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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 Announcements Units & Symbols Quiz on Wednesday Circular Motion Quiz on Thursday My contact information: • Email: [email protected] • Cell: 859 466 6402 Centripetal Force & Newton’s Universal Law of Gravitation Circular Motion Physics Mr. McCallister Recall: ac = vt2/r ac = rω2 Newton’s Second Law: F = ma Centripetal Force Centripetal Force is the force responsible for circular motion • Symbol: Fc • Unit: N Recall: F = ma Fc = mvt2/r Fc = mrω2 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 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 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. 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 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 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) 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 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 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? 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 Defying Gravity Imagine a cannon on a tall mountain: vesc, Earth = 11,200 m/s 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. 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. Gravitational Lensing Massive objects can bend light, just like a lens. Homework due Tomorrow: By end of class tomorrow: • p. 261 7H #2-4 • p. 265 7I #2-3 • p. 265 Section Review #1-5