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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 - PowerPoint PPT Presentation
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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
zzzyyyxxx iBAiBAiBABAC
)()()(
)/(cos)/(sin)/(tan,tan
,sin,cos,,:2
111
22
AAAAAAA
A
AAAAAAAiAiAAvectorsD
xyxyx
y
yxyxyyxx
Calculus:
2
1
11
12
1
1,
)(t
t
nnnnn
ttn
adtatant
dt
atd
2
1
2
1
)cos(cossinsincos
sinsincoscossin
12
12
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
dttavtvdt
xd
dt
dva
tt
vvaa
dttvxtxdt
dxv
tt
xxvv
0
0
)()(,,
)()(,,
02
2
0
00
0
3- or 2-Dimensional Kinematics:
t
t
oav
t
t
zyxav
dttavtvdt
rd
dt
vda
tt
vvaa
dttvrtridt
dzi
dt
dyi
dt
dx
dt
rdv
tt
rrvv
0
0
)()(,,
)()(,,
02
2
0
00
0
Equations of 1-D and 3-D kinematics for constant acceleration:
)(2,)(2
2
1)(,
2
1)(
2,,
2,
)(,)(
020
20
20
2
200
200
00
00
00
rravvxxavv
tatvrtrtatvxtx
vvvtvrr
vvvtvxx
tavtvtavtv
xxx
xx
xxxx
xxx
Circular motion: RvTdt
vda
R
vaa radc 2,, tan
2
Chapters 4 – 7,
jjFam
)'(,,, max lawsHookekxFFfFffmgw xNkkNsss
Energy, work, and power: 2
1
2
1
;cosx
x
P
P
x ldFdxFWconstFifFssFW
,;;2
112
2 vFdt
dWPKKKWmvK tot
;2
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
mGmF
E
Egg
rGMvrGMv
GM
aTconstrv
aaxismajorSemi
FocustoCenterelawssKepler
orbitcircularescape /,/2
2)3(,sin)2(,)1(:'
2/3
0
02
202
02
2
2
tan,),sin(),cos(0x
vvxAtAvtAxx
dt
xd
)();(;2
2 pendulumd
gor
I
mgdoscillatorspring
m
k
Tf
)(),cos(02 022
0202
2
iftAexxdt
dx
dt
xd t
222220
maxmax
202
2
4)()cos(
12
dd
dm
FAtF
mx
dt
dx
dt
xd
2
,)],(cos[)cos(02
22
2
2
kkT
vvtxkAkxtAyx
yv
t
y
2,;
2,
2);sin()sin(;
22
21
22
2
1 AFP
Fv
n
L
L
nvftkxAy
r
r
I
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.