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Initials: Problem 2 - 25 points A thin rod of mass M, and length L is mounted on a frictionless vertical axle going through its midpoint and can rotate in a horizontal plane. The rod is initially at rest when a small sticky ball of mass my is thrown horizontally with speed v. The ball strikes the rod at an angle 0 from the normal and sticks to the rod at a distance L/4 from the axle. A top view is shown in the figure. Top view a- Determine the net torque acting on the rod-ball system during the collision, with respect this collision? to the vertical axle of the rod. Based on your result, what quantity is conserved during b- Determine the rotational inertia of the rod-ball system after the collision.Initials: c- Determine the angular speed of the system immediately after the collision. Does it rotate clockwise or counterclockwise? Explain. Alo sbam Praybetter olgas "ends blan Pi eabend gift to note Now let's consider that the rod initially rotates in the counterclockwise direction with angular speed wo. You still throw the ball horizontally from the left and it sticks to the rod at a distance L/4 from the vertical axle, but this time the direction of the ball's velocity is along the normal to the rod (0=0). The goal is to get the system to be at rest after the collision. d- Draw a figure showing on what side of the axle the ball needs to land on the rod, and determine the speed of the ball that would get the rod-ball system to come to a full stop 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 )