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Transcript
Center of Mass The point located at the object’s average position of mass. The Center of Mass and the Center of Gravity Are located at the same point AS LONG AS gravity is consistent throughout the object. The center of mass does not always lie within the object itself…. When an object is in motion, its center of mass will follow a smooth line. Locating the Center of Gravity, CG Method One: It’s a balance point along the object. Method Two: If you suspend any object, its cg will be located along a vertical line drawn from the suspension point. Center of Gravity, cg, vs Center of Mass, cm In this course, we virtually ALWAYS use cm rather than cg, and for almost all situations, they are located at the same place. The cg is the average location of the weight- the cm is the average location of the mass. As long as the gravitational force is the same throughout the body, these two will be at the same place. (Near the event horizon of a black hole, the gravitational force could be very different in an extended body depending on which portion of the object is closer to the black hole! In this case the cg would NOT be at the cm.) The center of mass for multiple objects of total mass M is given by rcm 1 M m r i i i Finding the x-coordinate for the center of mass for 2 separate masses. xcm m1 x1 m2 x2 m1 m2 And the same for ycm! Newton’s First Law: When there’s no net external force acting on a system of masses (or a continuous mass), if the center of mass was at rest, it will remain at rest, if it was in motion, it continues that motion, regardless of what the individual masses are doing! Newton’s Second Law: If there is a net external force acting upon a system of masses, the center of mass will accelerate. When we draw free-body diagrams, we generally draw forces acting at the cm to determine the acceleration of the center of mass (which may NOT be the acceleration of every point of the body or every particle in the system) SF = mtacm Newton’s Third Law- we will revisit later when we look at recoil and collisions from a center of mass reference frame. The center of mass of a system moves like a particle of mass M = S mi under the influence of the net external force acting on the system. Let’s look at an example…. A projectile is fired into the air over level ground with an initial velocity of 24.5m/s at 36.9° to the horizontal. At its highest point, it explodes into two fragments of equal mass. One fragment falls straight down to the ground. Where does the other fragment land? Since the only external force acting on the system is gravity, the center of mass continues on its parabolic path as if there had been no explosion. The cm lands at R, where R is the range. The first fragment lands at 0.5R. The other fragment of equal mass must land at 1.5R. What if the fragment were 0.3M? At what location X will the other fragment land? The Range (for cm) = (vo sinq)/g x 2 x vo cos q = 58.8m (the cm will fall at ½ that range) cm (m1 )( x 1 ) (m 2 ) x 2 ) M 1 (0.3M )( x 58.8) (0.7M ) X ) 2 58.8 M X = 71.4m The Calculus Connection To find the center of mass of a continuous object, we replace the sum rcm rcm with an integral 1 m i ri M i 1 r dm M Where dm is a differential element of mass located at position r. MEMORIZE this equation, but you probably won’t have to evaluate the integral! The AP objective states that the student should “use integration to find the center of mass of a thin rod of non-uniform density”. I have seen questions where the student was asked to write the integration expression, but not to evaluate it, however… we’ll look at how to do it…