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by: Ray

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# LECT 15 Spring 15 PHYS 172

Marketplace > Purdue University > Physics 2 > PHYS 172 > LECT 15 Spring 15
Ray
Purdue

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Physic 172
COURSE
PHYS 172
PROF.
TYPE
Class Notes
PAGES
15
WORDS
KARMA
25 ?

## Popular in Physics 2

This 15 page Class Notes was uploaded by Ray on Tuesday April 5, 2016. The Class Notes belongs to PHYS 172 at Purdue University taught by in Spring2015. Since its upload, it has received 30 views. For similar materials see PHYS 172 in Physics 2 at Purdue University.

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Date Created: 04/05/16
, Spring 2015 PHYS 172: Modern Mechanics Lecture 8 – Momentum Principle Midterm AverxgeMrgeerGestaalscorawootaboutt3fout of 100. net Fnet  Ft    p  d dt   1 bob m         , , ,  rrb f ptob i pFtpFt , , , For ebob fe,bob ill springecitation 3: p  know the net force acting on it. The TMhosundnauentSlrriccplereletsmtphasized using the Cartesian xyz components  Fnet  Ftet    p  p d dt  detoraiie2xnohderfbheuoe.etoeisct For example, wncateueextmpMlP,woe will also see that it x  F2   mg  F1 m Static Equilibrium The WMeocanenlouusePtrsnfiplamental”sufscit ,0,0    0  0 mg  tan   g 2 tan  dp dt m  F  1 mg mg .2   F    F  F 2  2   F  F   F  Fet  Reasoning:  Geometrical dp dt WeNeedwttinmassfotee block, our system. x  F2    mg tan , ,0 1  1 m  F mg mg F  Static equiyibrium 1  F What is ? because its momentum is 3 3 =90 cm = (water density x cork volume) x g ) under watert is the 3 force on the cork is zero magnitude = (water mass displaced) x g 10 m/s ≈ arth: magnitude mg = 0.3N downward ) as shown in this diagram. 3 are acting on the cork. , in the string if the cork’s mass is 30 grams? T cork sphere (density 1/3 gram/cm wheso,hde,fatly,tasbuisplntfdrie9is0r.as/ cm A) 0.)0N)D0)E0.91.2 N TheTghreiyofycetfxrctexebttdeby the water: * The MP Three significant forcess us that the net on the cork is 0.6 N downward. Q1. A string attached to the bottom of a cup holds a y T T θ θ θ T TTTT sin T o T θ θ T θ θ θ o T mg ≤g for all Another example involving an unknown constraint force … Q2.nChneiperC.*=.EgTsin/thncenter figure. What is the tension T y T θ so . x m mg T θ =T sin T T g  m 2sin T he knot’s mass is so small that the tension in the upper strings we must o T   θ θ T The masses of the strings are small compared to Note:     net y FTT m T T mg So 2is 2is 0 apply it to somebythegraviotiosulficrmenerachostit For thApply it tThe three significant forces acting on the knot are the tension forces exerted 2  Fnet on m m1 to find . 2 m  2  ? ,0,0 F m  1 2   1 2 ou begincityand, therefre, the momentum of dp dt dv dp dt 2 2 dv dt  dv dt mm m 2 2 ,0,0  F mt on    2 2 net on  FFet  Fm m F on the lower box. As you do, the boxes begin to move together F hoakeoeqmagnrtssblock 2 was the MP to the two-box system tells us that A N YonmstatpcsboxeIf we knew thSince the block’s move together, their accelerations are the same, so, applying g 0, ,0 2 m  m m1  F to prod2ce this m2 m  F is zero, the normal components of m g by block 1 is, therefore, equal to the 0, ,02 2 m 2  ,0,0 m mF mm 2 N  F    F 1 2 2 2  dt m m dv m yomponent of the net force on  2 2  Fmet on net The Earth is eThe frictionThe buoyant, aerodynamic drag and other forces acting on m2 in this situation A Nonsta W iatobjacmopicecheare interacting significantly with ,0,0 m  mm2  2 1 2 m m  F F on F in this situation? 2 m as in the preceding example. 1 m and the floor, there is no way to tell. A) It is the same as in the previous example,of friction between box 1 theawsamsexconstant velocity.tent RQ. In thisIn thicaset,stweenet,hecboactngmove across the floor at a  0    Fet net   F         Fnet   d dt For example, here there is a change of momentum  p i The momentum principleomrkmiefilmwpdteihhtnchigngaggdiedetofthe   f p Derivative form of tcoordinate system but working with the components the direction of xyzthe system’s momentum vector and in the perpendicular direction lets us We’ve often worked with the components of each side of this equation in an   Fnet    dp dt   net  F 0  spechmngmllnthdinspentenftiusrrtcotsihaagetnfthes On the other hand, in a circnetr motion atconstant  F dp dt  Direction Change      Fnet    dp dt The General CasMe:BothnMtasniudtla Review sAessinsm5.nt95s7dufttetghtbook Read beforebthsniextLecturo,kne week from today,

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