Problem Solving Steps 1. Geometry & drawing: trajectory, vectors, coordinate axes free-body diagram, … 2. Data: a table of known and unknown quantities, including “implied data”. 3. Equations ( with reasoning comments ! ), their solution in algebraic form, and the final answers in algebraic form !!! 4. Numerical calculations and answers. 5. Check: dimensional, functional, scale, sign, … analysis of the answers and solution.

Formula Sheet – PHYS 218 Mathematics π = 3.14…; 1 rad = o = 360 o /2π; volume of sphere of radius R: V = (4π/3)R 3 Quadratic equation: ax 2 + bx + c = 0 → Vectors and trigonometry: Calculus:

Chapters Constants: g = 9.80 m/s 2, M earth = 6·10 24 kg, c = km/s, 1 mi = 1.6 km 1-Dimensional Kinematics: 3- or 2-Dimensional Kinematics: Equations of 1-D and 3-D kinematics for constant acceleration: Circular motion:

Chapters 4 – 7 Energy, work, and power: Dynamics: Chapters 8 – 11 Momentum: Equilibrium:Pressure: p = F ┴ /A Rotational kinematics: ω = dθ/dt, α = dω/dt, s = r θ, v tan = rω, a tan = rα, a rad = a c = rω 2 Constant acceleration: ω = ω 0 +αt; θ = θ 0 +(ω 0 +ω)t/2, θ = θ 0 +ω 0 t+αt 2 /2, ω 2 = ω αΔθ I = Σ i m i r i 2, I=I cm +Mr cm 2, K R =Iω 2 /2, E=Mv cm 2 /2+I cm ω 2 /2+U, W R = ∫τdθ, P R =dW R /dt=τ·ω Rotational dynamics: τ = Fl= Fr sinφ, Rigid body rotating around a symmetry axis: Iα z = τ z,

Exam Example 18: The Ballistic Pendulum (example 8.8, problem 8.43) A block, with mass M = 1 kg, is suspended by a massless wire of length L=1m and, after completely inelastic collision with a bullet with mass m = 5 g, swings up to a maximum height y = 10 cm. Find: (a) velocity v of the block with the bullet immediately after impact; (b) tension force T immediately after impact; (c) initial velocity v x of the bullet. Solution: V top =0 (a) Conservation of mechanical energy K+U=const (c) Momentum conservation for the collision (b) Newton’s second law yields y L

Exam Example 19: Collision of Two Pendulums

Exam Example 20: Head-on elastic collision (problems 8.48, 8.50) X V 02x V 01x m1m1 m2m2 0 Data: m 1, m 2, v 01x, v 02x Find: (a) v 1x, v 2x after collision; (b) Δp 1x, Δp 2x, ΔK 1, ΔK 2 ; (c) x cm at t = 1 min after collision if at a moment of collision x cm (t=0)=0 Solution: In a frame of reference moving with V 02x, we have V’ 01x = V 01x - V 02x, V’ 02x = 0, and conservations of momentum and energy yield m 1 V’ 1x +m 2 V’ 2x =m 1 V’ 01x → V’ 2x =(m 1 /m 2 )(V’ 01x -V’ 1x ) m 1 V’ 2 1x +m 2 V’ 2 2x =m 1 V’ 2 01x → (m 1 /m 2 )(V’ 2 01x -V’ 2 1x )=V’ 2 2x = (m 1 /m 2 ) 2 (V’ 01x - V’ 1x ) 2 → V’ 01x +V’ 1x =(m 1 /m 2 )(V’ 01x –V’ 1x )→ V’ 1x =V’ 01x (m 1 -m 2 )/(m 1 +m 2 ) and V’ 2x =V’ 01x 2m 1 /(m 1 +m 2 ) (a) returning back to the original laboratory frame, we immediately find: V 1x = V 02x +(V 01x – V 02x ) (m 1 -m 2 )/(m 1 +m 2 ) and V 2x = V 02x +(V 01x – V 02x )2m 1 /(m 1 +m 2 ) y’ X’ (a) Another solution: In 1-D elastic collision a relative velocity switches direction V 2x -V 1x =V 01x -V 02x. Together with momentum conservation it yields the same answer. (b) Δp 1x =m 1 (V 1x -V 01x ), Δp 2x =m 2 (V 2x -V 02x ) → Δp 1x =-Δp 2x (momentum conservation) ΔK 1 =K 1 -K 01 =(V 2 1x -V 2 01x )m 1 /2, ΔK 2 =K 2 -K 02 =(V 2 2x -V 2 02x )m 2 /2→ΔK 1 =-ΔK 2 (E=const) (c)x cm = (m 1 x 1 +m 2 x 2 )/(m 1 +m 2 ) and V cm = const = (m 1 V 01x +m 2 V 02x )/(m 1 +m 2 ) → x cm (t) = x cm (t=0) + V cm t = t (m 1 V 01x +m 2 V 02x )/(m 1 +m 2 )

Exam Example 21: Head-on completely inelastic collision (problems 8.86) Data: m 2 =2m 1, v 10 =v 20 =0, R, ignore friction Find: (a) velocity v of stuck masses immediately after collision. (b) How high above the bottom will the masses go after colliding? Solution: (a) Momentum conservation y x h m1m1 m2m2 Conservation of energy: (i) for mass m 1 on the way to the bottom just before the collision (ii) for the stuck together masses on the way from the bottom to the top (b)

Exam Example 22: Throwing a Discus (example 9.4)

Exam Example 23: Blocks descending over a massive pulley (problem 9.83) R m1m1 m2m2 x y ayay ω 0 ΔyΔy Data: m 1, m 2, μ k, I, R, Δy, v 0y =0 Find: (a) v y ; (b) t, a y ; (c) ω, α; (d) T 1, T 2 Solution: (a) Work-energy theorem W nc = ΔK + ΔU, ΔU = - m 2 gΔy, W nc = - μ k m 1 g Δy, since F N1 = m 1 g, ΔK=K=(m 1 +m 2 )v y 2 /2 + Iω 2 /2 = (m 1 +m 2 +I/R 2 )v y 2 /2 since v y = Rω (b) Kinematics with constant acceleration: t = 2Δy/v y, a y = v y 2 /(2Δy) (c) ω = v y /R, α = a y /R = v y 2 /(2ΔyR) (d) Newton’s second law for each block: T 1x + f kx = m 1 a y → T 1x = m 1 (a y + μ k g), T 2y + m 2 g = m 2 a y → T 2y = - m 2 (g – a y )

Combined Translation and Rotation: Dynamics Note: The last equation is valid only if the axis through the center of mass is an axis of symmetry and does not change direction. Exam Example 24: Yo-Yo has I cm =MR 2 /2 and rolls down with a y =R α z (examples 10.4, 10.6; problems 10.20, 10.75) Find: (a) a y, (b) v cm, (c) T Mg-T=Ma y τ z =TR=I cm α z a y =2g/3, T=Mg/3 ayay y

Exam Example 25: Race of Rolling Bodies (examples 10.5, 10.7; problem 10.22, problem 10.29) β Data: I cm =cMR 2, h, β Find: v, a, t, and min μ s preventing from slipping y x Solution 1: Conservation of EnergySolution 2: Dynamics (Newton’s 2 nd law) and rolling kinematics a=R α z x = h / sin β v 2 =2ax fsfs FNFN

Equilibrium, Elasticity, and Hooke’s Law Conditions for equilibrium: Static equilibrium: State with is equilibrium but is not static. Strategy of problem solution: (0) (i)Choice of the axis of rotation: arbitrary - the simpler the better. (ii) Free-body diagram: identify all external forces and their points of action. (iii) Calculate lever arm and torque for each force. (iv) Solve for unknowns. Exam Example 26: Ladder against wall (example 11.3, problem 11.10) (c) y man when ladder starts to slip Data: m, M, d, h, y, μ s Find: (a) F 2, (b) F 1, f s, h θ x d/2 y Solution: equilibrium equations yield (a) F 2 = Mg + mg ; (b) F 1 = f s Choice of B-axis (no torque from F 2 and f s ) F 1 h = mgx + Mgd/2 → F 1 = g(mx+Md/2)/h = f s (c) Ladder starts to slip when μ s F 2 = f s, x = yd/h → μ s g (M+m) = g (my man d/h+Md/2)/h → d