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10/14/2013PHY 113 C Fall Lecture 141 PHY 113 C General Physics I 11 AM-12:15 PM MWF Olin 101 Plan for Lecture 14: Chapter 12 – Static equilibrium 1.Balancing forces and torques; stability 2.Center of gravity 3.Will discuss elasticity in Lecture 15 (Chapter 15)

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10/14/2013 PHY 113 C Fall Lecture 142

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10/14/2013PHY 113 C Fall Lecture 143 Newton’s law of gravitation: Earth’s gravity: Stable circular orbits of gravitational attracted objects: R ES F a v M sat Summary of gravity: RERE m

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10/14/2013PHY 113 C Fall Lecture 144 From Webassign Assignment #12: When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration due to the Earth's gravitation? m/s 2 towards earth r=4.4R E m

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10/14/2013PHY 113 C Fall Lecture 145 From Webassign Assignment #12: An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.79 m/s 2. Determine the orbital period of the satellite. r

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10/14/2013PHY 113 C Fall Lecture 146 From Webassign Assignment #12: How much work is done by the Moon's gravitational field as a 1090 kg meteor comes in from outer space and impacts on the Moon's surface? RMRM iclicker question A.W>0 B.W<0

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10/14/2013PHY 113 C Fall Lecture 147 From Webassign Assignment #12: A space probe is fired as a projectile from the Earth's surface with an initial speed of m/s. What will its speed be when it is very far from the Earth? Ignore atmospheric friction and the rotation of the Earth. vivi

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10/14/2013PHY 113 C Fall Lecture 148 From Webassign Assignment #12: Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is v = 190 km/s and the orbital period of each is 10.7 days. Find the mass M of each star. (For comparison, the mass of our Sun is kg.)

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10/14/2013PHY 113 C Fall Lecture 149 From Webassign Assignment #12: Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is v = 190 km/s and the orbital period of each is 10.7 days. Find the mass M of each star. (For comparison, the mass of our Sun is kg.) iclicker exercise: Who might pose a question like this? A.A mean professor. B.A puzzle master. C.An observational astronomer.

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10/14/2013PHY 113 C Fall Lecture 1410 Meanwhile – back on the surface of the Earth: Conditions for stable equilibrium

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10/14/2013PHY 113 C Fall Lecture 1411 Stability of “rigid bodies” N migmig

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10/14/2013PHY 113 C Fall Lecture 1412 Center-of-mass Torque on an extended object due to gravity (near surface of the earth) is the same as the torque on a point mass M located at the center of mass. mimi riri r CM

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10/14/2013PHY 113 C Fall Lecture 1413 Notion of stability: mg(-j) r T F=ma T- mg cos mg sin ma =I r mg sin = mr 2 mra Notion of equilibrium: Example of stable equilibrium for

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10/14/2013PHY 113 C Fall Lecture 1414 Unstable equilibrium: mg(-j) r T Support above CM: Support below CM:

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10/14/2013PHY 113 C Fall Lecture 1415 Nik Wallenda walking on high wire across Grand Canyon

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10/14/2013PHY 113 C Fall Lecture 1416 Analysis of stability:

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10/14/2013PHY 113 C Fall Lecture 1417 * X

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10/14/2013PHY 113 C Fall Lecture 1418

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10/14/2013PHY 113 C Fall Lecture 1419 * X F g1 mg R CM

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10/14/2013PHY 113 C Fall Lecture 1420 iclicker question: F1F1 F2F2 Consider the above drawing of the two supports for a uniform plank which has a total weight Mg and has a weight mg at its end. What can you say about F 1 and F 2 ? (a)F 1 and F 2 are both up as shown. (b)F 1 is up but F 2 is down. (c)F 1 is down but F 2 is up. L/3 L Mg mg

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10/14/2013PHY 113 C Fall Lecture 1421 F1F1 F2F2 L/3 L Mg mg * X

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10/14/2013PHY 113 C Fall Lecture 1422 iclicker question: The fact that we found F 1 <0 means: A.We set up the problem incorrectly B.The analysis is correct, but the direction of F 1 is opposite to the arrow C.Physics makes no sense iclicker question: What would happen if we analyzed this problem by placing the pivot point at F 1 ?: A.The answer would be the same. B.The answer would be different. C.Physics makes no sense

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10/14/2013PHY 113 C Fall Lecture 1423 T Mg mg * X

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10/14/2013PHY 113 C Fall Lecture 1424 d * X

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10/14/2013PHY 113 C Fall Lecture 1425 mg Mg F wall N T Mg = 120 N mg = 98 N T < 110 N * X

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10/14/2013PHY 113 C Fall Lecture 1426 x A ladder of weight Mg and of length L is supported by the ground with static friction force f and by a frictionless wall as shown. The firefighter has weight mg and is half-way up the ladder. Find the force that the ladder exerts on the wall. * X

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