PRACTICE EXAM 2 SOLUTIONS

Transcription

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Deprtment of Phyic Phyic 8.01x Fll Term 00 PRACTICE EXAM SOLUTIONS Proble: Thi i reltively trihtforwrd Newton Second Lw problem. We et up coordinte ytem which i poitive rihtwrd nd downwrd (to nticipte the motion of the me. Coniderin the force blnce in the verticl direction for the firt m, we imply hve: N + = 1y = 0 N = In the horizontl direction, we hve the tenion in the trin nd kinetic friction: T µ k N = 1x T µ k ( = 1x For the econd m, we only need to concern ourelve with the verticl direction, which involve jut rvity nd the tenion in the rope (which i till the me T ince the trin i mle, lthouh now directed verticlly due to the pulley: T + m = m y We note tht ince the trin i of fixed lenth, the ccelertion of the two me mut be connected. With our choice of poitive in the coordinte ytem, they re in fct the me (if you choe different in convention in your coordinte ytem, they my hve oppoite in. So we cn collect our eqution toether in term of common ccelertion : 1x = y T µ k = T + m = m. Addin thoe two eqution llow u to eliminte T nd olve for : m µ k = m + b. Pluin in thi ccelertion into one of the eqution llow u to olve for T:

2 T = m m T = m m m µ k + m m (1+ µ k T = m + c. We know the ccelertion nd the ditnce b, o thi prt i jut imple ppliction of kinemtic: 1 x = t b b( + m t = = m µ k d. Once we know the time nd the ccelertion, the velocity i eily obtined by jut pluin in our previouly obtined vlue: v = t v = b (m µ k m + Problem :. A free-body force dirm for the m how the two force ctin on it: rvity downwrd, nd the norml force perpendiculr to the urfce of the cone. Note tht you hould not drw n dditionl centripetl force on your dirm: thi i jut lbel which we pply to force which re directed rdilly inwrd: in thi ce the norml force horizontl component. N m

3 b. We cn find the peed of the m by uin the fct tht it i movin in contnt horizontl circle. Set up coordinte ytem with poitive upwrd nd rdilly outwrd, nd conider the force eqution in the two direction: F y = m y N in m = 0 F = m x x N co = m v r Now by dividin our two eqution (fter rerrnin the m onto the other ide in the firt one, we cn eliminte N nd olve for the velocity: N co mv = N in rm r v = tn v = r cot c. In one circle, the m will trvel ditnce equl to the circumference, πr. So, ince we know the velocity, we cn find the time of complete circle eily: r tn T = πr r = π = π v cot d. It w little uncler in the oriinl problem etup whether the m w lidin round in circle on fixed frictionle urfce, or ttionry on rottin frictionle urfce. For thi prt of the problem, thi ditinction i importnt: the m i ttionry on the inner urfce of the cone tht i rottin, o we cn hve ttic friction between the m nd the urfce. N µn m

4 At the mximum peed before lippin, the ttic friction force will be directed downwrd lon the urfce of the cone to keep the m from lidin upwrd nd outwrd. We cn dd it to our force blnce eqution nd re-olve for the velocity. The trickiet prt i ettin the nle nd the in of the frictionl force correct: F y = m y N in µ N co m = 0 F = m x x N co µ N in = m v r A before, we cn move m to the other ide nd divide the two eqution to cncel N nd olve for the velocity: N co + µ N in mv = N in µ N co rm v = r co + µ in in µ co Problem 3: For thi problem, we will hve to ue ll of the condition of ttic equilibrium: tht the force re blnced in ll direction, nd tht the totl torque i zero for ny choice of pivot point.. Since the force F i completely unknown, it mke ene to plce our pivot point t tht loction, to eliminte it from our torque eqution. Since thi i lo lon the line of ction of the rvittionl weiht of the foot (t it center of m, we eliminte two force for the price of one. A uul we will dopt coordinte ytem which i poitive rihtwrd nd upwrd, nd with the poitive ene of rottion bein counterclockwie for computin cro product. The torque bout our choen pivot i: τ = 0 (T in d + Nb = 0 We note tht the norml force i equl to hlf the weiht of the peron, ince the problem tte the weiht i evenly ditributed on both feet. Pluin tht in for N, we cn olve for T: mb T = d in b. To et hndle on the force F nd it nle, we now need to conider force blnce in the x nd y direction. Do not nelect the weiht of the foot, even thouh tht force i not indicted in the dirm provided in the problem.

5 F = 0 x T co F in β = 0 F = 0 y T in F co β + N m f = 0 By rerrnin term nd dividin the two eqution to eliminte F, we cn olve for the unknown nle: in β T co = co β T in + N m f mb co tn β = d in mb in + m m f d in mb cot β = rctn mb + md m f d c. We cn plu thi bck into one of our force eqution to find the mnitude of the force: T co F in β = 0 T co F = inβ Actully pluin in the torque nd nle don t do much for the lebr, o we cn jut leve our nwer in thi form, ll the quntitie re known from previou prt of the problem. Problem 4A:. Coniderin the force blnce on the pennie, we hve their rvittionl weiht blnced by the prin force of the rubber bnd. Tkin upwrd to be poitive: m p + k(l l 0 = 0 m p k = l l0 b. For the purpoe of thi problem, we cn nelect the m of the rubber bnd nd the effect of rvity. In thi ce, we hve only the prin force of the rubber bnd providin the centripetl force for the circlin nut. Tkin rdilly outwrd to be poitive: F = m n k(d l 0 = m ( 4π rf n Now we cn plu in our previou expreion for k nd olve for the frequency f:

6 f = m p d l 0 l l 0 m 4π r n f = m p d l 0 4π m r l l 0 n c. The frequency i the number of cycle per econd, o the period (the number of econd per cycle i jut the invere of it: 1 T = f d. The nulr frequency i the number of rdin per econd. Since there re π rdin per complete cycle, the nulr frequency i jut: ω = πf For implicity, we cn leve our nwer in thi form inted of ubtitutin in n explicit expreion for f, which i lredy olved for. Problem 4B: Strtin from our umed power-lw force eqution, we cn tke the lo of both ide to iolte the power-lw exponent: F = ( S b m c lo F m = lo + b(lo S c We ee tht under the lo trnform, the dt hould follow triht line with lope b nd intercept lo. Exminin the dt, the triht line fit pe pproximtely throuh the firt nd the lt dt point, o we cn extrct the lope from thoe two: lo y lo y 1 b = lo x lo x 1 lo0.055 lo 0.34 b = =.0 lo 5 lo10 Note tht the prticulr unit of the meurement do not mtter, ince difference of lo i imply the lo of the rtio of the quntitie.