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,

Chapters 13, 14, and 15

Exam Example 1 : Coin Toss V y =0 y 0 y0y0 v 0y yvyvy a y t 0+6m/s??-g=-9.8m/s 2 ? Questions: (a) How high does the coin go? (b) What is the total time the coin is in the air? Total time T= 2 t = 1.2 s (c) What is its velocity when it comes back at y=0 ? for y=0 and v y <0 yields v y 2 = v 0 2 → v y = -v 0 = - 6m/s (problem 2.85)

Exam Example 2: Accelerated Car (problems 2.7 and 2.17) Data: x(t)= αt+βt 2 +γt 3, α=6m/s, β=1m/s 2, γ = -2 m/s 3, t=1s Find: (a) average and instantaneous velocities; (b) average and instantaneous accelerations; (c) a moment of time t s when the car stops. Solution: (a) v(t)=dx/dt= α+2βt +3γt 2 ; v 0 =α; (b) a(t)=dv/dt= 2β +6γt; a 0 =2β; (c) v(t s )=0 → α+2βt s +3γt s 2 =0 x 0 V(t) t 0 tsts α a(t) 2β2β

Exam Example 3: Truck vs. Car (problem 2.34) x 0 Data: Truck v=+20 m/s Car v 0 =0, a c =+3.2 m/s 2 Questions: (a) x where car overtakes the truck; (b) velocity of the car V c at that x; (c) x(t) graphs for both vehicles; (d) v(t) graphs for both vehicles. Solution: truck’s position x=vt, car’s position x c =a c t 2 /2 (a) x=x c when vt=a c t 2 /2 → t=2v/a c → x=2v 2 /a c (b) v c =v 0 +a c t → v c =2v x t 0 truck car truck car t V(t) v 0 t=2v/a c v/a c v c =2v t=2v/a c

Exam Example 4: Free fall past window (problem 2.84) Data: Δt=0.42 s ↔ h=y 1 -y 2 =1.9 m, v 0y =0, a y = - g Find: (a) y 1 ; (b) v 1y ; (c) v 2y y 0 y1y1 y2y2 V 0y =0 V 1y V 2y ayay h 1 st solution: (b) Eq.(3) y 2 =y 1 +v 1y Δt – gΔt 2 /2 → v 1y = -h/Δt + gΔt/2 (a) Eq.(4) → v 1y 2 = -2gy 1 → y 1 = - v 1y 2 /2g = -h 2 /[2g(Δt)] 2 +h/2 – g(Δt) 2 /8 (c) Eq.(4) v 2y 2 = v 1y 2 +2gh = (h/Δt + gΔt/2) 2 2 nd solution: (a)Free fall time from Eq.(3): t 1 =(2|y 1 |/g) 1/2, t 2 =(2|y 2 |/g) 1/2 → Δt+t 1 =t 2 (b) Eq.(4) → (c) Eq.(4) →

Exam Example 5: Relative motion of free falling balls (problem 2.94) y 0 H Data: v 0 =1 m/s, H= 10 m, a y = - g Find: (a) Time of collision t; (b) Position of collision y; (c) What should be H in order v 1 (t)= Solution: (a) Relative velocity of the balls is v 0 for they have the same acceleration a y = –g → t = H/v 0 (b) Eq.(3) for 2 nd ball yields y = H – (1/2)gt 2 = H – gH 2 /(2v 0 2 ) (c) Eq.(1) for 1 st ball yields v 1 = v 0 – gt = v 0 – gH/v 0, hence, for v 1 =0 we find H = v 0 2 /g

Projectile Motion a x =0 → v x =v 0x =const a y = -g → v oy = v oy - gt x = x 0 + v ox t y = y o + v oy t – gt 2 /2 v 0x = v 0 cos α 0 v 0y = v 0 sin α 0 tan α = v y / v x Exam Example 6: Baseball Projectile Data: v 0 =22m/s, α 0 =40 o x0x0 y0y0 v 0x v 0y axax a y xyvxvx vyvy t 00 ? ?0-9.8m/s 2 ????? Find: (a) Maximum height h; (b) Time of flight T; (c) Horizontal range R; (d) Velocity when ball hits the ground Solution: v 0x =22m/s·cos 40 o =+17m/s; v 0y =22m/s·sin40 o =+14m/s (a)v y =0 → h = (v y 2 -v 0y 2 ) / (2a y )= - (14m/s) 2 / (- 2 · 9.8m/s 2 ) = +10 m (b)y = (v 0y +v y )t / 2 → t = 2y / v 0y = 2 · 10m / 14m/s = 1.45 s; T = 2t =2.9 s (c)R = x = v 0x T = 17 m/s · 2.9 s = + 49 m (d)v x = v 0x, v y = - v 0y (examples , problem 3.12)

Exam Example 7: Ferris Wheel (problem 3.29) Data: R=14 m, v 0 =3 m/s, a || =0.5 m/s 2 Find: (a) Centripetal acceleration (b) Total acceleration vector (c) Time of one revolution T Solution: (a) Magnitude: a c =a ┴ = v 2 / r Direction to center: (b) θ (c)

Exam Example 8: Relative motion of a projectile and a target (problem 3.56) Data: h=8.75 m, α=60 o, v p0 =15 m/s, v tx =-0.45 m/s 0 y x Find: (a) distance D to the target at the moment of shot, (b) time of flight t, (c) relative velocity at contact. Solution: relative velocity (c) Final relative velocity: (b) Time of flight (a) Initial distance

How to measure friction by meter and clock? Exam Example 9: d) Find also the works done on the block by friction and by gravity as well as the total work done on the block if its mass is m = 2 kg (problem 6.68).

d) Work done by friction: W f = -f k L = -μ k F N L = -L μ k mg cosθ max = -9 J ; work done by gravity: W g = mgH = 10 J ; total work: W = mv || 2 /2 = 2 kg (1m/s) 2 /2 = 1 J = W g + W f = 10 J – 9 J = 1 J

Exam Example 10: Blocks on the Inclines (problem 5.92) m 1 m2m2 X X α1α1 α2 α2 Data: m 1, m 2, μ k, α 1, α 2, v x <0 Solution: Newton’s second law for block 1: F N1 = m 1 g cos α 1, m 1 a x = T 1x +f k1x -m 1 g sinα 1 (1) block 2: F N2 = m 2 g cosα 2, m 2 a x = T 2 x +f k2x + m 2 g sinα 2 (2) Find: (a) f k1x and f k2x ; (b) T 1x and T 2x ; (c) acceleration a x. (a) f k1x = sμ k F N1 = s μ k m 1 g cosα 1 ; f k2x = sμ k F N2 = sμ k m 2 g cosα 2 ; s = -v x /v (c) T 1x =-T 2x, Eqs.(1)&(2)→ (b)

Exam Example 11: Hoisting a Scaffold Y 0 m Data: m = 200 kg Find: (a) a force F y to keep scaffold in rest; (b) an acceleration a y if F y = N; (c) a length of rope in a scaffold that would allow it to go downward by 10 m Solution Newton’s second law: (a) Newton’s third law: F y = - T y, in rest a y = 0→ F(a=0)= W/5= mg/5 =392 N (b) a y = (5T-mg)/m = 5 (-F y )/m – g = 0.2 m/s 2 (c) L = 5·10 m = 50 m (pulley’s geometry)

Data: L, β Find: (a) tension force F; (b) speed v; (c) period T. Solution: Newton’s second law Centripetal force along x: Equilibrium along y: Two equations with two unknowns: F, v The conical pendulum (example 5.20) or a bead sliding on a vertical hoop (problem 5.119) Exam Example 12: R

Exam Example 13: Stopping Distance (problems 6.29, 7.29) x 0 Data: v 0 = 50 mph, m = 1000 kg, μ k = 0.5 Find: (a) kinetic friction force f kx ; (b)work done by friction W for stopping a car; (c)stopping distance d ; (d)stopping time T; (e)friction power P at x=0 and at x=d/2; (f)stopping distance d’ if v 0 ’ = 2v 0. Solution: (a)Vertical equilibrium → F N = mg → friction force f kx = - μ k F N = - μ k mg. (b) Work-energy theorem → W = K f – K 0 = - (1/2)mv 0 2. (c) W = f kx d = - μ k mgd and (b) yield μ k mgd = (1/2)mv 0 2 → d = v 0 2 / (2μ k g). Another solution: second Newton’s law ma x = f kx = - μ k mg → a x = - μ k g and from kinematic Eq. (4) v x 2 =v a x x for v x =0 and x=d we find the same answer d = v 0 2 / (2μ k g). (d) Kinematic Eq. (1) v x = v 0 + a x t yields T = - v 0 /a x = v 0 / μ k g. (e) P = f kx v x → P(x=0) = -μ k mgv 0 and, since v x 2 (x=d/2) = v 0 2 -μ k gd = v 0 2 /2, P(x=d/2) = P(x=0)/2 1/2 = -μ k mgv 0 /2 1/2. (f) According to (c), d depends quadratically on v 0 → d’ = (2v 0 ) 2 /(2μ k g) = 4d

Exam Example 14: Swing (example 6.8) Find the work done by each force if (a) F supports quasi-equilibrium or (b) F = const, as well as the final kinetic energy K. Solution: (a) Σ F x = 0 → F = T sinθ, Σ F y = 0 → T cosθ = w = mg, hence, F = w tanθ ; K = 0 since v=0. W T =0 always since Data: m, R, θ

Exam Example 15: Riding loop-the-loop (problem 7.46) Data: R= 20 m, v 0 =0, m=100 kg Find: (a) min h such that a car does not fall off at point B, (b) kinetic energies for that h min at the points B, C, and D, (c) if h = 3.5 R, compute velocity and acceleration at C. D Solution: (a)To avoid falling off, centripetal acceleration v 2 /R > g → v 2 > gR. Conservation of energy: K B +2mgR=mgh → (1/2)mv B 2 =mg(h-2R). Thus, 2g(h-2R) > gR → h > 5R/2, that is h min = 5R/2. (b) K f +U f =K 0 +U 0, K 0 =0 → K B = mgh min - 2mgR = mgR/2, K C = mgh min - mgR = 3mgR/2, K D = mgh min = 5mgR/2. (c) (1/2)mv C 2 = K C = mgh – mgR = 2.5 mgR → v C = (5gR) 1/2 ; a rad = v C 2 /R = 5g, a tan = g since the only downward force is gravity.

Exam Example 16: Spring on the Incline (Fig.7.25, p.231) Data: m = 2 kg, θ = 53.1 o, y 0 = 4 m, k = 120 N/m, μ k = 0.2, v 0 =0. 0 ysys yfyf y0y0 y θ Solution: work-energy theorem W nc =ΔK+ΔU grav +ΔU el (a)1 st passage: W nc = -y 0 μ k mg cosθ since f k =μ k F N = =μ k mg cosθ, ΔK=K 1, ΔU grav = - mgy 0 sinθ, ΔU el =0 → K 1 =mgy 0 (sinθ-μ k cosθ), v 1 =(2K 1 /m) 1/2 =[2gy 0 (sinθ–μ k cosθ)] 1/2 2 nd passage: W nc = - (y 0 +2|y s |) μ k mg cosθ, ΔK=K 2, ΔU grav = -mgy 0 sinθ, ΔU el =0 → K 2 =mgy 0 sinθ-(y 0 +2|y s |) μ k mgcosθ, v 2 =(2K 2 /m) 1/2 (b) (1/2)ky s 2 = U el = ΔU el = W nc – ΔU grav = mg (y 0 +|y s |) (sinθ-μ k cosθ) → αy s 2 +y s –y 0 =0, where α=k/[2mg (sinθ-μ k cosθ)], → y s =[-1 - (1+4αy 0 ) 1/2 ]/(2α) W nc = - (y 0 +|y s |) mgμ k cosθ (c) K f =0, ΔU el =0, ΔU grav = -(y 0 –y f ) mg sinθ, W nc = -(y 0 +y f +2|y s |) μ k mg cosθ → Find: (a) kinetic energy and speed at the 1 st and 2 nd passages of y=0, (b)the lowest position y s and friction energy losses on a way to y s, (c) the highest position y f after rebound. m

Exam Example 17: Proton Bombardment (problem 6.76) Data: mass m, potential energy U=α/x, initial position x 0 >0 and velocity v 0x <0. Find : (a) Speed v(x) at point x. (b) How close to the repulsive uranium nucleus 238 U does the proton get? (c) What is the speed of the proton when it is again at initial position x 0 ? Solution: Proton is repelled by 238 U with a force Newton’s 2 nd law, a x =F x /m, allows one to find trajectory x(t) as a solution of the second order differential equation: (a)Easier way: conservation of energy (b)Turning point: v(x min )=0 (c)It is the same since the force is conservative: U(x)=U(x 0 ) v(x)=v(x 0 ) 238 U 0 x m proton x0x0 x min

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

Exam Example 27: Motion in the gravitational field of two bodies (problem 13.62) 0 X1X1 X2X2 X0X0 X ΔX=X-X 0 M1M1 M2M2 m Data: masses M 1, M 2, m; positions x 1, x 2, x 0, x; v(t=0)=0 Find: (a) change of the gravitational potential of the test particle m; (b) the final speed of the test particle at the final position x; (c) the acceleration of the test particle at the final position x. Solution : (a) (b) Energy conservation: (c) Newton’s 2 nd law:

Exam Example 28: Satellite in a Circular Orbit R E =6380 km r MEME m Data: r = 2R E, R E = 6380 km Find: (a) derive formula for speed v and find its value; (b) derive formula for the period T and find its value; (c) satellite’s acceleration. Solution: use the value g = GM E /R E 2 = 9.8 m/s 2 (a) The only centripetal force is the gravitational force: (b) The period T is a time required for one orbital revolution, that is (c) Newton’s second law with the central gravitational force yields a tan = 0 and a rad = a c = F g /m = GM E /r 2 = (GM E /R E 2 ) (R E /r) 2 = g/4 = 2.45 m/s 2

Exam Example 29: Satellite in an Elliptical Orbit (problem 13.77) (perigee) (apogee) hphp haha 2R E Data: h p, h a, R E = 6380 km, M E = 6·10 24 kg Find: (a) eccentricity of the orbit e; (b) period T; (c) a rad; (d) ratio of speed at perigee to speed at apogee v p /v a ; (e) speed at perigee v p and speed at apogee v a ; (f) escape speeds at perigee v 2p and at apogee v 2a. Solution: (a) r p =h p +R E, r a = h a +R E, a =(r p +r a )/2, ea = a – r p, e = 1 – r p /a = 1- 2r p /(r p +r a ) = = (r a - r p )/(r a + r p ) = (h a -h p )/(h a +h p +2R E ) (d) Conservation of angular momentum (L a = L p ) or Kepler’s second law: r a v a = r p v p, v p /v a = r a /r p (b) Period of the elliptical orbit is the same as the period of the circular orbit with a radius equal to a semi-major axis R = a, i.e., (e) Conservation of mechanical energy K + U = const : (f) Conservation of mechanical energy for an escape from a distance r (the second space speed) : (c) Newton’s 2 nd law and law of gravitation: a rad = F grav / m = GM E /r 2.

Exam Example 30: A Ball Oscillating on a Vertical Spring (problems 14.38, 14.83) y 0 y0y0 y 1 =y 0 -A v0v0 v1=0v1=0 Data: m, v 0, k Find: (a) equilibrium position y 0 ; (b) velocity v y when the ball is at y 0 ; (c)amplitude of oscillations A; (d) angular frequency ω and period T of oscillations. Unstrained→ Equilibrium Lowest position Solution: F y = - ky (a) Equation of equilibrium: F y – mg = 0, -ky 0 = mg, y 0 = - mg/k (b) Conservation of total mechanical energy (c) At the extreme positions y 1,2 = y 0 ± A velocity is zero and y 2 =y 0 + A (d)

Applications of the Theory of Harmonic Oscillations Oscillations of Balance Wheel in a Mechanical Watch Newton’s 2 nd law for rotation yields Exam Example 31 : SHM of a thin-rim balance wheel (problems 14.41,14.97) Data: mass m, radius R, period T R Questions: a) Derive oscillator equation for a small angular displacement θ from equilibrium position starting from Newton’s 2 nd law for rotation. (See above.) b) Find the moment of inertia of the balance wheel about its shaft. ( I=mR 2 ) c) Find the torsion constant of the coil spring. (mass m)

Exam Example 32: Physical Pendulum (problem 14.99, 14.54) Data: Two identical, thin rods, each of mass m and length L, are joined at right angle to form an L-shaped object. This object is balanced on top of a sharp edge and oscillates. Find: (a) moment of inertia for each of rods; (b) equilibrium position of the object’s center of mass; (c) derive harmonic oscillator equation for small deflection angle starting from Newton’s 2 nd law for rotation; (d) angular frequency and period of oscillations. Solution : (a) dm = m dx/L, d cm (b) geometry and definition x cm =(m 1 x 1 +m 2 x 2 )/(m 1 +m 2 )→ y cm = d= 2 -3/2 L, x cm =0 mm (c) Iα z = τ z, τ z = - 2mg d sinθ ≈ - 2mgd θ X y 0 (d) Object’s moment of inertia θ

Exam Example 33: Sound Intensity and Delay A rocket travels straight up with a y =const to a height r 1 and produces a pulse of sound. A ground-based monitoring station measures a sound intensity I 1. Later, at a height r 2, the rocket produces the same second pulse of sound, an intensity of which measured by the monitoring station is I 2. Find r 2, velocities v 1y and v 2y of the rocket at the heights r 1 and r 2, respectively, as well as the time Δt elapsed between the two measurements. (See related problem )

(a) Derivation of the wave equation: y(x,t) is a transverse displacement. Restoring force exerted on the segment Δx of spring: F is a tension force. μ = Δm/Δx is a linear mass density (mass per unit length). Newton’s 2 nd law: μΔx a y = F y, a y = ∂ 2 y/∂t 2 Slope= F 2y /F=∂ y/∂x Slope = -F 1y /F=∂y/∂x Exam Example 34: Wave Equation and Transverse Waves on a Stretched String (problems – 15.53) Data: λ, linear mass density μ, tension force F, and length L of a string 0<x<L. Questions: (a) derive the wave equation from the Newton’s 2 nd law; (b) write and plot y-x graph of a wave function y(x,t) for a sinusoidal wave traveling in –x direction with an amplitude A and wavelength λ if y(x=x 0, t=t 0 ) = A; (c) find a wave number k and a wave speed v; (d) find a wave period T and an angular frequency ω; (e) find an average wave power P av. Solution: (b) y(x,t) = A cos[2π(x-x 0 )/λ + 2π(t-t 0 )/T] where T is found in (d); y X 0 L A (c) k = 2π / λ, v = (F/μ) 1/2 as is derived in (a); (d)v = λ / T = ω/k → T = λ /v, ω = 2π / T = kv (e)P(x,t) = F y v y = - F (∂y/∂x) (∂y/∂t) = (F/v) v y 2 P av = Fω 2 A 2 /(2v) =(1/2)( μF) 1/2 ω 2 A 2.