ex3 2210 2016 - EXAM 3 Physics 2210 Summer 2016 Unid Name Discussion TA Dr Ant lony Pantziris\/J ^ Discussion Section 2 Show all Work Circle your

ex3 2210 2016 - EXAM 3 Physics 2210 Summer 2016 Unid Name...

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Physics 2210 - Summer 2016 EXAM 3 D Name: Discussion TA: /J ; ^ - ' ; Unid: Discussion Section: r. Ant lony Pantziris 2 Show all Work!! Circle your answer(s). A block of mass m\ 1.00 kg sits atop a plane with coefficient of friction Uk = 0.140 and is connected to a block of mass 7M 2 = 2.00 kg through a string that goes over a pulley of mass M= 4.00 kg and radius R = 0.200 m. The pulley rotates about its axis without friction and the string moves over the pulley without slipping. The system starts at rest and block m falls through a height H = 1.50m. The moment of inertia of the pulley is / = MR 2 /2. (a) (b) (c) [3 pts.J How much mechanical energy is lost due to friction? [13 pts.J Use energy methods to find the speed of the mass after it has fallen the distance H. [4 pts.J What is the acceleration of the blocks? M,R (*) H
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Physics 2210 - Summer 2016 EXAM 3 Dr. Anthony Pantziris Name: Unid: Discussion TA: Discussion Section: Show all Work!! Circle vour answer(s). Arodofmass M:0.400 kgandlength L:0.700 misaffachedtoafixedpointP. AtpointBtherodmakesanangleof 40.0o with the vertical and its angular speed is ro = 3.00 rad/s. The moment of inertia of the rod about its center of mass is Iv[L2112.
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Unformatted text preview: When the rod is at point B, find: (a) (b) (c) (d) (e) (0 [3 pts.J the speed of the center of mass of the rod. [4 pts.J the torque of the weight of the rod about the point P. [3 pts.J the angular acceleration of the rod. [3 pts.J the tangential acceleration of the center of mass of the rod. [3 pts.] the centripetal acceleration of the center of mass of the rod. [4 pts.] the magnitude of the net force on the rod. A) Vr,nr = rl)f = N' = .J i.r;vtg \.?rL 1=12-7-t--/-B-5 tJ r) c) ';inA girt (4ol = o,ttt lv. UE;,Y il"L a,bon'.4. T {-2zr g(* *^t +.l,* I nL'= d) ftt= e) ftc= = h'd ng"t.s;ne, *^L' f - d AV: .T a z 2-0,7 L z D(. vz r ^e) | / Fnr+= (Y1'&c^ Acyt = 4JLr'+l.ttr =f. * nfs' p*tlef nrlf {"r,*,n k f,"l 4,\ Wthe'r+ o{ ;rrsr*tq 1o;"rt P M-(r)'= Tc^ * I n,r L'+ >E: YJ= nJ. *.r,no = *mLL.n q,8'. sln c+oo) !V.5 rad f y = 11.5 -l.0tz-f 0,1/ z 9'9tn o L 0.7 4.lzr w f,1*.' \ " 6c,rnr '^.,n,\ iv. u w,r,^...
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