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Unformatted text preview: Physics 41 Lecture 3 Lecture 3: Free-fall; Vectors Goals of this Lecture ∙ Apply the results obtained in the previous lecture to a specific example of an object that is freely falling and being accelerated by gravity. ∙ Review the properties of vectors including a discussion of polar and cartesian components. ∙ Apply the kinematics of free-fall, and the use of vectors, to the example of an object sliding down a slope. Summary of Results from Previous Lecture In the previous lecture we derived the equations that describe position u1D465 , velocity u1D463 , and acceler- ation, u1D44E , as a function of time for the special case of constant acceleration. The results that we derived are shown below. Furthermore we showed that positions, velocities and the acceleration are related via: u1D465 u1D453 = u1D465 u1D456 + u1D463 2 u1D453 − u1D463 2 u1D456 2 u1D44E (1.1) We now apply these results to a particular application – the motion of an object falling under the inﬂuence of gravity. 1 Last updated January 7, 2010 Physics 41 Lecture 3 1.1 Constant Acceleration: Object Falling Under the Inﬂuence of Gravity An object falling towards the earth’s surface experiences a constant downward acceleration 1 of 9 . 8 m/s 2 due to gravity. We start by defining some notation: ∙ It is conventional to use the variable u1D466 to describe position in the direction perpendicular to the earth’s surface, while the variable u1D465 is usually reserved for position parallel to the earth’s surface. ∙ It is also conventional to assume that u1D466 decreases as the object falls towards the earth, as represented graphically opposite. ∙ With this definition of direction, the velocity u1D463 = u1D451u1D466/u1D451u1D461 is positive if the object is moving upward and negative if the object is falling back to earth. Let’s consider two cases – an object thrown upward and an object falling downward. Both of these objects are said to be in free-fall , that is moving freely under the inﬂuence of gravity only. Case 1: The motion diagram for an object thrown upwards into the air is shown opposite. The velocity, u1D463 , decreases in magnitude as time increases, until u1D463 = 0. (After that the object falls back to earth, but we consider that in the next example). Then: 1. u1D463 u1D453 < u1D463 u1D456 2. Δ u1D463 = u1D463 u1D453 − u1D463 u1D456 < Consequently the acceleration due to gravity is negative : u1D44E = Δ u1D463 Δ u1D461 = u1D451u1D463 u1D451u1D461 < What does this mean physically ? It means that if we represent the acceleration as a vector, that vector has a magnitude of u1D454 = 9 . 8 m/s 2 and that the vector points towards the earth’s surface. Case 2: Now suppose that u1D463 starts off non-zero and negative as shown in the motion diagram opposite (i.e. you notice that an object is falling and start recording the characteristics of its motion). Then u1D463 becomes more negative as time increases. Then: 1. The magnitude of u1D463 u1D453 is greater than the magnitude of...
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