Problem Solving Steps1. 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, … analysisof the answers and solution.

,...,,, jFavr

,,, zyx

Formula Sheet – PHYS 218 Mathematics π = 3.14…; 1 rad = 57.30o= 360o/2π; volume of sphere of radius R: V = (4π/3)R3 Quadratic equation: ax2 + bx + c = 0 → a

acbbx

2

42

2,1

Vectors and trigonometry: 222, zyxzzyyxx AAAAiAiAiAA

sin,cos ABBABABABAABBA zzyyxx

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)/(cos)/(sin)/(tan,tan

,sin,cos,,:2

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Calculus:

2

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adtatant

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atd

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dd

d

dd

d

Chapters 1 - 3Constants: g = 9.80 m/s2, Mearth = 6·1024 kg, c = 300 000 km/s, 1 mi = 1.6 km

1-Dimensional Kinematics:

t

t

xxxx

xoxx

xavx

t

t

xxxavx

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dva

tt

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dttvxtxdt

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t

t

oav

t

t

zyxav

dttavtvdt

rd

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vda

tt

vvaa

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dzi

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dyi

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Equations of 1-D and 3-D kinematics for constant acceleration:

)(2,)(2

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1)(,

2

1)(

2,,

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)(,)(

020

20

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200

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00

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rravvxxavv

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vvvtvrr

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Circular motion: RvTdt

vda

R

vaa radc 2,, tan

2

Chapters 4 – 7,

jjFam

)'(,,, max lawsHookekxFFfFffmgw xNkkNsss

Energy, work, and power: 2

1

2

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;cosx

x

P

P

x ldFdxFWconstFifFssFW

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1;; 2

21 kxUmghUUUUW elasticgrav

otherWUKUK 1122

zyxx iz

Ui

y

Ui

x

UF

dx

dUxF

;)(

Dynamics:

Chapters 8 – 11Momentum:

ft

t

fi i

iicmcmii dtFPPF

dt

Pd

m

rmrvMvmPvmp

0

0,,,,

20210122110 vmvmvmvmFifconstvm ffj jj

Equilibrium: )(0,0 axisanyF ii

Pressure: p = F┴/A

Rotational kinematics: ω = dθ/dt, α = dω/dt, s = r θ, vtan = rω, atan= rα, arad= ac = rω2

Constant acceleration: ω = ω0+αt; θ = θ0+(ω0+ω)t/2, θ = θ0+ω0t+αt2/2, ω2 = ω02+2αΔθ

I = Σimiri2, I=Icm+Mrcm

2, KR=Iω2/2, E=Mvcm2/2+Icmω2/2+U, WR= ∫τdθ, PR=dWR/dt=τ·ω

Rotational dynamics: τ = Fl= Fr sinφ,

dt

LdvmrprLFr

i iiiii ,,

Rigid body rotating around a symmetry axis: Iαz = τz ,

IL0000 IILLifconstL fff

Chapters 13, 14, and 15

2

211

2221

221 1067.6,8.9,,,

kg

NmG

s

m

R

Gmgmgw

r

mGmU

r

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orbitcircularescape /,/2

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t

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2);sin()sin(;

22

21

22

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1 AFP

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n

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L

nvftkxAy

r

r

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IavnnSWSW

Exam Example 1 : Coin Toss Vy=0y

00v

0vv

ay0 v0y y vy ay t

0 +6m/s ? ? -g=-9.8m/s2 ?Questions:(a) How high does the coin go?

)4( )(2 020

2 yyavv yyy msm

sm

a

vvy

y

y 8.1/8.92

)/6(

2 2

220

2

(b) What is the total time the coin is in the air?

)1(t

vva yyy

0 s

sm

sm

a

vvt

y

y 6.0/8.9

/62

0

Total time T= 2 t = 1.2 s(c) What is its velocity when it comes back at y=0 ?

)4( )(2 020

2 yyavv yyy for y=0 and vy<0 yields vy2 = v0

2 → vy = -v0 = - 6m/s

(problem 2.37)

Exam Example 2: Accelerated Car (problems 2.7 and 2.17)

Data: x(t)=αt+βt2+γt3, α=6m/s, β=1m/s2, γ = -2 m/s3, t=1sFind: (a) average and instantaneous velocities;(b) average and instantaneous accelerations;(c) a moment of time ts when the car stops.Solution: (a) v(t)=dx/dt= α+2βt +3γt2 ; v0=α;

(b) a(t)=dv/dt= 2β +6γt; a0=2β;

(c) v(ts)=0 → α+2βts +3γts2=0

2/)( ttttxv

ttaatvtva 322/))((/))(( 00

ssm

smtt ss 17.1

/6

/)61(

3

33

22

x0

v

a

V(t)

t0ts

α

a(t)

2β

Exam Example 3: Truck vs. Car (problem 2.34)

x0

c o n s tv

cv

ca

Data: Truck v=+20 m/sCar v0=0, ac=+3.2 m/s2

Questions: (a) x where car overtakes the truck;(b) velocity of the car Vc 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 xc=act2/2 (a) x=xc when vt=act2/2 → t=2v/ac → x=2v2/ac (b) vc=v0+act → vc=2v

x

t0

truckcar

truckcar

t

V(t)

v

0

t=2v/ac

v/ac

vc=2v

t=2v/ac

Exam Example 4: Free fall past window (problem 2.74)Data: Δt=0.42 s ↔ h=y1-y2=1.9 m, v0y=0, ay= - g

Find: (a) y1 ; (b) v1y ; (c) v2y

y0

y1

y2

V0y=0

V1y

V2y

ayh

1st solution: (b) Eq.(3) y2=y1+v1yΔt – gΔt2/2 → v1y= -h/Δt + gΔt/2(a) Eq.(4) → v1y

2 = -2gy1 →

y1 = - v1y2 /2g = -h2/[2g(Δt)]2 +h/2 – g(Δt)2 /8

(c) Eq.(4) v2y2 = v1y

2 +2gh = (h/Δt + gΔt/2)2

2nd solution:(a)Free fall time from Eq.(3): t1=(2|y1|/g)1/2 , t2=(2|y2|/g)1/2 → Δt+t1=t2

2

11

2

11 22

1||||2

2

)(||||

2

tg

t

h

gyhygt

tghyy

gt

(b) Eq.(4) → ||2 11 ygv y (c) Eq.(4) → )|(|2 12 hygv y

Exam Example 5: Relative motion of free falling balls (problem 2.82) y

0

H

0v aData: v0=1 m/s, H= 10 m, ay= - g

Find: (a) Time of collision t;(b) Position of collision y;(c) What should be H in order v1(t)=0.

1

2

Solution: (a) Relative velocity of the balls is v0 for they have the same acceleration ay= –g → t = H/v0

(b) Eq.(3) for 2nd ball yields y = H – (1/2)gt2 = H – gH2/(2v02)

(c) Eq.(1) for 1st ball yields v1 = v0 – gt = v0 – gH/v0 , hence, for v1=0 we find H = v0

2/g

Projectile Motionax=0 → vx=v0x=constay= -g → voy= voy- gtx = x0 + vox ty = yo + voy t – gt2/2v0x= v0 cos α0 v0y= v0 sin α0 tan α = vy / vx Exam Example 6: Baseball Projectile Data: v0=22m/s, α0=40o

x0 y0 v0x v0y ax ay x y vx vy t

0 0 ? ? 0 -9.8m/s2 ? ? ? ? ?Find: (a) Maximum height h;(b) Time of flight T;(c) Horizontal range R; (d) Velocity when ball hits the ground

Solution: v0x=22m/s·cos40o=+17m/s; v0y=22m/s·sin40o=+14m/s

(a)vy=0 → h = (vy2-v0y

2) / (2ay)= - (14m/s)2 / (- 2 · 9.8m/s2) = +10 m(b)y = (v0y+vy)t / 2 → t = 2y / v0y= 2 · 10m / 14m/s = 1.45 s; T = 2t =2.9 s (c)R = x = v0x T = 17 m/s · 2.9 s = + 49 m (d)vx = v0x , vy = - v0y

(examples 3.7-3.8, problem 3.12)

Exam Example 7: Ferris Wheel (problem 3.27)

Data: R=14 m, v0 =3 m/s, a|| =0.5 m/s2

Find:(a) Centripetal acceleration(b) Total acceleration vector(c) Time of one revolution T

Solution:(a) Magnitude: ac =a┴ = v2 / rDirection to center: rr /

(b)

)/(tan/tan ||1

||

22||||

aaaa

aaaaaa

||a

a

a

θ

(c)

||

||200

02||

||0

4

022

)2/(2

a

RavvT

RTvTa

TavTvTR

Exam Example 8: Relative motion of a projectile and a target (problem 3.52)

Data: h=8.75 m, α=60o, vp0 =15 m/s, vtx =-0.45 m/s

0

y

x

0pv

tv

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

tp vvv

)(60cos 0

00 txptxxpx vvvvv ghvvghvvEqKinematic ppyyppy 2sin2)4.( 22

02

02

gvvgvt ppypy /)sin(/ 0 tvD x

Exam Example 9:

How to measurefriction by meter and clock?

d) Find also the works done on the block by friction and by gravityas well as the total work done on the block if its mass is m = 2 kg (problem 6.66).

(example 5.17)

d) Work done by friction: Wf = -fkL = -μk FN L = -L μk mg cosθmax = -9 J ; work done by gravity: Wg = mgH = 10 J ;

total work: W = mv||2 /2 = 2 kg (1m/s)2 /2 = 1 J = Wg + Wf = 10 J – 9 J = 1 J

Exam Example 10: Blocks on the Inclines (problem 5.90)

m1

m2

X

X

α1 α2

1W

1NF

2NF

2W

XW1

XW2

1kf

2kf

Data: m1, m2, μk, α1, α2, vx<0 a

Solution:Newton’s second law for

block 1: FN1 = m1g cosα1 , m1ax= T1x+fk1x-m1g sinα1 (1)

block 2: FN2 = m2g cosα2 , m2ax= T2 x+fk2x+ m2g sinα2 (2)

Find: (a) fk1x and fk2x ;(b) T1x and T2x ;(c) acceleration ax .

1T 2T

(a) fk1x= sμkFN1= sμkm1g cosα1 ; fk2x= sμkFN2= sμkm2g cosα2; s = -vx/v (c) T1x=-T2x, Eqs.(1)&(2)→

(b)

21

222111 )sincos()sincos(

mm

sgmsgma kkx

21

12212111121

))cos(cossin(sin)sincos(

mm

sgmmggsamTT k

kxxx

v

Exam Example 11: Hoisting a Scaffold

Ya

T

TT

TT

0

m F

gmW

Data: m = 200 kg

Find: (a) a force Fy to keep scaffold in rest;(b) an acceleration ay if Fy = - 400 N;(c) a length of rope in a scaffold that would allow it to go downward by 10 m

SolutionNewton’s second law: WTam

5

(a) Newton’s third law: Fy = - Ty , in rest ay = 0→ F(a=0)= W/5= mg/5 =392 N

(b) ay= (5T-mg)/m = 5 (-Fy)/m – g = 0.2 m/s2

(c) L = 5·10 m = 50 m (pulley’s geometry)

(problem 5.62)

Data: L, β Find: (a) tension force F;(b) speed v;(c) period T.

Solution:Newton’s second law

j

cj amF

Centripetal force along x: RmvmaF c /sin 2Equilibrium along y: cos/)(cos mgFamgF

cos/sincos/sin

sin,tansin)/()(

2

2

LgLgv

LRRgmFRvb

g

LTLgLvRTc

cos2cos//2/2)(

Two equations with two unknowns: F,v

The conical pendulum (example 5.20)

or a bead sliding on a vertical hoop (problem 5.107)

Exam Example 12:

R

FF

gm

gm ca

ca

Exam Example 13: Stopping Distance (problems 6.31, 7.27)

x

v

0a

Data: v0 = 50 mph, m = 1000 kg, μk = 0.5 Find: (a) kinetic friction force fkx ;(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 v0’ = 2v0 .

NF

gm

kf

Solution:(a)Vertical equilibrium → FN = mg → friction force fkx = - μk FN = - μk mg .(b) Work-energy theorem → W = Kf – K0 = - (1/2)mv0

2 .

(c) W = fkxd = - μkmgd and (b) yield μkmgd = (1/2)mv02 → d = v0

2 / (2μkg) . Another solution: second Newton’s law max= fkx= - μkmg → ax = - μkg and from kinematic Eq. (4) vx

2=v02+2axx for vx=0 and x=d we find

the same answer d = v02 / (2μkg) .

(d) Kinematic Eq. (1) vx = v0 + axt yields T = - v0 /ax = v0 / μkg . (e) P = fkx vx → P(x=0) = -μk mgv0 and, since vx

2(x=d/2) = v02-μkgd = v0

2 /2 , P(x=d/2) = P(x=0)/21/2 = -μkmgv0 /21/2 .(f) According to (c), d depends quadratically on v0 → d’ = (2v0)2/(2μkg) = 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 Fx = 0 → F = T sinθ , Σ Fy = 0 → T cosθ = w = mg , hence, F = w tanθ ; K = 0 since v=0 .

WT =0 always since ldT

Rddsdl

0000

)cos1(sincostancos wRdwRRdwdsFldFW F

0

0 0

)cos1(sin

)sin(

wRdwR

RdwldwW grav

0 0

sincos)( FRRdFldFWb F

Data: m, R, θ

)cossin( wwFRWWWWK TgravF

Exam Example 15: Riding loop-the-loop (problem 7.42)

Data: R= 20 m, v0=0, m=100 kg

Find: (a) min h such that a car does not fall off at point B,(b) kinetic energies for that hmin at the points B, C, and D,(c) if h = 3.5 R, compute velocity and acceleration at C.

D

Solution: v

rada

tana

a

(a)To avoid falling off, centripetal acceleration v2/R > g → v2 > gR.Conservation of energy: KB+2mgR=mgh → (1/2)mvB

2=mg(h-2R) . Thus, 2g(h-2R) > gR → h > 5R/2 , that is hmin = 5R/2.

(b) Kf+Uf=K0+U0 , K0=0 → KB = mghmin- 2mgR = mgR/2 ,

KC = mghmin- mgR = 3mgR/2 , KD = mghmin = 5mgR/2.

(c) (1/2)mvC2 = KC= mgh – mgR = 2.5 mgR → vC = (5gR)1/2 ;

arad = vC2/R = 5g, atan = g since the only downward force is gravity.

g

Exam Example 16: Spring on the Incline (Fig.7.25, p.227)Data: m = 2 kg, θ = 53.1o, y0 = 4 m, k = 120 N/m, μk = 0.2, v0 =0.

0

ys

yf

y0

y

gm

NF

kf

θ

Solution: work-energy theorem Wnc=ΔK+ΔUgrav+ΔUel

(a)1st passage: Wnc= -y0μkmg cosθ since fk=μkFN==μkmg cosθ, ΔK=K1 , ΔUgrav= - mgy0 sinθ, ΔUel=0 → K1=mgy0(sinθ-μkcosθ), v1=(2K1/m)1/2 =[2gy0(sinθ–μkcosθ)]1/2 2nd passage: Wnc= - (y0+2|ys|) μkmg cosθ, ΔK=K2, ΔUgrav= -mgy0sinθ, ΔUel=0 →K2=mgy0sinθ-(y0+2|ys|) μkmgcosθ, v2=(2K2/m)1/2 (b) (1/2)kys

2 = Uel = ΔUel = Wnc – ΔUgrav = mg (y0+|ys|) (sinθ-μkcosθ) →αys

2 +ys –y0 =0, where α=k/[2mg (sinθ-μkcosθ)], → ys =[-1 - (1+4αy0)1/2]/(2α) Wnc = - (y0+|ys|) mgμkcosθ(c) Kf =0, ΔUel=0, ΔUgrav= -(y0–yf) mg sinθ, Wnc= -(y0+yf+2|ys|) μkmg cosθ →

k

ks

k

ksf

yyy

yyyy

tan

|)|(2

cossin

cos|)|(2 00

00

Find: (a) kinetic energy and speed at the 1st and 2nd passages of y=0, (b) the lowest position ys and friction energy losses on a way to ys, (c) the highest position yf after rebound. m

Exam Example 17: Proton Bombardment (problem 6.72)

Data: mass m, potential energy U=α/x,initial position x0>0 and velocity v0x<0.

Find: (a) Speed v(x) at point x.(b) How close to the repulsive uranium nucleus 238U does the proton get?(c) What is the speed of the proton when it is again at initial position x0?

Solution: Proton is repelled by 238U with a force

Newton’s 2nd law, ax=Fx/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(xmin)=0

(c)It is the same since the force is conservative: U(x)=U(x0) v(x)=v(x0)

238U

0 x

mproton

x0

F

xmin

0v

02

xdx

dUFx

22

2

mxdt

xd

xxmvxUxU

mvxvxUxUmvmv

112)()(

2)()()(

2

1

2

1

0

200

200

20

2

020

0min

20

0min 2

2

2

11

xmv

xx

mv

xx

Exam Example 18: The Ballistic Pendulum (example 8.8, problem 8.43)

A block, with mass M = 1 kg, is suspended by amassless 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 vx of the bullet.

Solution:

gMm

)(

T

T

v

Vtop=0

(a) Conservation of mechanical energy K+U=const

gyvvMmgyMmvMmUK toptop 2)(2

1)()(

2

1 220

ca

(c) Momentum conservation for the collision

(b) Newton’s second law Lvaaan dTgMmaMm c /2)()( 2

yields

L

ygMm

L

vg

MmT

21

2

)(

2

2

gym

Mmv

m

MmvvMmmv xx 2)(

yL

Exam Example 19: Collision of Two Pendulums

Exam Example 20: Head-on elastic collision (problems 8.48, 8.50)

X

V02x V01x

m1

m2

0

Data: m1, m2, v01x, v02x Find: (a) v1x, v2x after collision;(b) Δp1x, Δp2x , ΔK1, ΔK2 ;(c) xcm at t = 1 min after collisionif at a moment of collision xcm(t=0)=0Solution: In a frame of reference moving

with V02x, we have V’01x= V01x- V02x, V’02x = 0, and conservations of momentum and energy yieldm1V’1x+m2V’2x=m1V’01x → V’2x=(m1/m2)(V’01x-V’1x)

m1V’21x+m2V’2

2x=m1V’201x→ (m1/m2)(V’2

01x-V’21x)=V’2

2x = (m1/m2)2(V’01x- V’1x)2 →

V’01x+V’1x=(m1/m2)(V’01x–V’1x)→ V’1x=V’01x(m1-m2)/(m1+m2) and V’2x=V’01x2m1/(m1+m2)

(a) returning back to the original laboratory frame, we immediately find:

V1x= V02x+(V01x– V02x) (m1-m2)/(m1+m2) and V2x = V02x +(V01x– V02x)2m1/(m1+m2)

y’

X’

(a) Another solution: In 1-D elastic collision a relative velocity switches directionV2x-V1x=V01x-V02x. Together with momentum conservation it yields the same answer.

(b) Δp1x=m1(V1x-V01x), Δp2x=m2(V2x-V02x) → Δp1x=-Δp2x (momentum conservation)

ΔK1=K1-K01=(V21x-V2

01x)m1/2, ΔK2=K2-K02=(V22x-V2

02x)m2/2→ΔK1=-ΔK2 (E=const)

(c)xcm = (m1x1+m2x2)/(m1+m2) and Vcm = const = (m1V01x+m2V02x)/(m1+m2)

→ xcm(t) = xcm(t=0) + Vcm t = t (m1V01x+m2V02x)/(m1+m2)

Exam Example 21: Head-on completely inelastic collision (problems 8.82)

Data: m2=2m1, v10=v20=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

1v

y

x

h

m1

m2

v

121

112111 3

1)( v

mm

vmvvmmvm

Conservation of energy: (i) for mass m1 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)

gRvgRvvmgRm 23

12

2

11

2111

9182)()(

2

1 21

2

212

21

R

g

v

g

vhghmmvmm

Exam Example 22: Throwing a Discus (example 9.4)

Exam Example 23: Blocks descending over a massive pulley (problem 9.75)

Rm1

m2

gm

2

gm

1

1NF

2T

1T

2T

1T

x

y

F

ay

ω

0

Δy

Data: m1, m2, μk, I, R, Δy, v0y=0

Find: (a) vy; (b) t, ay; (c) ω,α; (d) T1, T2

kf

Solution: (a) Work-energy theoremWnc= ΔK + ΔU, ΔU = - m2gΔy,

Wnc = - μk m1g Δy , since FN1 = m1g,ΔK=K=(m1+m2)vy

2/2 + Iω2/2 = (m1+m2+I/R2)vy2/2 since vy = Rω

221

1212

2221 /

)(2)(

2

1

RImm

mmygvmmygv

R

Imm k

yky

(b) Kinematics with constant acceleration: t = 2Δy/vy , ay = vy2/(2Δy)

(c) ω = vy/R , α = ay/R = vy2/(2ΔyR)

(d) Newton’s second law for each block:

T1x + fkx = m1ay → T1x= m1 (ay + μk g) ,

T2y + m2g = m2ay → T2y = - m2 (g – ay)

Combined Translation and Rotation: Dynamics

i

zcmizi

cmi IandaMF

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 Icm=MR2/2 and

rolls down with ay=Rαz (examples 10.4, 10.6; problems 10.20, 10.67)

Find: (a) ay, (b) vcm, (c) T Mg-T=May τz=TR=Icmαz

ay=2g/3 , T=Mg/3

ay

3

42

gyayv cm

y

Exam Example 25: Race of Rolling Bodies (examples 10.5, 10.7; problem 10.22, problem 10.27)

β v a

Data: Icm=cMR2, h, βFind: v, a, t, and min μs

preventing from slipping

y

xSolution 1: Conservation of Energy Solution 2: Dynamics(Newton’s 2nd law) androlling kinematics a=Rαz

RvandcMRIfor

MvcIMvMgh

UKUKUK

/

)1(2

1

2

1

2

1

0,0,

2

222

212211

c

ghv

1

2

x = h / sinβ

v2=2axc

ga

1

sin

g

ch

v

x

v

xt

)1(2

sin

12

fs

c

c

Mg

Mg

c

c

F

fMg

c

cMaMgf

N

sss

1

tan

cos

sin

1minsin

1sin

FN

cMafcMRaIRf

c

gaMafMgF

szcmsz

sx

1

sinsin

c

ghh

c

gaxv

1

2

sin1

sin22

Equilibrium, Elasticity, and Hooke’s Law

00)2(

00)1(

i

ii aF

Conditions for equilibrium:

Static equilibrium: 0v

State with

is equilibrium but is not static.0 c o n s tv

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.

0F

0

Exam Example 26: Ladder against wall (example 11.3, problem 11.10)

(c) yman when ladder starts to slip

Data: m, M, d, h, y, μs

Find: (a) F2, (b) F1, fs,

1F

2F

sf

gm

gM h

θ

x

d/2

y

Solution: equilibrium equations yield(a) F2= Mg + mg ; (b) F1 = fs Choice of B-axis (no torque from F2 and fs) F1h = mgx + Mgd/2 → F1= g(mx+Md/2)/h = fs

(c) Ladder starts to slip when μsF2 = fs, x = yd/h→ μsg (M+m) = g (mymand/h+Md/2)/h →

d

h

d

md

Mh

d

h

m

Mh

md

hmMy s

ssman 22

)( 222

Exam Example 27: Motion in the gravitational field of two bodies (problem 13.58)

0X

1

X2

X0

X

ΔX=X-X0M

1M

21F

2Fm

Data: masses M1

, M2

, m; positions x1

, x2

, x0

, 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 2nd

law:

2202

11010

1111

xxxxmGM

xxxxmGMUUU

mUvUmvUK /22

1 2

2

1

12

2

212 /

xx

M

xx

MGmFFa xxx

Exam Example 28: Satellite in a Circular Orbit

RE=6380 km

r

v

cg ra v FF

ME

mData: r = 2RE , RE = 6380 kmFind: (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 = GME/RE2 = 9.8 m/s2

ca

(a) The only centripetal force is the gravitational force:

skmmsmgR

r

Rg

r

R

R

GM

r

GMv

r

mv

r

mGMmaFlawndsNewton

EEE

E

E

EEcg

/6.51019.3/8.92

2'

6222

2

2

2

(b) The period T is a time required for one orbital revolution, that is

hsm

m

g

R

R

r

RGM

R

r

GM

r

rGM

r

v

rT E

E

E

EE

EE

4/8.9

1038.6222

22222

62/3

2/3

2

2/32/3

(c) Newton’s second law with the central gravitational force yields

atan = 0 and arad = ac = Fg/m = GME/r2 = (GME/RE2) (RE/r)2 = g/4 = 2.45 m/s2

Exam Example 29: Satellite in an Elliptical Orbit (problem 13.67)

(perigee) (apogee)

hp ha2RE

Data: hp , ha , RE= 6380 km, ME= 6·1024 kg

av

pv

rada

gravFFind: (a) eccentricity of the orbit e; (b) period T; (c) arad;

(d) ratio of speed at perigee to speed at apogee vp/va;(e) speed at perigee vp and speed at apogee va;(f) escape speeds at perigee v2p and at apogee v2a.

pv2

Solution: (a) rp =hp+RE, ra= ha+RE, a =(rp+ra)/2,ea = a – rp, e = 1 – rp/a = 1- 2rp/(rp+ra) = = (ra- rp)/(ra+ rp) = (ha-hp)/(ha+hp+2RE)

(d) Conservation of angular momentum (La= Lp) or Kepler’s second law: rava= rpvp, vp/va= ra/rp

(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.,

EGM

aT

2/32

(e) Conservation of mechanical energy K + U = const :

;)(

2

222 2

2222

app

aEp

a

E

a

pp

a

Ea

p

Ep

rrr

rGMv

r

mGM

r

rmv

r

mGMmv

r

mGMmv

)(

2

apa

pE

a

ppa rrr

rGM

r

rvv

(f) Conservation of mechanical energy for an escape from a distance r (the second space speed) :

a

Ea

p

Ep

EE

r

GMv

r

GMv

r

GMv

r

mGMmv 2,

22

2 222

22

(c) Newton’s 2nd law and law of gravitation: arad= Fgrav/ m = GME/r2.

Exam Example 30: A Ball Oscillating on a Vertical Spring (problems 14.69, 14.77)

y

0

y0

y1=y0-A

v0

v1=0

Data: m, v0 , k Find: (a) equilibrium position y0;(b) velocity vy when the ball is at y0;(c)amplitude of oscillations A; (d) angular frequency ω and period T of oscillations.

Unstrained→

Equilibrium

Lowest position

Solution: Fy = - ky

(a) Equation of equilibrium: Fy – mg = 0, -ky0 = mg , y0 = - mg/k (b) Conservation of total mechanical energy

20

200

20 2

1

2

1

2

1mvkymgymvEUUKE yela s t icg rav

kmgvmkygyvv y /)/2( 22000

20

(c) At the extreme positions y1,2 = y0 ± A velocity is zero and

2

20

202

0

20

2

2,120

22,12,1 1

2

1

2

1

mg

kv

k

mg

k

mvyA

k

mv

k

mg

k

mgymvkymgy

y2=y0+A

(d) k

mT

m

k 2

2,

Applications of the Theory of Harmonic Oscillations

Oscillations of Balance Wheel in a Mechanical Watch

ITft

Idt

dI zzz

22,)cos(

0,,2

2

Newton’s 2nd law for rotation yields

Exam Example 31: SHM of a thin-rim balance wheel (problems 14.41-14.43)

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 2nd law for rotation. (See above.)b) Find the moment of inertia of the balance wheel about its shaft. ( I=mR2 )c) Find the torsion constant of the coil spring.

2)/2(/2 TRmIT

(mass m)

Exam Example 32: Physical Pendulum (problem 14.88, 14.53)Data: Two identical, thin rods, each of mass m and length L,are joined at right angle to form an L-shaped object. Thisobject 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 smalldeflection angle starting from Newton’s 2nd law for rotation;(d) angular frequency and period of oscillations.

Solution: (a) dm = m dx/L ,

d

cm

2

0

21 )3/1()/( mLdxxLmI

L

(b) geometry and definition xcm=(m1x1+m2x2)/(m1+m2)→ ycm= d= 2-3/2 L, xcm=0

m m

(c) Iαz = τz , τz = - 2mg d sinθ ≈ - 2mgd θ I

mgd

dt

d 2,02

2

2

X

y

0

(d) Object’s moment of inertia 2

,22

32

3

22 2

1 TL

g

I

mgdmLII

θ

gM

Exam Example 33:Sound Intensity and Delay

A rocket travels straight upwith ay=const to a height r1

and produces a pulse of sound. A ground-based monitoringstation measures a soundintensity I1. Later, at a heightr2, the rocket produces thesame second pulse of sound,an intensity of which measuredby the monitoring station is I2.Find r2, velocities v1y and v2y ofthe rocket at the heights r1 andr2, respectively, as well as the time Δt elapsed between the two measurements.(See related problem 15.25.)

(a) Derivation of the wave equation: y(x,t) is a transverse displacement. Restoring force exerted on the segment Δx of spring:

xxx

yyy x

y

x

yFFFF 12

F is a tension force.μ = Δm/Δx is a linear massdensity (mass per unit length).

Newton’s 2nd law: μΔx ay= Fy , ay= ∂2y/∂t2

F

vx

yv

t

yEquationWave

,02

22

2

2

Slope=F2y/F=∂ y/∂xSlope = -F1y/F=∂y/∂x

Exam Example 34: Wave Equation and Transverse Waves on a Stretched String (problems 15.49, 15.63)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 2nd law;(b) write and plot y-x graph of a wave function y(x,t) for a sinusoidal wave travelingin –x direction with an amplitude A and wavelength λ if y(x=x0, t=t0) = 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 Pav .

Solution: (b) y(x,t) = A cos[2π(x-x0)/λ + 2π(t-t0)/T] where T is found in (d);

y

X0

LA

(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) = Fyvy = - F (∂y/∂x) (∂y/∂t) = (F/v) vy

2 Pav = Fω2A2 /(2v) =(1/2)(μF)1/2ω2A2.