soln2a - Problem Set 2 Solutions 1(a Area of plate without...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem Set 2 Solutions 1. (a) Area of plate without hole = (0.2)2 = 0.04 m2 Area of hole = “(0.05)2 = 0.00735 m2 Area of plate with hole = 0.03215 m2 M1 = mass of plate without hole = 1.5(0.4)i’0.03215 = 1-.8? kg M2 = mass of material of hole = 1.5(0.00785)f0.03215 = 0.3? kg s-—- ---s’ Moment of inertia about axis aa‘ through the CM ICM= M1a2f12 - M21234 = 1.3?(02fl12 — 0.37(0.05)2;4 = 0.0060 kg-m2 Moment of inertia about axis bb' (from parallel axis theorem) 1 I ICM + (MI'MZXWQ) = 0.0060 +1.5g0.1)2 = 0.0210 kg-m (’0) Set 10: = F, with torque produced by the weight of the plate. 1 gram:2 = — Mg(af2)sin8 a -Mg(af2)9 i9 diam2 = —(0.1Mg/I)0 = {.3020 Q9 ' m02=0.1(1.5)(9.81)f0.0210 (00 = 8.37 radfs M3 Period T = 211me = 0.75 s 2. Moment of inertia (by parallel axis theorem] I = mL2f12 + mxz, and 1a = F. I 'n Ix 03 :“ .MS (111sz 12 + m2)d28fdt2 = -mgx sine as -mngl 0302 = gX/(LZI‘IZ + x2) To minimise the period T=2nfwm maximise {002 or minimise U0002 dfdx(L2f12x + x) = -L2f(12x2) + 1 = 0 x2 = sz12 x = 0.29L = 29 cm 3. (a) If the springs are put under a tension T, their total extension should equal that of a single spring of force constant k. T = klxl = k2X1 5 1((X1 + X2) (k1 - 10X; 2 lCXz and (k2 '- k)X2 = log mix; = k!(k1 — k) = (k; - k)!k - k2 = (k: - k)(k2 - k) = klkz + k2 - (k. + kak k = klkflkl + k2) = 10(15)f25 = 6 Nr’m (b) Moment of inertia of a rectangular plate of dimensions a x b about a perpendicular axis through its centre is Mtia2 + b )!12, and of a rod about its centre is 11112;" 12. .‘.Moment of inertia of rod-plate system about its centre of mass is P 1CM = 0.5[(0.3)2 + (0.3)21r12 + 0.25{0.3)2r12 = 0.0203 kg-m2 Moment of inertia about P is IP =1CM + 0350141,)2 = 0.1408 kg-ml When the rod is displaced through angle 8, the spring is extended by LG, where L is the distance from the pivot to the springs; the springs therefore exert a force kLB and a torque —kL29. I; When the system is pivoted at its CM the weight produces no torque, but when pivoted at P the weight gives a torque -Mg(0.4sin9) z -0.4Mg9 (i) When pivoted at the CM: [CM dzefdtz = -(0.4)2k6 (139th = -(0.16kflcM)e = @029 (002 = 0.16(6)r'0.0208 = 46.15 Period = 27:10:10 = 0.92 5 (ii) When pivoted at P: 1P (Prams2 = -(0.4Mg + 0.64109 (of = (0.4Mg + 0.6%)pr = [O.4(0.?5)(9.81) + 0.64(6)]!0.1403 = 48.2 030 = 6.94 rad/s Period T = 21mg = 0.91 s 4. (a) At equilibrium, net torque on the rod is zero. Let the upward force exerted by the and be P. senes- mg{Lf2) — FL = 0 F = tug/2 If the spring is compressed by x0, F = kxo x{J = 05ka (b) If spring is compressed by x, torque acting is {ka - mgLQ] = -k(x - xo)L But x - x0 = LB F =10: then gives Idinr'dt2 = -kL26 diam) = -(kL2tI)B = «9.326 frequency = (00/21: = (l!2n)\}(kL2H) 5. Tensions in the strings to the two masses are TI and T2. The string between the two pulleys remains almost horizontal, so the tensions act on the rod in a horizontal direction. Torque is produced by these tensions, and by the welght of the rod. IP = MLZB _ Pp: -Mg(U2) sine -T1L c059 + T21. c059 : -Mgu2 - TIL + TzL Ipa = n (MLZB) cFa/dt2 = —Mg(uz)a + ('1‘2-T1)L .................. (1) Consider the two masses m. If the rod is displaced to the right, the right-hand mass will have an acceleration upward, while the left-hand mass will have an equal acceleration downward. mg - T1 = ma T2 — mg = ma T2 - T1 = Zma Acceleration of the masses is related to the angular acceleration of the rod by a = -Lot (note the Sign) T2 - T. = -2mL dzfifdtl ...................... (2) From (1) and (2) (MLZIS + 2mm diam? = -Mg(Lr'2)9 dzefdtz = mole (1302 = MgKZMLB + 4mL) = 0.6(9.81)l[2(0.6)(1)f3 + 4(0f1)(l)] (on = 2.?1 radfs Period T = 2mm, = 2.3 s mum—1m The CM of the rod moves up by h = (Ll2)~(L;’2)c056. _ The potential energy terms of the two masses cancel, because one moves up the same distance as the other moves down. E = Ipmlrz + Mgh + 2(mv2f2) dEfdt = o = 1pm dmldt + mg(L:’2)sin9 :19de + 2m dwdt v=Lm, dBI'dt = a), and dw’dt = Ldmfdt 1pm dmfdt + Mg(Ll2) sinflm + 2mL2m dwfdt = 0 (MLEB + 2mL2) dwfdt = ~Mg(Lf2) sine z -Mg(L:’2)B as above ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern