This took a lot of time and effort and I understand if you wish to skip past everything and just read my questions about it in the The too long didn't read summary (TL;DR) at the bottom. 1. The problem statement, all variables and given/known data The 10-g bullet having a velocity of v = 750 m/s is fired into the edge of the 6-kg disk as shown. The disk is originally at rest. The rod AB has a mass of 3 kg. Determine the magnitude of the angular velocity of the disk just after the bullet becomes embedded into its edge. Calculate the angle θ the disk will swing when it stops. Express your answers using three significant figures. Assumptions: No air resistance No friction No other environmental forces or influences not mentioned in problem statement Knowns: vb = 750⋅m⋅s-1 (velocity of the bullet) mb = 0.01⋅kg (mass of the bullet) md = 6⋅kg (mass of the disk) mr = 3⋅kg (mass of the rod) Lr = 2⋅m (length of the rod) rd = 0.4⋅m (radius of the disk) Unknowns: ω = ??? [rad s-1] (Angular velocity of the system just after the bullet embeds in the disk) [radians per second, 3 significant figures] θ = ??? [°] (Change in the angle the pendulum makes from rest to its highest position before coming back after collision) [In degrees, 3 significant figures] 2. Relevant equations Moment of Inertia for a slender rod about one end Ir = (1/3)mr⋅Lr2 Moment of Inertia for a disk Id = (1/2)md⋅rd2 Parallel Axis Theorem Io = Ia + m⋅ro2 Conservation of Angular Momentum HA,1 = HA,2 ∑Ii⋅ω1 + ∑∫Mi⋅dt + ∑m⋅r⋅ivi,1 = ∑Ii⋅ω2 (this is probably where I'm getting confused) 3. The attempt at a solution Find the mass moment of inertia of rod and disk system about point A IA = Ir + Id + md⋅ro2 IA = (1/3)mr⋅L2 + (1/2)md⋅rd2+ md(Lr+ rd)2 IA = (1/3)(3⋅kg)(2⋅m)2 + (1/2)(6⋅kg)(0.4⋅m)2 + (6⋅kg)(2⋅m + 0.4⋅m)2 IA = 39.04⋅kg⋅m2 Apply conversation of angular momentum before impact to just after the impact. HA,1 = HA,2 mb⋅vb(Lr+ rd) = [IA+mb(Lr+ rd)2]⋅ω ∴ ω = [mb⋅vb(Lr+ rd)]/[IA+mb⋅(Lr + rd)2] ω = [(0.01⋅kg)(750⋅m⋅s-1)(2⋅m + 0.4⋅m)]/[(39.04⋅kg⋅m2) + (0.01⋅kg)(2⋅m + 0.4⋅m)2] ω ≈ 0.460386315 rad s-1 ω ≅ 0.460 rad s-1 The confusion (part 1) This part seems intuitive and correct to me (finding ω). It's also the right answer. The next part, where θ (Change in the angle the pendulum makes from rest to its highest position before coming back after collision) however I am confused as to how the procedure is intuitive or even "sensical" What Chegg and other solution manual/websites do is "apply angular momentum after the impact" I would probably have used conservation of energy or something of the sort but I want to know how to do it with the conservation of angular momentum because I'm not sure I fully understand how to do it and it's important for other applications plus it's the only way to get the answer they're looking for on the homework and multiple choice no-partial-credit exam. So they start off saying something I would of figured "Apply conservation of angular momentum after the impact" HA,1 = HA,2 So far so good. The angular momentum of the system needs to be the same as it is right after the bullet embeds into the disk and the system starts moving with angular velocity ω .. (which is HA,1), as compared to the angular momentum it does when the pendulum has reached its max sway (HA,2). Now the part that sends me off the deep end of what is going on and how it makes any sense what so ever They say -Wd(Lr + r) - Wr(1/2)L + (1/2)IAω2 = [-Wd(Lr + r) - Wr(1/2)L]cos(θ) None of this is angular momentum? This is kinetic energy and potential energy. So I'm wondering how they came up with this equation because even when I do conservation of energy I wind up with Relevant Equations (part 2) Conservation of Energy E1 = E2 Kinetic energy (linear) KE = (1/2)m⋅v2 Kinetic energy (angular) KE = (1/2)I⋅ω2 Potential energy (gravitational) PE = m⋅g⋅h The attempt at the solution (part 2) Esys,1 = Esys,2 KEsys,1 + PEsys,1 = KEsys,2 + PEsys,2 (1/2)Isys⋅ω12 + msys⋅g⋅h1 = (1/2)Isys⋅ω22 + msys⋅g⋅h2 ω2 will be zero since the max height will have decelerated the angular velocity completely. h1 will be zero since this is the initial potential energy and will be referenced as zero and h2 will be the increase in vertical height when the pendulum reaches its highest point. Δh will be used to describe the change in gravitational potential energy. Consequently the gravitational potential energy term will drop out of the left hand side of the equation or first state and the kinetic energy equation will drop out of the right hand side of the equation or second state. (1/2)Isys⋅ω2 = msys⋅g⋅Δh2 Since we are interested in the angle θ and not Δh, Δh must be rewritten in terms of Lr, rd and θ L being the distance from the center of mass of the part of the system to the axis o located at point a. Applying Δh = L(1 - cos(θ)) to our energy balance equation (1/2)Isys⋅ω2 = [md(Lr+rd) + mr(1/2)Lr]g(1-cos(θ)) ∴ θ = arccos(1 - [(1/2)Isys⋅ω2]/[md(Lr+rd) + mr(1/2)Lr]g]) θ = arccos(1 - [(1/2)(IA + mb(Lr + rd)2)ω2]/[md(Lr+rd) + mr(1/2)Lr]g]) θ = arccos(1 - [(1/2)((39.04⋅kg⋅m2) + (0.01⋅kg)(2⋅m + 0.4⋅m)2)((0.460386315 rad s-1)2]/[(6⋅kg)(2⋅m + 0.4⋅m) + (3⋅kg)(1/2)(2⋅m)](9.81⋅m⋅s-2)]) θ ≈ 12.650008408° θ ≅ 12.7° The confusion (part 2) This is the correct answer and is the same value produced by the solution manuals energy balance equation yet a few major questions remain: The too long didn't read summary (TL;DR) 1) How do you do this (find the angle θ) without energy balance and with momentum balance, or is it even possible given the information? (most important question) 2) How did the solution manual's author produce the equation for energy balance on intuition since I took a much different route as shown above. (not that important of a question) 3) What happened to the massive amount of kinetic energy in the bullet? (somewhat important) Answered, thank you. The energy is lost due to heat from the inelastic collision. If you consider the energy of the bullet to be Eb = KEb = (1/2)mb⋅vb2 Eb = (1/2)(0.01⋅kg)(750⋅m⋅s-1)2 Eb = 2812.5⋅kg⋅m2s-2 Logically the energy is conserved since there is no friction and air resistance etc. etc. So shouldn't the energy in the bullet, the energy in the pendulum after impact, and the energy during the highest point in the swing all be equal? Eb = Esys,1 = Esys,2 (1/2)mb⋅vb2 = (1/2)Isys⋅ω2 = msys⋅g⋅Δh2 However Esys,1 = (1/2)[(39.04⋅kg⋅m2) + (0.01⋅kg)(2⋅m + 0.4⋅m)2](0.460386315 rad s-1)2 Esys,1 ≈ 4.143476833⋅kg⋅m2s-2 Esys,2 = [md(Lr+rd) + mr(1/2)Lr]g(1-cos(θ)) Esys,2 = [(6⋅kg)(2⋅m+0.4⋅m) + (3⋅kg)(1/2)(2⋅m)](9.81⋅m⋅s-2)(1 - cos(12.650008408°) Esys,2 ≈ 4.143427465 ⋅kg⋅m2s-2 Eb ≠ Esys,1 ≈ Esys,2 2812.5⋅kg⋅m2s-2 >> 4.143476833⋅kg⋅m2s-2 ≈ 4.143427465 ⋅kg⋅m2s-2 This would be a major problem if I had decided to set the energy balance to the bullet and the state where the pendulum is at its highest and then solve for θ. The answer would be drastically different and presumably wrong and I'm not sure why exactly Eb = Esys,2 (1/2)mb⋅vb2 = [md(Lr+rd) + mr(1/2)Lr]g(1-cos(θ)) θ would be huge, also where is over 2800 J of energy going when we are ending up with only about 4 J ?