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UT Arlington PHYS 1443 - PHYS 1443 Lecture Notes

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PHYS 1443 – Section 001 Lecture #12AnnouncementsCenter of MassMotion of a Diver and the Center of MassExample 9-12Center of Mass of a Rigid ObjectExample for Center of Mass in 2-DExample of Center of Mass; Rigid BodyWednesday, June 21, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu1PHYS 1443 – Section 001Lecture #12Wednesday, June 21, 2006Dr. Jaehoon Yu•Center of Mass•CM and the Center of GravityWednesday, June 21, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu2Announcements•Reading assignments: CH. 9 – 10.•Quiz tomorrow–Early in the class–Covers Ch. 8.5 – Ch. 9•Mid-term grade discussions today•Exam results–Average: 70/102–Top score: 100/102•Grade proportions–Exam constitutes 22.5% each, totaling 45%–Homework: 25%–Lab: 20%–Quizzes: 10%–Extra credit: 10%Wednesday, June 21, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu3Center of MassWe’ve been solving physical problems treating objects as sizeless points with masses, but in realistic situation objects have shapes with masses distributed throughout the body. Center of mass of a system is the average position of the system’s mass and represents the motion of the system as if all the mass is on the point. Consider a massless rod with two balls attached at either end.CMx �The total external force exerted on the system of total mass M causes the center of mass to move at an acceleration given by as if all the mass of the system is concentrated on the center of mass./a F M=�rrWhat does above statement tell you concerning forces being exerted on the system?m1m2x1x2The position of the center of mass of this system is the mass averaged position of the systemxCMCM is closer to the heavier object1 1 2 2m x m x+1 2 m m+Wednesday, June 21, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu4Motion of a Diver and the Center of MassDiver performs a simple dive.The motion of the center of mass follows a parabola since it is a projectile motion.Diver performs a complicated dive.The motion of the center of mass still follows the same parabola since it still is a projectile motion.The motion of the center of mass of the diver is always the same.Wednesday, June 21, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu5Example 9-12Thee people of roughly equivalent mass M on a lightweight (air-filled) banana boat sit along the x axis at positions x1=1.0m, x2=5.0m, and x3=6.0m. Find the position of CM. Using the formula for CMiiiiiCMmxmx1.0M �12.03MM= =M M M+ +4.0( )m=5.0M+ �6.0M+ �Wednesday, June 21, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu6Center of Mass of a Rigid ObjectThe formula for CM can be expanded to Rigid Object or a system of many particles A rigid body – an object with shape and size with mass spread throughout the body, ordinary objects – can be considered as a group of particles with mass mi densely spread throughout the given shape of the objectCMx =iiiiiCMmymyThe position vector of the center of mass of a many particle system is CMrrMxmxiiiCMCMx1CMr rdmM=�r rmirirCMiiiiiCMmzmzCM CM CMx i y j z k= + +rr r i i i i i ii i iiim x i m y j m z km+ +=� � ��rr ri iiCMm rrM=�rrMxmiiimi 0lim xdmM11 1 2 2 n nm x m x m x+ +���+i iim x� iim�1 2 nm m m=+ +���+Wednesday, June 21, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu7Example for Center of Mass in 2-DA system consists of three particles as shown in the figure. Find the position of the center of mass of this system.Using the formula for CM for each position vector componentiiiiiCMmxmxOne obtainsCMxCMrrIfkgmmkgm 1;23213 40.754CMi jr i j+= = +r rr rrm1y=2(0,2)m2x=1(1,0)m3x=2(2,0)(0.75,4)rCMiiiiiCMmymyiiiiimxm321332211mmmxmxmxm321322mmmmmCMyiiiiimym321332211mmmymymym32112mmmm CMx i=r( )2 3 11 2 32 2m m i m jm m m+ +=+ +r rCMy j+rWednesday, June 21, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu8Example of Center of Mass; Rigid BodyThe formula for CM of a continuous object isLxxCMxdmMx01ThereforeLxdxdm=dxSince the density of the rod () is constant;CMxShow that the center of mass of a rod of mass M and length L lies in midway between its ends, assuming the rod has a uniform mass per unit length.Find the CM when the density of the rod non-uniform but varies linearly as a function of x,  xCMxMdxdmLM /The mass of a small segmentLxxxdxM01LxxxM022112211LM MLM 2112LLxxdx0Lxxxdx0Lxxx0221221LLxxxdxM01LxxdxxM021LxxxM033113311LM MLM


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UT Arlington PHYS 1443 - PHYS 1443 Lecture Notes

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