Why newtons 3rd law a fundamental result 2 1 v r v r

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Unformatted text preview: mo: two masses connected by spring. Why? Newton’s 3rd Law. A fundamental result. 2 1 v r v r 2 1 v r CM O CM 3 v r 3 Consider a collection of particles or pieces of a continuous object, with masses m1 , m2 .... and positions r1 , r2 .... Sum Fi = mi ai over all particles i=1,2,3..... “Internal Forces” cancel in pairs due to Newton’s 3rd law. The final result is a simple relation between the net external force and the acceleration of the center of mass. (transparency) of mass : rcm = mi ri , M M = tota TEX for keynote a fundamental result........... et external force : i Fnet,ext = 1 Define center of mass : rcm = mi ri , Mi TEX for keynote net external force : Fnet,ext = Fi,ext. iM = total mass 2 d rcm = acceleration of center of mass = i 2 dt 2 d rcm Fi,ext. acm = acceleration of 1 center of mass = dt2 mass Define center of mass : rcm = mi ri , M = total M net,ext i = M a cm Then : Fnet,ext cm net external force : Fnet,ext = Fi,ext. Then : F =Ma i d2 rcm T ∝ L/g acm = acceleration of center of mass = dt2 π Thus: Center of Mass=: 3.141592653589793.... Thenbehaves like= point particle in response Fnet,ext a M acm T∝ to the net external forces. L/g r Ramifications........ If there are no external forces, the center of mass is either stationary or moves with constant velocity. If there are no external forces, “ total momentum” is conserved. (we’ll see.) The cancelation of the effect of internal forces is sometimes surprising, and powerful for predictions. Intuitions about Center of Mass: The center of mass is a weighted average of the positions of the particles, weighted by mass. For symmetrical objects, like a disk, it is at the geometrical center. Later we will see: it is the point where the object can be “balanced”......(has to do with torque...) m1 + m2 m1 + m2 Example Where is the center of mass in the following mass distribution? y 2m (12,12) m (0,0) m (24,0) x Solution: X CM YCM ∑mx = i i m0 + (2 m)12 + m24 = = 12 4m i M ∑my = i M i i m0 + (2 m)12 + m0 = =6 4m y 2m (12,12) m (0,0) € RCM = (12,6) m (24,0) x of mass : rcm = mi ri , M M = tota TEX for keynote review of THE fundamental result.. et external force : i Fnet,ext = 1 Define center of mass : rcm = mi ri , Mi TEX for keynote net external force : Fnet,ext = Fi,ext. iM = total mass 2 d rcm = acceleration of center of mass = i 2 dt 2 d rcm Fi,ext. acm = acceleration of 1 center of mass = dt2 mass Define center of mass : rcm = mi ri , M = total M net,ext i = M a cm Then : Fnet,ext cm net external force : Fnet,ext = Fi,ext. Then : F =Ma i d2 rcm T ∝ L/g acm = acceleration of center of mass = dt2 π Thus: C...
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This note was uploaded on 09/29/2012 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell University (Engineering School).

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