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Initials: Problem 1 - 25 points For gange - ABEDIaYHT Isband Vineet ham string central A yo-yo is made of two identical disks of seong cylinder mass M and radius R, glued together to a central cylinder of negligible mass and disks radius R/2. A side view and a top view are shown to the right. A string of negligible thickness is wound tightly around the central cylinder, so that the string does not R slip around it. The yo-yo is released from rest and drops vertically down, while R/ R/2 . . ... being held by the upper end of the string. The acceleration due to gravity has magnitude M g. M Side view Top view a- Determine the rotational inertia of the yo-yo with respect to its rotational axis of symmetry. ion nevis Tedlion sved I der tonot you no mate to Us yd bebids I ,noHibbs ulmsxe earli no bun havigoat "aveda bendmo es asiorlog motionimaxs art b- Write down the equations describing the rotational motion of the yo-yo about its rotational axis of symmetry and the translational motion of its center of mass.elfilial Initials: c- Determine the magnitude of the acceleration of the center of mass of the yo-yo as it unwinds downward. grub boviseron ei ylionsup lady alnest wow no trend boy bill to plies Isathov end of Now let's consider that the rod initially rotates in tow angular speed on You still throw the balthorizontally man wednesday a distance (/4 from the vertical axle, but thiis the the Union of the heatavelartest plea The normal to the rod (2-0). The goal is to get the sentewit to bert d- Determine the magnitude of the tension force exerted by the string on the yo-yo. after the collision.Initials: ing to the 139 ~4 Initials: Simple Marmonk Physics 8A Final Equation Sheet 7 (t ) = x(t) i + y(t)j AT = 12-11 =4xi+ 4yj Newton's 1s' Law AT B = constant = > F = 0 VAV = At - = VAU-x1 + VAD-y) B = lim 47 = vxi + vyj Newton's 2nd Law At - 0 4t [F = ma aAv = = QAV-xi + aAV-yl a = lim 40 = Newton's 3rd Law At to at = axi + ay ) FMe/You = -Fyou/Me B (t ) = vo + alt - tol Tension: r (t) = ro + volt - to] + salt - to]2 T Ax = X2 -X1 Ax X2 - X1 VAV-x At t2 - t1 Normal Force Ax dx Vx = lim - N At-0 At dt Avx Vx2 - Vx1 aAV-x - At Gravity t2 - t1 AVX dux Fo = mg ax = lim - At-0 At dt Ux (t) = Vox + axlt - tol Static Friction fs S HSN x (t) = xo + Vox[t - tol+ 5 ax[t - to]2 v2 = Vox + 2ax[x - xol Kinetic Friction Ay = y2 - y1 fx = HKN Ay _ yz - y1 VAV-y At tz - t1 Spring Force Ay dy Fap = - kx (x measured from relaxed position) Vy = lim At-0 4t dt AVy Vy2 - Vy1 Uniform Circular Motion aAv-y At t2- 1 a = Avy dvy r ay = lim At-0 At dt v = rw Vy (t) = Voy + dy[t - tol 2 TT 1 T = W y(t) = yo + Voy[t - to] + 5ay[t - to]2 vy = Voy vay + 2ayly - yolInitials: Fext = 0 = AP = 0 Work and Energy 2mz B mi - miz unit - Uzi W = F . di mi + mz my + mz 2m1 m2 - m1 W = FL cose (constant angle and magnitude) V26 mi + mz m + mz W. = -mgAh (height axis pointing upward) WSp = - =(X3 - XA) (x measured from relaxed position) K = =mv Rotational Kinematics Wnet = > Wi = 4K U + K =E e (t) = angular position AE = WNC 40 = 02 - 01 Ug = mgh (height axis pointing upward) 40 02 - 01 WAD Usp = = kx2 (x measured from relaxed position) At t2- t1 40 do w = lim - Universal Gravitation At-0 At dt AW W 2 - W 1 GMm aAV =- At t2 FG = 12 Aw dw GM a = lim g = At-0 At dt GMm w (t ) = wo+ alt- to] UG = r 0 (t) = 0 0 + wolt - to] += alt- to]2 GM W2 = wo+ 2a[0 - Oo] Vorb = s = ro (arc length) v = rw 2TY 3/2 atan = ra T = 12 2 VGM arad = -= rw2 Rotational Dynamics 2GM Vesc = 1 = Et=1 miri (point masses) Ip = ICM + Md2(parallel axis theorem) Linear Momentum It| = rF sin 0 p = mv Text = la P = P L = rp sino; L = Iw (@=angular velocity) Fext = dp W = | Tdo JO1 Krot = =102Initials: nitials: Cu. Cp and Y = Cp/Cy for an Ideal Gas Simple Harmonic Oscillator dzx 2+ w 2 x = 0 1. For a monoatomic gas, d=3 dt2 x(t) = A cos(wt + ) Cy = 38, Cp = 38 and Y = = 1.67 w = 2nf = ~ (@=angular frequency) 2. For a diatomic gas, d=5 Mass-spring: w = Cy = SR, Cp = 12 and Y = = = 1.4 3. For a triatomic or polyatomic gas, d=6 Simple pendulum: w = Cy = 3R, Cp = 4R and y = = = 1.33 Physical pendulum: w = mgLem Quadratic Equation Torsional pendulum: w = ax2 + bx + c = 0 has the solutions 20 (-b + vb2 - 4ac Fluids Derivatives P = V; P = perp Pgauge = Pabs - Patm d(xn) - = nxn-1 P = Po + pgh dx FB = PgVsub d(cos (ax + b)) -a sin(ax + b) Qm = PVA = pQv = const. dx 1' P + pgy + 5 puz = const. Lengths, areas and volumes Circumference of circle: 2nR Thermodynamics Area of disk: TR Surface area of sphere: 4TR PV = nRT = NKBT Lateral area of cylinder: 2TRh Eint = nCUT Volume of cylinder: TRZh AEint = Q - W R = NAKB; Co = AR Volume of sphere: 4TR3/3 2Cp Cp = CD + R ; Y = Cy Trigonometry e =- Wnet = 1 - Lout (efficiency) sin (90) = cos (09) = - cos (180) = 1 Qin Qin sin (09) = cos (909) = sin (180) = 0 TL eideal = 1 - TH cos (60) = sin (30) = 1/2 sin (60) = cos (309) = V3/2 sin (450) = cos (450) = 12/2 W Isobaric CPRAT PAV cos (180- 0) = - cose CVRAP P=const. Isochoric V= const. sin (180- 0) = sine Isothermal nRT In ( vo) nRT In T=const. cos (90 + 0) = - cos (90 - 0) = - sine Adiabatic 0 - # ( P, VJ - POVO) PV=const. sin (90 + 0) = sin (90 - 0) = cose cos2 0 + sin- 0 = 1Initials: Table of rotational inertias Axis Axis Axis Solid cylinder Hoop about Annular cylinder (or disk) about central axis (or ring) about central axis central axis 1=MR2 ( @ ) 1 = +M(R, 2 + R2? ) ( b ) 1 = 1MR (c) Axis Axis Axis Solid cylinder Thin rod about Solid spherea (or disk) about axis through center bout any central diameter perpendicular to diameter length 2R 1=!MR + $ ML2 (d ) 1= AML? (e) 1=3MR (D) Axis Axis Axis Thin spherical shella Hoop about any bout any R diameter Slab about 2 F perpendicular diameter axis through center 1 = AMR (8) 1= ;MR? (h ) 1 = 12 M(2 + 6? ) (1 )