BOBO?PdHZHZHZ ?,H;%4?`-Z(Charles E. Miller, Jr.xHH(FG(HH(d'@>p@  Savel@/cP8 Ejec@-?PŀZ1 j$1g}=c>'Z <1 ?nP44 }DSETr7?Q??P?|?P?m?R\<  6 7   1 2    N O               : ;     ] ^  ~                      ! "  Q S  T U  V X              E F G  J K  k n  o p     @xx@>hh@>?= NEWTON'S SECOND LAW Part II: Circular Motion In this experiment we study F = ma as it applies to an object moving in a circle.  Before attaching the spring and the string, adjust R so that M is directly above the pointer. Now attach the spring and the string and adjust m until M is directly above the pointer. From the force diagrams above, convince yourself that, since T2 has no horizontal component, the spring force T3 = mg because you experimentally arranged for the spring and string forces to be horizontal. Remove the string connecting M and m. Now, with your fingers rotate the vertical rod at such a rate as to obtain M directly above the pointer. If, as you maintain the rate of rotation which keeps M directly above the pointer, the unit rocks as it rotates, you will need to adjust the counter weight so that the center of mass is shifted to the axis of rotation. With the electric timer, find the time, t, for 50 revolutions*, determine the period of revolution, and the centripetal acceleration of M using the following equations. The period T = t/50 Speed v = 2R/T Acceleration a = v2/R = 42R/T2  Because we have arranged experimentally for T2 to be vertical and the spring force to be horizontal, the resultant horizontal force is T'3 = Ma . If we neglect the mass of the spring, T'3 = T3 = mg, the above value. Since the mass of the spring, Ms, participates in the motion, an approximate correction for this effect is to add Ms/2 to M. Using MKS units, make a graph of T3 on the vertical axis versus a, as obtained from 42R/T2. Note that the origin (a = 0, T3 = 0) is also a legitimate point on the graph. Why? Recall that y = mx + b to help you interpret the graph. What is the percentage difference between the value of your slope and the value predicted by Newton's Second Law? *When counting a periodic motion, it is helpful to count 5-4-3-2-1-0-1-2-3, etc. and start the time on the count of zero.ZNDSET.H% l?mx@x?6*@z @ 1## 8 D"D"1*(< 8 1BuI 8 "D"D1JBbI 8"""3"3 "33 "? "K' "L @ 1u%z 8"O @ 1Ou 8"1"1 "1 "= # P7CP<="O"O "O%"O&"<0Vb ffff1i<uO 8"O'#"X""K"W"4H"@H":":" F" F" E  +lR" J +RNm0@~L (Im"P#"Y":4":":3#"/-"0'#"%! (>T(T+2&T +1(C3(&2 (>M+M (hM g(T^Pointer" """y"Q (CCounter + Weight">","2 .D@ PJ.DD  c1 )JR )JR )JR )JR? )JR )JR )JR B!@ { @"7(+1 ('T *AmgDSET.H Do=(@h?6*@ -q @"5"@"@"A "M "_"_"Z"<"<"<"!"!"' (@T(T (E3(2 (BM (jM g (>'((3?Repeat the above procedure, measurements, and calculations for * 7four other values of R ranging from the minimum to the * Cmaximum possible values of R. Make a data table which includes * Dquantities you will need in your analysis. These should include t, * BT, R, a, and mg, the spring tension. Using a double pan balance, * measure M.DSET.H?<?6*?DSET.H @2=???6*?FNTMCUTSDSUM/Charles E. Miller, Jr.HDNISTYLD@STYL?l=???D??`?S      G              |HASH $%x%p& ( ( ) *  (  ( (& L:Qhjazo}Bod  CHAR?D   "     @ @            ^ HASH         R CELL?"HASH  GRPH=fn.HASHogl RULR=@4HASH@@ LKUP    $NAMEDefault Default SSHeaderBodyFooterFootnoteFootnote IndexDFNTM HelveticaGenevaNew YorkETBL@FNTMCUTSDSUMHDNISTYL"ETBL)