fg = G m1 m2 / r2 • Substitution: • 8.92 x 10-11 = 6.673 x 10-11 x (.8 x .8)/r2 • Solve: • r = 0.692 m

  16. Defying Gravity Imagine a cannon on a tall mountain: vesc, Earth = 11,200 m/s

  17. Black Holes In 1916, Karl Schwarzschild predicted an object so massive and dense that something travelling at even the speed of light near the dense object could not escape its gravity.

  18. Black Holes The radius at which light cannot escape the object’s gravity is called the Schwarzschild radius. The physical space at the Schwarzschild radius is called the event horizon. The physical space inside the event horizon is called a black hole.

  19. Gravitational Lensing Massive objects can bend light, just like a lens.

  20. Homework due Tomorrow: • By end of class tomorrow: • p. 261 7H #2-4 • p. 265 7I #2-3 • p. 265 Section Review #1-5

More Related