Problem Solving Steps
   
1. Geometry & drawing: trajectory, vectors r , v , a , F j ,... ,
  
coordinate axes  x ,  y ,  z , 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
π = 3.14…; 1 rad = 57.30o= 360o/2π;
volume of sphere of radius R: V = (4π/3)R3
 b  b 2  4ac
Quadratic equation: ax2 + bx + c = 0 → x1, 2 
2a




Vectors and trigonometry: A  Ax i x  Ay i y  Az i z , A  Ax2  Ay2  Az2
Mathematics
 
A
  B AB cos Ax B x  A y B y Az Bz ,


C  A  B  ( Ax  Bx )i x  ( Ay  B y )i y  ( Az  Bz )iz
 
A  B  AB sin 



2  D vectors : A  Ax i x  A y i y , A  Ax2  A y2 , Ax  A cos , A y  A sin  ,
tan 
Ay
  tan1 ( A y / Ax )  sin 1 ( A y / A)  cos1 ( Ax / A)
Ax
t2
d (at n )
a
n 1
n
n 1
n 1



ant
,
at
dt

t

t
Calculus:
2
1

dt
n 1
2
t1
,
d sin 
 cos   cosd  sin  2  sin 1
d
1
2
d cos
  sin    sin d  (cos 2  cos1 )
d
1
Chapters 1 - 3
Constants: g = 9.80 m/s2, Mearth = 6·1024 kg, c = 300 000 km/s, 1 mi = 1.6 km
t
x

x
dx
0
1-Dimensional Kinematics:
v x  vav  x 
,
vx  ,
x (t )  x0   v x (t )dt
t  t0
dt
t0
a x  aav  x
3- or 2-Dimensional
 Kinematics:


v v
 x ox ,
t  t0
dvx d 2 x
ax 
 2 ,
dt dt
t
v x (t )  v0 x   a x (t )dt
t0
 dr dx  dy  dz 

 t
v

i x  i y  iz ,
r (t )  r0   v (t )dt
dt dt
dt
dt
t0


t
 dv d 2 r



a
 2 ,
v (t )  v 0   a (t )dt
dt dt
t0
Equations of 1-D and 3-D kinematics for constant acceleration:

 
v x (t )  v 0 x  a x t ,
v ( t )  v 0  at
 
v  vx
  
 v v
x  x0  v x t, v x  0 x
,
r  r0  v t , v  0
2
2

 
1
1
x (t )  x 0  v 0 x t  a x t 2 ,
r (t )  r0  v 0 t  at 2
2
2
2 2
  
2
2
v x  v 0 x  2a x ( x  x 0 ) ,
v  v 0 2a  (r  r0 )
dv
v2
a c  a rad 
,
a tan 
,
T  v  2R
Circular motion:
R
dt
r  r0
 
v  v av 
,
t  t0
 
v  vo
 
a  a av 
,
t  t0
 Chapters 4 – 7

Dynamics: ma   F j , w  mg, f s  f smax  s FN , f k  k FN , Fx  kx ( Hooke' s law)
x2
P2
 

 
Energy, work, and power: W  F  s  Fs cos  if F  const ; W   Fx dx   F  dl
x1
P1
1
W  U1  U 2  U ; U grav  mgh ; U elastic  kx 2 ;
2

1
dW
2
K  mv ; Wtot  K 2  K1  K ; P 
 F  v , K 2  U 2  K1  U1  Wother
2
dt

 U  U  U  
dU
Fx ( x )  
; F  
ix 
iy 
i z 
dx

x

y

z



Chapters 8 – 11

tf




m
r





dP
Momentum: p  mv , P   mi v i  Mvcm , rcm   i i ,
 F , Pf  P0   Fdt
m
dt
i
 i
t0
j






m
v

const
if
F

0

m
v

m
v

m
v

m
v
j j j
1 1f
2 2f
1 10
2 20
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+Mrcm2, KR=Iω2/2, E=Mvcm2/2+Icmω2/2+U, WR= ∫τdθ, PR=dWR/dt=τ·ω
 

 dL
   

Rotational dynamics: τ = Fl= Fr sinφ,   r  F , L   ri  pi   ri  mvi ,
 
dt


i
i

L  const if
  0
 L f  L0  I f  f  I 00


Rigid body rotating around a symmetry axis: Iαz = τz , L  I


Equilibrium:  Fi  0,  i  0 (any axis)
Pressure: p = F┴/A
Chapters 13, 14, and 15
2
Gm1m 2
Gm1m 2
GmE
m
11 Nm
Fg 
, Ug  
, w  mg, g 
 9.8 2 , G  6.67  10
2
2
r
r
RE
s
kg2
Center  to  Focus
2a 3 / 2
Kepler' s laws : (1) e 
, (2) rv sin   const, (3) T 
Semi  major axis a
v escape  2GM / r ,
v circular
orbit
GM
 GM / r
v 02
v0
d2x
2
2


x

0

x

A
cos(

t


),
v



A
sin(

t


),
A

x

,
tan



0
 x0
dt 2
2
mgd
g
2
k
  2 f  ;  
( spring oscillator);  
or  
( pendulum )
T
m
I
d
d2x
dx
2
t
2
2

2



x

0

x

Ae
cos(

t


),





(if    0 )
0
0
2
dt
dt
Fmax
d2x
dx
1
2
 2
 0 x 
Fmax cos( d t )  A 
2
dt
m
dt
m ( 02   d2 ) 2  4 2 d2
2
2 y
 
2
2  y
v
 0  y  A cos( t  kx)  A cos[k ( x  vt )], v   , k 
2
2
t
x
T k

F  2 A 2
I1 r22
nv
2L
F
 2 ; y SW  ASW sin(kx) sin( t ); f n 
, n 
; v
, Pav 
I 2 r1
2L
n

2
Exam Example 1 : Coin Toss
y
(problem 2.85)
y0 v0y
y vy
ay
t
0 +6m/s ? ? -g=-9.8m/s2 ?
Questions:
(a) How high does the coin go?
( 4) v
2
y
 v  2a y ( y  y0 )  y 
2
0y
v v
2
y
2a y
2
0
Vy=0
0

v0
 (6 m / s )

 1.8m
2
 2  9.8m / s
2
(b) What is the total time the coin is in the air?
v y  v0
v y  v0 y
 6m / s

t


 0.6 s
(1) a y 
2
ay
 9.8m / s
t

a
 
v  v0
Total time T= 2 t = 1.2 s
(c) What is its velocity when it comes back at y=0 ?
(4) v 2y  v 02 y  2 a y ( y  y 0 )
for y=0 and vy<0 yields vy2 = v02 → vy = -v0 = - 6m/s
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=1s
Find:
(a) average and instantaneous velocities;
0
(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;
v  x (t ) / t     t   t 2
V(t)
(b) a(t)=dv/dt= 2β +6γt; a0=2β;

v

a
x
a  ( v ( t )  v 0 ) / t  ( a 0  a ( t )) / 2  2   3 t α
2β
(c) v(ts)=0 → α+2βts +3γts2=0
0
a(t)
    2  3 
(1  6)m / s 2
ts 
 ts 
 1 .17 s
3
3
 6m / s
ts
t
Exam Example 3: Truck vs. Car (problem 2.34)

v  c o n st

x
vc 
Data: Truck v=+20 m/s
Car v0=0, ac=+3.2 m/s2
0
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.
ac
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
V(t)
vc=2v
car
v
0
v/ac
x
truck
car
truck
t=2v/ac
0
t
t=2v/ac
t
Exam Example 4: Free fall past window (problem 2.84)
Data: Δt=0.42 s ↔ h=y1-y2=1.9 m, v0y=0, ay= - g
y
Find: (a) y1 ; (b) v1y ; (c) v2y
0
V0y=0
1st solution:
(b) Eq.(3) y2=y1+v1yΔt – gΔt2/2 → v1y= -h/Δt + gΔt/2
y1
(a) Eq.(4) → v1y2 = -2gy1 →
y1 = - v1y2 /2g = -h2/[2g(Δt)]2 +h/2 – g(Δt)2 /8
y2
(c) Eq.(4) v2y2 = v1y2 +2gh = (h/Δt + gΔt/2)2
2nd solution:
h
V1y
ay
V2y
(a)Free fall time from Eq.(3): t1=(2|y1|/g)1/2 , t2=(2|y2|/g)1/2 → Δt+t1=t2
g
g(t )
1  h gt 
t  | y1 |  | y1 | h 
 t 2 g | y1 |  h  | y1 |  

2
2
2 g  t 2 
2
(b) Eq.(4) →
(c) Eq.(4) →
v1 y  2 g | y1 |
v2 y  2 g(| y1 | h)
2
Exam Example 5: Relative motion of free falling balls
(problem 2.94)
y
Data: 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.
Solution:
H
0
2
1

v0

a
(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 = v02/g
Projectile Motion
ax=0 → vx=v0x=const
ay= -g → voy= voy- gt
x = x0 + vox t
y = yo + voy t – gt2/2
v0x= v0 cos α0
v0y= v0 sin α0
tan α = vy / vx
Exam Example 6: Baseball Projectile
(examples 3.7-3.8, problem 3.12)
x0
Find: (a) Maximum height h;
0
(b) Time of flight T;
(c) Horizontal range R;
(d) Velocity when ball hits the ground
Solution:
y0 v0x v0y ax
0
?
?
Data: v0=22m/s, α0=40o
ay
x y vx vy t
0 -9.8m/s2 ? ? ?
?
?
v0x=22m/s·cos40o=+17m/s; v0y=22m/s·sin40o=+14m/s
(a) vy=0 → h = (vy2-v0y2) / (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
Exam Example 7: Ferris Wheel (problem 3.29)
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
(a) Magnitude: ac =a┴ = v2 / r
Direction to center:  r / r
  
(b) a  a ||  a   a  a ||2  a 2
tan   a || / a     tan 1 ( a || / a  )
(c) 2R  T  v  T (v 0  a||T / 2) 
T 
 v0 
v 02  4Ra||
a||
a||
2

a
θ

a||
T 2  v 0T  2R  0
Exam Example 8: Relative motion of a projectile and a target
Data: h=8.75 m, α=60o, vp0 =15 m/s,
y
(problem 3.56)
 vtx =-0.45 m/s
v p0

vt
Find: (a) distance D to the target
at the moment of shot,
(b) time of flight t,
(c) relative velocity at contact.
  
x
0
Solution: relative velocity v  v p  vt
0
v

v

v

v
cos
60
 (vtx )
(c) Final relative velocity: x
p0 x
tx
p0
Kinem aticEq.(4)  v 2py  v 2p 0 y  2 gh  v py   v 2p 0 sin 2   2 gh
(b) Time of flight t  v py / g  (v py  v p0 sin  ) / g
(a) Initial distance D  vxt
Exam Example 9:
How to measure
friction by
meter and clock?
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).
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
d) Work

a

T2 
F
 N2
Exam Example 10: Blocks on the Inclines (problem 5.92)
Data: m1, m2, μk, α1, α2, vx<0
Find: (a) fk1x and fk2x ;
(b) T1x and T2x ;
(c) acceleration ax .
m1
F
v N1

W1 X
Solution:
Newton’s second law for
α1
T1
 X
f k1

W1

f
k
2
m
W
2
α2

W2
2X
X
block 1: FN1 = m1g cosα1 , m1ax= T1x+fk1x-m1g sinα1 (1)
block 2: FN2 = m2g cosα2 , m2ax= T2 x+fk2x+ m2g sinα2 (2)
(a) fk1x= sμkFN1= sμkm1g cosα1 ; fk2x= sμkFN2= sμkm2g cosα2; s = -vx/v
(c) T1x=-T2x, Eqs.(1)&(2)→ a
x
m 1 g ( s  k cos  1  sin  1 )  m 2 g ( s  k cos  2  sin  2 )

m1  m 2
(b) T1x  T2 x  m1 (ax  sk g cos1  g sin 1 ) 
m1m2 g (sin 1  sin  2  sk (cos 2  cos1 ))
m1  m2


T  T 
T
T

T
m
Exam Example 11: Hoisting a Scaffold
a Y Data: m = 200 kg
Find:
0
(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
 

m a  5T  W
Solution
 Newton’s second law:
F(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


W  mg
(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
Exam Example 12:
The conical pendulum (example 5.20)
or a bead sliding
on a vertical hoop
(problem 5.119)

F

ac

F



F j  m ac
j

R

 a
Two equations with two unknowns: F,v
mg
mg c
2
Centripetal force along x: F sin   mac  mv / R
Equilibrium along y: F cos   mg  (a) F  mg / cos 
(b )
v 2  R ( F / m ) sin   Rg tan 
, R  L sin 
 v  Lg sin 2  / cos   sin  Lg / cos 
(c )
T  2 R / v  2 L /
Lg / cos  
T  2
L cos 
g
Exam Example 13: Stopping Distance (problems
 6.29, 7.29)
Data: v0 = 50 mph, m = 1000 kg, μk = 0.5
FN

fk
0
x
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 .

v
 
mg a
Solution:
(a)Vertical equilibrium → FN = mg → friction force fkx = - μk FN = - μk mg .
(b) Work-energy theorem → W = Kf – K0 = - (1/2)mv02 .
(c) W = fkxd = - μkmgd and (b) yield μkmgd = (1/2)mv02 → d = v02 / (2μkg) .
Another solution: second Newton’s law max= fkx= - μkmg → ax = - μkg
and from kinematic Eq. (4) vx2=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 vx2(x=d/2) = v02-μkgd = v02 /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.
Data: m, R, θ
Solution:
WT =0 always since

dl  ds  Rd
W grav


T  dl
  
  w  dl   (  w sin  ) Rd 
0
0

  wR  sin d   wR (1  cos  )
(a) Σ Fx = 0 → F = T sinθ ,
0
Σ Fy = 0 → T cosθ = w = mg , hence, F = w tanθ ; K = 0 since v=0 .



  
W F   F  d l   F cos  ds   w tan  cos  Rd   wR  sin  d   wR (1  cos  )
   0
(b) WF   F  dl   F cos Rd  FR sin 
0
0
0
0
0
K  W  WF  Wgrav  WT  R(F sin   w  w cos )
Exam Example 15: Riding loop-the-loop (problem 7.46)

g

a rad

atan D 
a
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.

Solution:
v
(a)To avoid falling off, centripetal acceleration v /R > g → v
2
2
> gR.
Conservation of energy: KB+2mgR=mgh → (1/2)mvB2=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.
Exam Example 16: Spring on the Incline (Fig.7.25, p.231)
Data: m = 2 kg, θ = 53.1o, y0 = 4 m, k = 120 N/m, μk = 0.2, v0 =0.
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,
m
(c) the highest position yf after rebound.
Solution: work-energy theorem Wnc=ΔK+ΔUgrav+ΔUel

fk
y
y0
(a)1st passage: Wnc= -y0μkmg cosθ since fk=μkFN=
yf
=μkmg cosθ, ΔK=K1 , ΔUgrav= - mgy0 sinθ, ΔUel=0
0
N
→ K1=mgy0(sinθ-μkcosθ),
ys
v1=(2K1/m)1/2 =[2gy0(sinθ–μkcosθ)]1/2
2nd passage: Wnc= - (y0+2|ys|) μkmg cosθ, ΔK=K2,
θ
ΔUgrav= -mgy0sinθ, ΔUel=0 →
K2=mgy0sinθ-(y0+2|ys|) μkmgcosθ, v2=(2K2/m)1/2
(b) (1/2)kys2 = Uel = ΔUel = Wnc – ΔUgrav = mg (y0+|ys|) (sinθ-μkcosθ) →
αys2 +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 →

F

mg
2( y0  | y s |)  k cos
2( y0  | y s |)  k
y f  y0 
 y0 
sin    k cos
tan   k
Exam Example 17: Proton Bombardment (problem 6.76)
Data: mass m, potential energy U=α/x,
238U
initial position x0>0 and velocity v0x<0.
0
 proton
v0
m
xmin

F
x0
x
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?
dU 
238
 2 0
Solution: Proton is repelled by U with a force Fx  
dx x
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
d 2x


2
dt
m x2
1 2 1 2
2
2
m v  m v0  U ( x0 )  U ( x)  v( x)  v02  U ( x0 )  U ( x)  v02 
2
2
m
m
 1 1
  
 x0 x 
2 x0
1
1 mv02
 
 xmin 
(b) Turning point: v(xmin)=0 
xmin x0 2
2  mv02 x0
(c) It is the same since the force is conservative: U(x)=U(x0) v(x)=v(x0)
Exam Example 18: The Ballistic Pendulum (example 8.8, problem 8.43)

ac L


T
T
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 vx of the bullet.
Solution:
(a) Conservation of mechanical energy K+U=const

( m  M2 ) g
y
Vtop=0

v
1
1
2
K 0  U to p  ( m  M ) v  ( m  M ) gy  ( m  M ) v to p  v  2 gy
2
2


(b) Newton’s second law ( m  M ) a  ( m  M ) g  2 T an d a  a c  v 2 / L
2

 (m  M ) g 
m

M
v
2y
yields T 
 g 
 
1 

2
L 
2
L 


(c) Momentum conservation for the collision
mM
mM
mv x  ( m  M ) v  v x 
v
m
m
2 gy
Exam Example 19: Collision of Two Pendulums
Exam Example 20: Head-on elastic collision (problems 8.48, 8.50)
Data: m1, m2, v01x, v02x
Find: (a) v1x, v2x after collision;
(b) Δp1x, Δp2x , ΔK1, ΔK2 ;
(c) xcm at t = 1 min after collision
if at a moment of collision xcm(t=0)=0
Solution: In a frame of reference moving
V01x
y’
V02x
m1
m2
X’
with V02x, we have V’01x= V01x- V02x, V’02x = 0, and
conservations of momentum and energy yield
m1V’1x+m2V’2x=m1V’01x → V’2x=(m1/m2)(V’01x-V’1x)
0
m1V’21x+m2V’22x=m1V’201x→ (m1/m2)(V’201x-V’21x)=V’22x = (m1/m2)2(V’01x- V’1x)2 →
X
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)
(a) Another solution: In 1-D elastic collision a relative velocity switches direction
V2x-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-V201x)m1/2, ΔK2=K2-K02=(V22x-V202x)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 8.86)
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 y
m1v1
1
m1v1  (m1  m2 )v  v 
 v1
m1  m2 3
Conservation of energy: (i) for mass m1 on the
m1

v1

v
h
way to the bottom just before the collision
1
1
x
2
m2
m1 gR  m1v1  v1  2 gR  v 
2 gR
2
3
(ii) for the stuck together masses on the way from the bottom to the top
(b)
2
2
1
v
v
R
2
1
(m1  m2 )v  (m1  m2 ) gh  h 


2
2 g 18g 9
Exam Example 22: Throwing a Discus (example 9.4)
Exam Example 23: Blocks descending over a massive
pulley (problem 9.83)


Data: m1, m2, μk, I, R, Δy, v0y=0
Find: (a) vy; (b) t, ay; (c) ω,α; (d) T1, T2 m1
FN 1
Solution: (a) Work-energy theorem

m1 g
Wnc= ΔK + ΔU, ΔU = - m2gΔy,

fk
Wnc = - μk m1g Δy , since FN1 = m1g,

T1
ΔK=K=(m1+m2)vy2/2 + Iω2/2 = (m1+m2+I/R2)vy2/2 since vy = Rω 0
2 gy(m2   k m1 )
1
I  2
 m1  m2  2 v y  gy(m2   k m1 )  v y 
2
R 
m1  m2  I / R 2

 T1
ωF
R

T2
T2
x
ay
m2
Δy 
y m2 g
(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


 Fi  Macm and
i

iz
 I cm z
i
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, 10.75)
Find: (a) ay, (b) vcm, (c) T
Mg-T=May
τz=TR=Icmαz
y
ay
ay=2g/3 , T=Mg/3
4 gy
v cm  2 ay 
3
Exam Example 25: Race of Rolling Bodies (examples 10.5, 10.7; problem 10.22,
y
Data: Icm=cMR2, h, β problem 10.29)
Find: v, a, t, and min μs
FN
preventing from slipping
β
 
v a
Solution 1: Conservation of Energy Solution 2: Dynamics
K1  U1  K 2  U 2 , K1  0, U 2  0  (Newton’s 2nd law) and
rolling kinematics a=Rαz
1
1
1
Mv 2  I 2  (1  c ) Mv 2
2
2
2
I  cMR2 and   v / R
Mgh 
for
2 gh
v
1 c
v2=2ax
x
fs
g sin 
 F x  Mg sin   f s  Ma  a  1  c
g sin    z  f s R  I cm z  cMRa  f s  cMa
a
1  c v  2 ax  2 g sin  h  2 gh
1  c sin 
1 c
x 2x
1
t 

v
v
sin 
2 h (1  c )
g
fs
c
c Mg sin  c tan 
f s  Mg sin   Ma 
Mg sin   min  s 


1 c
F N 1  c Mg cos 
1 c
Equilibrium, Elasticity, and Hooke’s Law
Conditions for equilibrium:
Exam Example 26: Ladder against wall



(1)  Fi  0  a  0 (example 11.3, problem 11.10)
F1
i


( 2)    0    0

Static equilibrium:
v0

i
State with v  c o n st  0
is equilibrium but is not static.
d/2

F2 x
y
Data: m, M, d, h, y, μs
Find: (a) F2, (b) F1, fs,

fs

Mg

mg
h
d
θ
Strategy of
problem
solution:

(c) yman when ladder starts to slip
(0)
F 0
Solution: equilibrium equations yield
(i) Choice of the axis of rotation: (a) F2= Mg + mg ; (b) F1 = fs
arbitrary - the simpler the better. Choice of B-axis (no torque from F2 and fs)
(ii) Free-body diagram:
F1h = mgx + Mgd/2 → F1= g(mx+Md/2)/h = fs
identify all external forces and
(c) Ladder starts to slip when μsF2 = fs, x = yd/h
their points of action.
→ μsg (M+m) = g (mymand/h+Md/2)/h →
(iii) Calculate lever arm and
 s ( M  m ) h 2 Mh  s h 2 Mh 2 
d 
y man 



s  
torque for each
 force.
md
2m
d
md 
2h 
(iv) Solve   0 for unknowns.


Exam Example 27: Motion in the gravitational field of two bodies
(problem 13.62)
M
1

F1
0
m
X
0
X
1
M
ΔX=X-X
0

F2
X
2
X
2
Data: masses M , M , m; positions x , x , x , x; v(t=0)=0
1 2
1 2 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.
 1
1

 x x
 0 1 x  x1
Solution: (a) U  U  U 0  GM 1m

 1
1
  GM 2 m


 x x

 0 2 x  x2




1 2

K



U

mv  U  v   2U / m
(b) Energy conservation:
2
 M2
nd
M1 


(c) Newton’s 2 law: a x  F2 x  F1x  / m  G
2
2 
 x2  x  x1  x  

v

Exam Example 28: Satellite in a Circular Orbit
Data: r = 2RE , RE = 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.
r
Solution: use the value g = GME/RE2 = 9.8 m/s2
(a) The only centripetal force is the gravitational force:
ac
GM E m mv 2
Newton' s 2 nd law  F g  ma c 

v
2
r
r

GM E R E2

2
RE r
R E2
g

r
m


F g ra v  F c
ME
RE=6380 km
GM E

r
gR E
 9.8 m / s 2  3.19 10 6 m  5.6 km / s
2
(b) The period T is a time required for one orbital revolution, that is
 r 
2 r
2 r
2 r
2 r
T



 2  
v
GM E
GM E
GM E
 RE 
RE
r
RE2
3/ 2
3/ 2
3/ 2
6
RE
6
.
38

10
m
 2 23 / 2
 4h
2
g
9.8 m / s
(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 13.77)
Data: hp , ha , RE= 6380 km, ME= 6·1024 kg
Find: (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;
hp 2RE
(f) escape speeds at perigee v2p and at apogee v2a.
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)
 (apogee)
a rad 
(perigee)

v p
v2 p

Fgrav
va
ha
(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., T
(c) Newton’s 2nd law and law of gravitation: arad= Fgrav/ m = GME/r2.
2a 3 / 2

GM E
(d) Conservation of angular momentum (La= Lp) or Kepler’s second law:
rava= rpvp, vp/va= ra/rp
(e) Conservation of mechanical energy K + U = const :
mv 2p GM E m mv a2 GM E m mv 2p r p2 GM E m
2GM E ra v  v r p  2GM E r p





 vp 
; a
p
2
ra
ra ( r p  ra )
2
rp
2
ra
2 ra
ra
r p ( r p  ra )
(f) Conservation of mechanical energy for an escape from a distance r
(the second space speed) :
mv 22 G M E m

 v2 
2
r
2G M E
 v2 p 
r
2G M E
,
rp
v 2a 
2G M E
ra
Exam Example 30: A Ball Oscillating on a Vertical Spring (problems 14.38, 14.83)
Data: m, v0 , k
y
Unstrained→
Equilibrium
y2=y0+
A
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.
0 v
0
y0
Solution: Fy = - ky
Lowest position y1=y0-A v =
1
0
(a) Equation of equilibrium:
Fy – mg = 0, -ky0 = mg , y0 = - mg/k
(b) Conservation of total mechanical energy
1
1 2 1
2
E  K  U g ra v  U ela stic  E 0  mv y  mgy 0  ky 0  mv 02
2
2
2
v y   v 02  y 0 ( 2 g  ky 0 / m )   v 02  m g 2 / k
(c) At the extreme positions y1,2 = y0 ± A velocity is zero and
2
2
1 2
1
mg
 mg  mv 0
2
mgy 1, 2  ky1, 2  mv 0  y1, 2  
 
 A
 
2
2
k
k
k


(d)  
k
2
, T 
 2
m

m
k
mv 02 mg
kv 02
y 

1
k
k
mg 2
2
0
Applications of the Theory of Harmonic Oscillations
Oscillations of Balance Wheel in a Mechanical Watch
Newton’s 2nd law for rotation yields
d 
I z   z ,  z   ,
  0
2
dt
I
2

    cos(t   ) ,   2 f 

T
I
2
(mass m)
R
Exam Example 31: SHM of a thin-rim balance wheel (problems 14.41,14.97)
Data: mass m, radius R , period T
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.
T  2 I /     m(2R / T )2
Exam Example 32: Physical Pendulum (problem 14.99, 14.54)
0
X
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;
m

Mg
(b) equilibrium position of the object’s center of mass;
(c) derive harmonic oscillator equation for small
deflection angle starting from Newton’s 2nd law for rotation;
(d) angular frequency and period of oscillations.
θ d m
cm
y
L

Solution: (a) dm = m dx/L , I 1  ( m / L ) x 2 dx  (1 / 3) mL 2
0
(b) geometry and definition xcm=(m1x1+m2x2)/(m1+m2)→ ycm= d= 2-3/2 L, xcm=0
(c) Iαz = τz , τz = - 2mg d sinθ ≈ - 2mgd θ
(d) Object’s moment of inertia I  2I1 
d 2
2m gd
2
   0 ,  
2
dt
I
2 2
2mgd
3g
2
mL   

,T 
3
I

2 2L
Exam Example 33:
Sound Intensity and Delay
A rocket travels straight up
with ay=const to a height r1
and produces a pulse of sound.
A ground-based monitoring
station measures a sound
intensity I1. Later, at a height
r2, the rocket produces the
same second pulse of sound,
an intensity of which measured
by the monitoring station is I2.
Find r2, velocities v1y and v2y of
the rocket at the heights r1 and
r2, respectively, as well as the
time Δt elapsed between
the two measurements.
(See related problem 15.25.)
Exam Example 34: Wave Equation and Transverse Waves
on a Stretched String (problems 15.51 – 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 2nd 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=x0, t=t0) = A;
(c) find a wave number k and a wave speed v;
y A
L
(d) find a wave period T and an angular frequency ω;
0
X
(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);
(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) vy2
Pav = Fω2A2 /(2v) =(1/2)(μF)1/2ω2A2.
(a) Derivation of the wave equation: y(x,t) is a transverse displacement.
Restoring force exerted on the segment
F is a tension force.
Δx of spring:
μ = Δm/Δx is a linear mass
 y
y 
density (mass per unit length).
Slope = -F1y/F=∂y/∂x
Slope=
F2y/F=∂ y/∂x
 
 
Fy  F2 y  F1 y  F  
  
 x  x  x  x  x 
Newton’s 2nd law: μΔx ay= Fy , ay= ∂2y/∂t2
2
2 y
2  y
 Wave Equation 2  v
0,v 
2
t
x
F
