final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
Mechanics - Basic Physical Concepts
3
Math: Circle: 2 π r, π r2 ; Sphere: 4 π r2 , (4/3)
√ πr
2 −4 a c
−b±
b
2
Quadratic Eq.: a x + b x + c = 0, x =
2a
Cartesian and polar coordinates:
y
x = r cos θ, y = r sin θ, r2 = x2 + y 2 , tan θ = x
Trigonometry: cos α cos β + sin α sin β = cos(α − β)
α−β
sin α + sin β = 2 sin α+β
2 cos 2
α−β
cos α + cos β = 2 cos α+β
2 cos 2
sin 2 θ = 2 sin θ cos θ, cos 2 θ = cos2 θ − sin2 θ
1 − cos θ = 2 sin2 2θ , 1 + cos θ = 2 cos2 2θ
~ = (Ax , Ay ) = Ax ı̂ + Ay ̂
Vector algebra: A
~ =A
~+B
~ = (Ax + Bx , Ay + By )
Resultant:
R
~·B
~ = A B cos θ = Ax Bx + Ay By + Az Bz
Dot: A
Cross product: ı̂ × ̂ = k̂, ̂ × k̂ = ı̂, k̂ × ı̂ = ̂
¯
¯
¯ ı̂
̂
k̂ ¯
¯
¯
~
~
~
¯
C = A × B = ¯ Ax Ay Az ¯¯
¯
¯ B
x By Bz
C = A B sin θ = A⊥ B = A B⊥ , use right hand rule
d xn = n xn−1 ,
d
1
Calculus: dx
dx ln x = x ,
d sin θ = cos θ,
d cos θ = − sin θ,
d
dθ
dθ
dx const = 0
Measurements
Dimensional analysis: e.g.,
2
F = m a → [M ][L][T ]−2 , or F = m vr → [M ][L][T ]−2
PN
PN
Summation:
i=1 (a xi + b) = a i=1 xi + b N
Motion
One dimensional motion: v = ddts , a = ddtv
s −s
v −v
Average values: v̄ = tff −tii , ā = tff −tii
One dimensional motion (constant acceleration):
v(t) : v = v0 + a t
s(t) : s = v̄ t = v0 t + 12 a t2 , v̄ = v02+v
v(s) : v 2 = v02 + 2 a s
Nonuniform acceleration: x = x0 + v0 t + 12 a t2 +
1 j t3 + 1 s t4 + 1 k t5 + 1 p t6 + . . ., (jerk, snap,. . .)
6
24
120
720
ttrip
v0y
2 = g
1
2
h = 2 g tf all , R = vox ttrip
2
Circular: ac = vr , v = 2 Tπ r , f = T1 (Hertz=s−1 )
q
Curvilinear motion: a = a2t + a2r
Relative velocity: ~v = ~v ′ + ~u
Projectile motion: trise = tf all =
Law of Motion and applications
Force: F~ = m ~a, Fg = m g, F~12 = −F~21
2
Circular motion: ac = vr , v = 2 Tπ r = 2 π r f
Friction: Fstatic ≤ µs N
Fkinetic = µk N
P
Equilibrium (concurrent forces):
~
i Fi = 0
Energy
Work (for all F): ∆W = WAB = WB − WA
1
RB
F~ · d~s (in Joules)
Fk s = F s cos θ = F~ · ~s → A
Effects due to work done: F~ext = m ~a − F~c − f~nc
Wext |A→B = KB − KA + UB − UA + Wdiss |A→B
RB
Kinetic energy: KB −KA = A
m ~a ·d~s, K = 12 m v 2
R
K (conservative F~ ): U − U = − B F~ · d~s
B
A
A
Ugravity = m g y,
Uspring = 12 k x2
∂
U
From U to F~ : Fx = − ∂x , Fy = − ∂∂yU , Fz = − ∂∂zU
Fgravity = − ∂∂yU = −m g,
U = −k x
Fspring = − ∂∂x
2
∂
U
∂
U
Equilibrium: ∂x = 0, ∂x2 > 0 stable, < 0 unstable
W = F v = F v cos θ = F
~ · ~v (Watts)
Power: P = ddt
k
Collision
Rt
Impulse: I~ = ∆~
p = p~f − p~i → tif F~ dt
Momentum: p~ = m ~v
x1 +m2 x2
Two-body: xcm = m1m
1 +m2
pcm ≡ M vcm = p1 + p2 = m1 v1 + m2 v2
Fcm ≡ F1 + F2 = m1 a1 + m2 a2 = M acm
K1 + K2 = K1∗ + K2∗ + Kcm
Two-body collision: p~i = p~f = (m1 + m2 ) ~vcm
vi′ = vi∗′ + vcm
vi∗ = vi − vcm ,
Elastic: v1 − v2 = −(v1′ − v2′ ),
vi∗′ = −vi∗ , vi′ = 2 vcm − vi
R
P
~r dm
m ~r
Many body center of mass: ~rcm = P i i = R
mi
mi
P
p
Force on cm: F~ext = d~
=
M~
a
,
p
~
=
p
~
cm
i
dt
Rotation of Rigid-Body
Kinematics: θ = rs , ω = vr , α = art
R
P
Moment of inertia: I =
mi ri2 = r2 dm
1
1
2
Idisk = 2 M R , Iring = 2 M (R12 + R22 )
1 M ℓ2 , I
1
2
2
Irod = 12
rectangle = 12 M (a + b )
Isphere = 25 M R2 , Ispherical shell = 23 M R2
I = M (Radius of gyration)2 , I = Icm + M D2
Kinetic energies: Krot = 12 I ω 2 , K = Krot + Kcm
Angular momentum: L = r m v = r m ω r = I ω
Torque: τ = ddtL = m ddtv r = F r = I ddtω = I α
Wext = ∆K +∆U +Wf ,
K = Krot + 12 m v 2 ,
P =τω
Rolling, angular
momentum
and
´
³
´
³ torque
Ic + M v 2
Rolling: K = 12 Ic + M R2 ω 2 = 12 R
2
~
Angular momentum: L = ~r × p~, L = r⊥ p = I ω
~
Torque: ~τ = d L = ~r × d~p = ~r × F~ , τ = r F = I α
dt
dt
1 dL = τ = mgh
Gyroscope: ωp = ddtφ = L
L
Iω
dt
Static equilibrium
P
P~
~τi = 0
Fi = 0, about any point
+mB ~rBcm
Subdivisions: ~rcm = mA ~rAcm
mA +mB
Elastic modulus = stress/strain
stress: F/A
strain: ∆L/L, θ ≈ ∆x/h, −∆V /V
⊥
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
Gravity
F~21 = −G m12m2 r̂12 ,
r12
for r ≥ R,
g(r) = G M
r2
G = 6.67259 × 10−11 N m2 /kg2
Rearth = 6370 km, Mearth = 5.98 × 1024 kg
³ ´2
2
Circular orbit: ac = vr = ω 2 r = 2Tπ r = g(r)
U = −G mrM ,
M
E = U + K = −Gm
2r
r0
r2 = 1−ǫ
Fluid mechanics
Pascal: P = FA⊥1 = FA⊥2 , 1 atm = 1.013 × 105 N/m2
1
2
Archimedes: B = M g, Pascal=N/m2
P = Patm + ρ g h, with P = FA⊥ and ρ = m
V
R
R
F = P dA −→ ρ g ℓ 0h (h − y) dy
Continuity equation: A v = constant
Bernoulli: P + 21 ρ v 2 + ρ g y = const,
P ≥0
Oscillation motion
f = T1 , ω = 2Tπ
2
2
S H M: a = ddt2x = −ω 2 x, α = ddt2θ = −ω 2 θ
x = xmax cos(ω t + δ), xmax = A
v = −vmax sin(ω t + δ), vmax = ω A
a = −amax cos(ω t + δ) = −ω 2 x, amax = ω 2 A
E = K + U = Kmax = 12 m (ω A)2 = Umax = 21 k A2
Spring: m a = −k x
Simple pendulum: m aθ = m α ℓ = −m g sin θ
Physical pendulum: τ = I α = −m g d sin θ
Torsion pendulum: τ = I α = −κ θ
Wave motion
Traveling waves: y = f (x − v t), y = f (x + v t)
In the positive x direction: y = A sin(k x − ω t − φ)
λ
T = f1 , ω = 2Tπ , k = 2λπ , v = ω
k = T
q
Along a string: v = F
µ
fixed end: phase inversion
open end: same phase
General: ∆E = ∆K + ∆U = ∆Kmax
1 ∆m
2
P = ∆E
∆t = 2 ∆t (ωA)
∆m ∆x
∆m
Waves: ∆m
∆t = ∆x · ∆t = ∆x · v
P = 21 µ v (ω A)2 , with µ = ∆m
∆x
∆m ∆A ∆r
∆m
Circular: ∆m
∆t = ∆A · ∆r · dt = ∆A · 2 π r v
∆m
2
Spherical: ∆m
∆t = ∆V · 4 π r v
v=
q
Sound
B,
ρ
s = smax cos(k x − ω t − φ)
1
2
Intensity: I = P
A = 2 ρ v (ω smax )
I
Intensity level: β = 10 log10 I , I0 = 10−12 W/m2
0
Plane waves: ψ(x, t) = c sin(k x − ω t)
Circular waves: ψ(r, t) = √c sin(k r − ω t)
Spherical: ψ(r, t) = rc sin(k r − ω t)
L
⊥
⊥
ii) L = r m ∆r
−→ ∆A = 21 r ∆r
const.
∆t = 2 m =
³ 2´
³∆t ´2 ∆t
r
1
4π
2
π
a
M
3
2
1 +r2
iii) G a2 =
a, a =
2 , T = GM r
T
Escape kinetic energy: E = K + U (R) = 0
Reflection of wave:
∂s
∆P = −B ∆V
V = −B ∂x
∆Pmax = B κ smax = ρ v ω smax
∆m A ∆x
Piston: ∆m
∆t = ∆V · ∆t = ρ A v
r
2
F = − ddrU = −m G M
= −m vr
r2
Kepler’s Laws of planetary motion:
r0
0
i) elliptical orbit, r = 1−ǫrcos
θ r1 = 1+ǫ ,
2
′
v
Doppler effect: λ = v T , f0 = T1 , f ′ = λ
′
′
Here v = vsound ± vobserver , is wave speed relative
to moving observer and λ′ = (vsound ± vsource )/f0 ,
detected wave length established by moving source of
frequency f0 . freceived = fref lected
Shock waves: Mach Number= vvsource
= sin1 θ
sound
Superposition of waves
Phase difference: sin(k x − ωt) + sin(k x − ω t − φ)
Standing waves: sin(k x − ω t) + sin(k x + ω t)
Beats: sin(kx − ω1 t) + sin(k x − ω2 t)
Fundamental modes: Sketch wave patterns
λ
String: λ
2 = ℓ, Rod clamped middle: 2 = ℓ,
Open-open pipe: λ
2 = ℓ,
Open-closed pipe: λ
4 =ℓ
Temperature and heat
Conversions: F = 95 C + 32◦ ,
K = C + 273.15◦
Constant volume gas thermometer: T = a P + b
Thermal expansion: α = 1ℓ ddTℓ , β = V1 ddTV
∆ℓ = α ℓ ∆T , ∆A = 2 α A ∆T , ∆V = 3 α V ∆T
Ideal gas law:
P V = nRT = N kT
R = 8.314510 J/mol/K = 0.0821 L atm/mol/K
k = 1.38 × 10−23 J/K, NA = 6.02 × 1023 , 1 cal=4.19 J
Calorimetry: ∆Q = c m ∆T, ∆Q = L ∆m
R
First law: ∆U = ∆Q − ∆W , W = P dV
−H ℓi
∆T
Conduction: H = ∆Q
∆t = −k A ∆ℓ , ∆Ti = A ki
Stefan’s law: P = σ A e T 4 , σ = 5.67 × 10−8 m2WK 4
Kinetic theory of gas
2
m vx
x
F = ∆p
∆t = d
N 2
Pressure: P = NAF = mVN vx2 = m
3V v
K
1
P = 23 N
V K, K x = 3 = 2 k T , T = 273 + Tc ,
P V = N k T , n = N/NA , k = 1.38×10−23 J/K,
NA = 6.02214199 × 1023 #/kg/mole
Constant V:
∆Q = ∆U = n CV ∆T
Constant P:
∆Q = n CP ∆T
Ideal gas: ∆px = 2 m vx ,
C
γ = CP , CP − CV = R
V
CV = d2 R, for transl.+rot+vib, d = 3 + 2 + 2
Adiabatic expansion: P V γ = constant
t
1
Mean free path: ℓ = (v vrms
= √ 12
)
t π d2 n
rel rms
V
2 π d nV
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
Question 1, chap 13, sect 4.
part 1 of 1
10 points
~ = (2 N)ı̂ + (3.4 N) ̂ is applied
A force F
to an object that is pivoted about a fixed axis
aligned along the z coordinate axis. The force
is applied at the point ~r = (4.1 m)ı̂+(5.5 m) ̂.
Find the z-component of the net torque.
Correct answer: 2.94.
Given: The radius of the earth is REarth =
6.37 × 106 m .
How high does a rocket have to go above the
Earth’s surface until its weight is 0.36 times
its weight on the Earth’s surface?
Correct answer: 4246.67.
Explanation:
Basic concepts
By Newton’s Universal Law of Gravitation,
Explanation:
Basic Concept:
W ∝
~
~τ = ~r × F
= [(4.1 m) (3.4 N) − (5.5 m) (2 N)]k̂
= (2.94 N m) k̂
In this case, the net torque only has a zcomponent.
REarth = 6.37 × 106 m .
The rocket will be at a distance of h + REarth
from the center of the Earth when it weighs
n W . Thus
1
2
REarth
nW
(h + REarth)2
=
=
1
W
(h + REarth)2
2
REarth
and
2
n (h + REarth)2 = REarth
Question 2, chap 14, sect 4.
part 1 of 1
10 points
1.7 m3 of concrete weighs 61000 N. The
compression strength of concrete is 1.3 ×
107 N/m2 .
What is the height of the tallest cylindrical
pillar, made from this amount of concrete,
that will not collapse under its own weight?
Correct answer: 362.295.
Explanation:
Stress ≡
So
S=
And
force
area
W
W
Wh
=
=
A
V /h
V
SV
h=
W
Question 3, chap 9, sect 99.
part 1 of 1
10 points
1
r2
Solution: The radius of the Earth is
Solution: From the definition of torque
~
~τ = ~r × F
= [xı̂ + y ̂] × [Fx ı̂ + Fy ̂]
= [(4.1 m)ı̂ + (5.5 m) ̂] × [(2 N)ı̂ + (3.4 N) ̂]
3
2
REarth
n
REarth
h + REarth = √
n
(h + REarth)2 =
REarth
√
− REarth
n
1
= REarth √ − 1.0
n
1
6
= 6.37 × 10 m √
− 1.0
0.36
= 4246.67 km .
h=
Question 4, chap 13, sect 99.
part 1 of 1
10 points
A thin hollow cylindrical rod has length
2.6 m, mass 0.05 kg, and diameter 0.3 cm .
The thin cylindrical rod is free to rotate about
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
0.3 cm
0.05 kg
when ω = 0
1m
Determine the value of the spring constant
k if the masses are located at 1 m from the
center of the cylinder when the cylindrical rod
rotates at 20 rad/s .
Correct answer: 720.
Explanation:
Let : Lc = 2.6 m ,
Rc = negligible ≪ Lc ,
m = 0.9 kg ,
M = 0.05 kg ,
x = 0.5 m ,
x + ∆x = 1 m ,
∆x = (1 m) − (0.5 m)
= 0.5 m , and
ω = 20 rad/s .
Applying Newton’s second law, the net inward force acting on each of the 0.9 kg masses
is
X
Fradial :
k ∆x = m (x + ∆x) ω 2 , thus
m (x + ∆x) ω 2
k=
∆x
(0.9 kg) (1 m) (20 rad/s)2
=
(0.5 m)
A uniform flat plate of metal is situated in
the reference frame shown in the figure below.
5
b
4
y
a vertical axis that passes through its center
and is perpendicular to the cylinder’s axis.
Inside the cylinder are two masses of 0.9 kg
each, attached to springs of spring constant k
and unstretched lengths 0.5 m . The springs
have negligible mass. The inside walls of the
cylinder are frictionless.
Assume the diameter of the cylindrical rod
is much smaller than its length.
2.6 m
0.9 kg
0.9 kg
4
3
2
1
0
0 1
2 3
b
4 5 6
x
7 8
Calculate the x coordinate of the center of
mass of the metal plate.
Correct answer: 8.
Explanation:
Basic Concept: The center of mass coordinate is
Z
x dm
,
x≡
m
Z
where m ≡ dm , and dm = σ y dx , where σ
mass of the plate.
is the areal density
area
Solution: Let
(x1 , y1 ) = (4, 0)
(x2 , y2 ) = (10, 0)
(x3 , y3 ) = (10, 4) .
The equation for the hypotenuse is
y − y1
y3 − y1
=
.
x − x1
x3 − x 1
The slope of the hypotenuse is
y3 − y1
x3 − x 1
2
4−0
=+ .
=
10 − 4
3
s=
Rewriting the equation, we have
= 720 N/m .
Question 5, chap 10, sect 2.
part 1 of 1
10 points
b
9 10
y = s (x − x1 ) + y1
2
(x − 4) + 0 .
= +
3
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
The x-coordinate of the center of mass is
Z x2
x y dx
σ
x1
x = Z x2
y dx
σ
x1
Z x2
x s (x − x1 ) dx
x1
= Z x2
s (x − x1 ) dx
x1
Z x2
x (x − x1 ) dx
x1
= Z x2
(x − x1 ) dx
x1
x2
1 3 1
2 x − (x1 ) x
2
= 3
1 2
x − (x1 ) x 2
x1
1 3
1
(x2 − x31 ) − (x1 ) (x22 − x21 )
3
2
=
1 2
2
(x − x1 ) − (x1 ) (x2 − x1 )
2 2
3 x1 (x22 − 2 x1 x2 + x21 )
=
3 (x22 − 2 x1 x2 + x21 )
2 (x2 − x1 ) (x22 − 2 x1 x2 + x21 )
+
3 (x22 − 2 x1 x2 + x21 )
2
= x1 + (x2 − x1 )
(1)
3
2
= 4 + (10 − 4)
3
= 8.
Alternate solution: The center of mass
1
of any triangle is of the height of the triangle
3
measured from its base. Therefore Eq. 1 is the
x-coordinate of the center of mass of the metal
plate.
The y-coordinate of the center of mass of
the metal plate is
1
(y3 − y2 )
3
1
= 0 + (4 − 0)
3
= 1.33333 .
y = y2 +
Note: This problem has a different triangle
for each student.
5
Question 6, chap 17, sect 3.
part 1 of 1
10 points
A vertical tube 1.12 m long is open at the
top. It is filled with 27 cm of water.
If the speed of sound is 344 m/s, what will
the fundamental resonant frequency be?
Correct answer: 101.176.
Explanation:
Let : lt = 1.12 m ,
lw = 27 cm = 0.27 m ,
v = 344 m/s .
and
The frequency of the nth harmonic in an air
column is given by
fn = n
v
4L
(n = 1, 3, 5, ...) .
n = 1 gives the fundamental frequency and
the height of the air column is the length of the
tube minus the length of the water column, so
f1 =
v
4L
v
4 (lt − lw )
344 m/s
=
4 (1.12 m − 0.27 m)
=
= 101.176 Hz .
Question 7, chap 7, sect 2.
part 1 of 1
10 points
To stretch a certain nonlinear spring by an
amount x requires a force F given by
F = 40 x − 6 x2 ,
where F is in Newtons and x is in meters.
What is the change in potential energy ∆U
when the spring is stretched 2 meters from its
equilibrium position?
1. ∆U = 56 J
2. ∆U = 64 J correct
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
6
agains a frictionless vertical wall. The angle
between the ladder and the horizontal is 59◦ .
3. ∆U = 80 J
4. ∆U = 16 J
5.8 m
5. ∆U = 28 J
From conservation of energy, this is the
change in potential energy.
Question 8, chap 16, sect 4.
part 1 of 1
10 points
A stretched string is fixed at both ends,
494.1 cm apart.
If the density of the string is 0.037 g/cm and
its tension is 600 N, what is the wavelength of
the first harmonic?
Correct answer: 988.2.
Explanation:
For a string with both ends fixed, its wavelength of the nth harmonic is
λn =
2d
,
n
which doesn’t depend on its other properties.
So the wavelength of its first harmonic is 2 d .
Question 9, chap 14, sect 2.
part 1 of 1
10 points
A 69 kg (676.2 N) person climbs up a
uniform 12 kg (117.6 N) ladder. The ladder is 5.8 m long; its lower end rests on a
rough horisontal floor (static friction coefficient µ = 0.28) while its upper end rests
b
µ=0
Explanation:
The work required for the spring to move 2
meters from its equilibrium position is
Z 2m
∆U =
F · dx
0
Z 2m
=
(40 x − 6 x2 ) · dx
0
2 m
= 20 x2 − 2 x3 0
2
= 20 (2 m) − 0 − 2 (2 m)3 − 0
= 64 J .
59◦
d
b
117.6 N
676.2 N
µ = 0.28
Note: Figure is not to scale.
Let d denote the climbing person’s distance
from the bottom of the ladder (see the above
diagram). When the person climbs too far
(d > dmax ), the ladder slips and falls down
(kaboom!).
Calculate the maximal distance dmax the
person will reach before the ladder slips.
Correct answer: 2.66849.
Explanation:
Let : θ = 59◦ ,
L = 5.8 m ,
Wℓ = 117.6 N ,
Wp = m g = 676.2 N ,
µ = 0.28 .
and
The ladder and the person, remain in static
equilibrium up until the critical distance dmax
where the ladder begins to slide. At this point,
the static friction reaches its maximum magnitude f = fmax , and any further displacement will cause the bottom of the ladder to
begin to slide. At any distance d ≤ dmax the
forces must balance and the torques must balance since the ladder is in static equilibrium.
The relevant forces acting on the ladder are
shown in the figure below.
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
Nw
7
maximal height dmax follows from the maximal magnitude of the static friction force,
thus
Wp dmax
Wℓ
+
.
2 tan θ
L tan θ
(6)
Solving this equation for the dmax , we finally
have
h
i
L
Wℓ
dmax =
µ Wℓ + Wp tan θ −
Wp
2
5.8 m
×
=
676.2
n Nh
i
fmax = µ (Wℓ + Wp ) =
b
Wℓ
f
θ
mg
b
P ivot
Nf
The forces in both the x and y directions
must balance separately to maintain equilibrium. Also, since the system is in equilibrium,
we can choose any point as the origin to define the net torque on the ladder. We will
choose the bottom of the ladder as this point
because it will simplify the calculation (there
are two forces acting at the bottom of the
ladder–if we choose the bottom of the ladder
to define the moment arms, these forces will
have a zero moment arm, so they will not enter into the torque balance equation). Thus,
at any distance d ≤ dmax we have
X
Fx =f − Nw = 0 ,
(1)
X
Fy =Nf − Wℓ − Wp = 0 , and (2)
X
L
τ◦ =Wp d cos θ + Wℓ cos θ
(3)
2
− Nw L sin θ = 0 ,
where f is the static friction force and Nf and
Nw are the normal forces from respectively
the floor and the wall. Solving for all the
forces we have:
Nf = Wℓ + Wp
(4)
Wp d
Wℓ
+
.
(5)
f = Nw =
2 tan θ L tan θ
The maximum magnitude of force that
static friction can exert is f = fmax = µNf .
According to eq. (5), the static friction increases as d increases. Consequently, the
× 0.28 117.6 N + 676.2 N tan 59◦
117.6 N o
−
2
= 2.66849 m .
Question 10, chap 17, sect 4.
part 1 of 1
10 points
Standing at a crosswalk, you hear a frequency of 556 Hz from the siren on an approaching police car. After the police car
passes, the observed frequency of the siren is
473 Hz. Assume the speed of sound in air to
be 343 m/s.
Determine the car’s speed from these observations.
Correct answer: 27.6667.
Explanation:
Let : f ′ =556 Hz ,
f ′′ =473 Hz , and
v =343 m/s .
When the police car is approaching, the
observed frequency is
f′ =
f
1−
vs .
v
After the police car passes, the observed frequency is
f
f ′′ =
vs .
1+
v
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
Thus
vs
1+
f′
v = v + vs
=
v
′′
s
f
v − vs
1−
v
f ′ (v − vs ) = f ′′ (v + vs )
f ′ − f ′′
vs = ′
v
f + f ′′
556 Hz − 473 Hz
=
(343 m/s)
556 Hz + 473 Hz
= 27.6667 m/s .
disk is carefully lowered onto a horizontal surface and released at time t0 with zero initial
linear velocity along the surface. The coefficient of friction between the disk and the
surface is 0.01 .
The kinetic friction force between the surface and the disk slows down the rotation of
the disk and at the same time gives it a horizontal acceleration. Eventually, the disk’s
linear motion catches up with its rotation,
and the disk begins to roll (at time trolling )
without slipping on the surface.
The acceleration of gravity is 9.8 m/s2 .
keywords: Doppler effect
Question 11, chap 4, sect 5.
part 1 of 1
10 points
8
6 kg
20 cm , radius
4 rad/s
2
I= m R2
3
A turntable is designed to acquire an angular velocity of 45 rev/s in 0.7 s, starting from
rest.
Find the average angular acceleration of the
turntable during the 0.7 s period.
Correct answer: 403.919.
Once the disk rolls without slipping, what
is its angular speed?
Correct answer: 1.6.
Explanation:
Explanation:
µ = 0.01
ω = 45 rev/s = 282.743 rad/s
Let : r = 20 cm = 0.2 m ,
ω0 = 4 rad/s ,
m = 6 kg , and
µ = 0.01 .
We have:
∆ω
∆t
282.743 rad/s
=
0.7 s
= 403.919 rad/s2
α=
From the perspective of the surface, let the
speed of the center of the disk be vsurf ace .
Using the frictional force f , we can determine
the acceleration
Question 12, chap 13, sect 2.
part 1 of 1
10 points
Assume: When the disk lands on the surface it does not bounce.
The disk has mass 6 kg and outer radius
20 cm with a radial mass distribution (which
may not be uniform) so that its moment of
2
inertia is m R2 .
3
The disk is rotating at angular speed
4 rad/s around its axis when it touches the
surface, as shown in the figure below. The
f = µmg,
X
and
Fsurf ace = m a , or
m a = µ m g , so
a = µ g , and
µg
α=
.
R
(1)
Since
ωsurf ace = α t , we have
µg
t.
=
R
(2)
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
After pure rolling begins at trolling there is no
longer any frictional force and consequently
no acceleration. From the perspective of the
center of the disk, let the tangential velocity
of the rim of the disk be vdisk and the angular
velocity be ω ; the angular acceleration is
X
τ = I α , so
τ
α=
I
µmgR
=
2
m R2
3
3 µg
=
(3)
2 R
3 (0.01) (9.8 m/s2 )
=
2
(0.2 m)
= 0.735 rad/s2 .
2
ω0 ,
5
2
= (4 rad/s)
5
= 1.6 rad/s .
=
3 µg
t
2 R 3 µg
2 R
= ω0 −
ω0
2 R
5 µg
2
= ω0 .
5
2
= (4 rad/s)
5
= 1.6 rad/s .
ωrot = ω0 −
Using Eqs. 1 and 5, we have
ωrot = α t
µg
t
=
R µ g 2 ω0 R
=
T
5 µg
(6)
Question 13, chap 8, sect 5.
part 1 of 1
10 points
(4)
When the disk reaches pure rolling, the velocity from the perspective of the surface will be
the same as the velocity from the perspective
of the center of the disk; that is, there will be
no slipping. Setting the velocity ωdisk from
Eq. 4 equal to ωsurf ace from Eq. 2 gives
ωdisk = ωsurf ace
3 µg
µg
ω0 −
t=
t , or
2 R
R
5
µ g t = R ω0 , so
2
2 R ω0
t=
.
5 µg
2 (0.2 m) (4 rad/s)
=
5 (0.01) (9.8 m/s2 )
= 3.26531 s .
(6)
or using Eqs. 4 and 5, we have
The time dependence of ω is
ω = ω0 − α t
3 µg
t.
= ω0 −
2 R
9
A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope
in 10 s.
The average power expended by the student
to overcome gravity is most nearly
1. P = 875 W.
2. P = 5, 600 W.
3. P = 560 W. correct
4. P = 1.1 W.
(5)
5. P = 87.5 W.
Explanation:
P =
W
Fd
(700 N) (8 m)
=
=
= 560 W .
t
t
10 s
Question 14, chap 18, sect 2.
part 1 of 1
10 points
A fountain sends a stream of water 26.4 m
up into the air.
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
The acceleration of gravity is 9.8 m/s2 .
If the base of the stream is 6.14 cm in
diameter, what power is required to send the
water to this height?
Correct answer: 17425.6.
10
Question 15, chap 9, sect 99.
part 1 of 1
10 points
Given:
G = 6.67 × 10−11 N m2 /kg2
Explanation:
Mearth = 5.98 × 1024 kg .
Let : h1 = 26.4 m ,
d = 6.14 cm ,
6.14 cm
r=
= 0.0307 m ,
2
g = 9.8 m/s2 .
and
Using conservation of energy, we may write
1
m v 2 = m g h , and
2
p
v = 2gh
q
= 2 (9.8 m/s2 ) (26.4 m)
= 22.7473 m/s .
is the velocity of the stream at its base. The
flow rate at the base is
π r 2 v = π (0.0307 m)2 (22.7473 m/s)
= 0.0673529 m3 /s .
Using the density of water as 1000 kg/m3 , we
see that 67.3529 kg of water emerges from the
fountain each second.
m
= ρ π r2 v
t
= (1000 kg/m3 ) π (0.0307 m)2 (22.7473 m/s)
= 67.3529 kg .
Thus, each second, sufficient work is done to
raise 67.3529 kg to a height of 26.4 m or
W
P =
t
= mg
h
t
h
t
= (1000 kg/m3 ) π (0.0307 m)2
(26.4 m)
×(22.7473 m/s) (9.8 m/s2 )
(1 s)
= ρ π r2 v g
= 17425.6 W .
An earth satellite remains in orbit at a
distance of r = 19930 km from the center of
the earth.
What is its period?
Correct answer: 27991.6.
Explanation:
The orbiting satellite is pulled towards the
GM m
earth with a force
, where G is the
r2
Gravitational constant, M is the mass of the
earth, m is the mass of the satellite and r is
the distance between the center of the earth
and the satellite.
The force serves as the centripetal force
m v2
.
r
Mm
m v2
G 2 =
r
r
r
GM
v=
r
(6.67 × 10−11 N m2 /kg2 )
=
1.993 × 107 m
1
(5.98 × 1024 kg) 2
×
1
= 4473.63 m/s
and
l = 2πr
= 2 (3.1415926) (1.993 × 107 m)
= 1.25224 × 108 m
So,
l
v
1.25224 × 108 m
=
4473.63 m/s
T =
= 27991.6 s
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
Question 16, chap 16, sect 99.
part 1 of 1
10 points
A pulse moves on a string at 1 m/s, traveling to the right. At point A, the string is
tightly clamped and cannot move.
1 m/s
A
1m
Which of the following shows how the string
would look soon after 2 seconds?
1.
2.
A
A
The pulse travels to the right and is reflected back from the fixed point. The wave
cannot propagate to the right beyond the
point A. Reflected pulses from fixed end
points are inverted.
Question 17, chap 7, sect 4.
part 1 of 1
10 points
A 6.7 kg bowling ball is lifted 2.2 m into a
storage rack.
The acceleration of gravity is 9.8 m/s2 .
Calculate the increase in the ball’s potential
energy.
Correct answer: 144.452.
Explanation:
Potential energy is
U = mgh
= (6.7 kg) (9.8 m/s2 ) (2.2 m)
= 144.452 J
correct
3.
4.
5.
A
A
Question 18, chap 18, sect 4.
part 1 of 1
10 points
The air pressure above the liquid in figure is
1.28 atm . The depth of the air bubble in the
liquid is h = 32 cm and the liquid’s density is
827 kg/m3 .
The acceleration of gravity is 9.8 m/s2 .
air
A
density
of liquid
827 kg/m3
32 cm
6.
7.
8.
Explanation:
A
11
air
A
Determine the air pressure in the bubble
suspended in the liquid.
Correct answer: 132257.
A
Explanation:
We use P = P0 + ρ g h, where ρ =
827 kg/m3 for the liquid, and P0 =
1.28 atm = 129664 Pa . Thus,
P = P0 + ρ g h
= (129664 Pa)
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
+ (827 kg/m3 ) (9.8 m/s2 ) (0.32 m)
= 132257 Pa .
Question 19, chap 17, sect 4.
part 1 of 1
10 points
A police car is traveling at a speed vc to
the left. A truck is traveling at a speed vt to
the left. When the police car is stationary its
siren’s has a wavelength of λc = 0.343 m and
a frequency of fc = 1000 Hz .
The speed of the observer in the truck,
and the speed of the source, the police car
are shown in the figure below.
41 m/s
44 m/s
Police
Truck
What is the frequency ft heard by an observer on the moving truck?
Correct answer: 1007.81.
Explanation:
Let : vc
vt
λc
fc
va
= 41 m/s ,
= 44 m/s ,
= 0.343 m ,
= 1000 Hz ,
≡λf
= (0.343 m) (1000 Hz)
= 343 m/s .
The Doppler shifted frequency f ′ heard in the
truck is
va ± v0
f′ =
f,
(1)
va ∓ vs
where va is the speed of sound in air, vo is
the speed of the observer, and vs is the speed
of the source. The upper sign is used when
the relative velocities are toward one another,
and vice versa.
The relative velocity of the observer is towards the source so the upper sign is used in
the numerator (± → +), and the relative velocity of the source is away from the observer
12
so the lower sign is used in the denominator
(∓ → +). Therefore Eq. 2 becomes
va + vt
ft =
fc
va + vc
(343 m/s) + (44 m/s)
(1000 Hz)
=
(343 m/s) + (41 m/s)
= 1007.81 Hz .
First of four versions.
Question 20, chap 18, sect 3.
part 1 of 1
10 points
The acceleration of gravity is 9.8 m/s2 .
What must be the contact area between
a suction cup (completely exhausted) and a
ceiling in order to support the weight of an
44.2 kg student?
Correct answer: 0.00427601.
Explanation:
Since the suction cup is exhausted, the force
on it given by the atmospheric pressure is F =
Patm A, where A is the area of the suction cup.
In order to support the student, this force
must be equal in magnitude to the student’s
weight, so
mg
A=
= 0.00427601 m2 .
Patm
Question 21, chap 13, sect 3.
part 1 of 1
10 points
A 28.0 kg turntable with a radius of 50
cm is covered with a uniform layer of dry
ice that has a mass of 7.39 kg. The angular
speed of the turntable and dry ice is initially
0.57 rad/s, but it increases as the dry ice
evaporates.
What is the angular speed of the turntable
once all the dry ice has evaporated?
Correct answer: 0.720439.
Explanation:
Basic Concepts:
Li = Lt,i + Ldi,i
= It ωi + Idi ωi
1
1
= Mt R2 ωi + mdi R2 ωi
2
2
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
1
Lf = Lt,f + Ldi,f = It ωf = Mt R2 ωf
2
Lf = Li
Given:
Mt = 28.0 kg
R = 50 cm
mdi = 7.39 kg
ωi = 0.57 rad/s
of the U-tube is in the liquid being tested,
and the other side is in water of density
1000 kg/m3 .
The air is partially removed at the upper
part of the tube and the valve is closed. The
height of the water above its pool surface is
0.68 m . The height of the liquid above its
pool surface is 0.45 m . The difference in the
heights of the pool surfaces is 0.08 m .
⊗ Valve
Solution:
Angular momentum is conserved, so
1
1
Mt R2 ωf = (Mt + mdi ) R2 ωi
2
2
Mt + mdi
ωf =
· ωi
Mt
28 kg + 7.39 kg
· (0.57 rad/s)
=
28 kg
= 0.720439 rad/s
Question 22, chap 6, sect 3.
part 1 of 1
10 points
A car is moving at 18 m/s along a curve
on a horizontal plane with radius of curvature
59 m .
The acceleration of gravity is 9.8 .
What is the required minimum coefficient
of static friction between the road and the
car’s tires to keep the car from skidding?
Correct answer: 0.56036.
Explanation:
The frictional force between the car and
the road must supply the centripetal acceleration necessary to keep the car on the curve.
Therefore, we have
m
2
vcar
= µ Fnormal = µ m g , so
r
v2
(18 m/s)2
µ = car =
= 0.56036 .
gr
(9.8) (59 m)
Question 23, chap 18, sect 4.
part 1 of 1
10 points
One method of measuring the density of a
liquid is illustrated in the figure. One side
13
0.68 m
0.45 m
0.08 m
test liquid
water
Figure: Not drawn to scale
Find the density of the liquid on the left.
Correct answer: 1511.11.
Explanation:
Let : ρw = 1000 kg/m3 ,
hw = 0.68 m ,
h = 0.45 m , and
∆h = 0.08 m .
Note: The difference in the heights of the
pools 0.08 m does not matter since atmospheric pressure is nearly the same at both
pool heights.
The pressure at the upper surface of each
liquid is given by
P = Patm − ρw g hw = Patm − ρ g h.
Therefore,
hw
ρw
h
(0.68 m)
(1000 kg/m3 )
=
(0.45 m)
ρ=
= 1511.11 kg/m3 .
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
Question 24, chap 15, sect 4.
part 1 of 1
10 points
Having landed on a newly discovered
planet, an astronaut sets up a simple pendulum of length 1.14 m and finds that it makes
190 complete oscillations in 338 s. The amplitude of the oscillations is very small compared
to the pendulum’s length.
What is the gravitational acceleration on
the surface of this planet?
Correct answer: 14.2213.
Explanation:
Basic Concept A simple pendulum oscillating with small amplitude has period
s
ℓ
T = 2π
g
where ℓ is the pendulum’s length and g is the
gravitational acceleration at the pendulum’s
location. The pendulum in question made
n = 190 complete oscillations in time t =
t
338 s, which implies the period of T = .
n
Thus
s
t
ℓ
= 2π
n
g
s
ℓ
t
=
2nπ
g
t2
ℓ
=
2
2
4n π
g
4 n2 π 2 ℓ
t2
Thus the planet’s surface gravity is
g=
g=
4 n2 π 2 ℓ
= 14.2213 m/s2 .
t2
Question 25, chap 8, sect 1.
part 1 of 1
10 points
A spring-loaded toy gun is used to shoot a
64 g projectile straight up in the air. When
the spring is compressed by 30 cm before the
14
shooting, the projectile reaches a maximum
height of 49 cm .
The acceleration of gravity is 9.8 m/s2 .
Assume there is no air resistance, no friction inside the gun’s barrel, and the spring is
ideal (i.e., spring force is exactly according to
Hooke’s law). The spring’s equilibrium position is at the end of the gun barrel and all
measurements are made from the end of the
gun barrel.
The figure below shows the toy gun
(a) when the projectile is pressed against the
spring, and
(b) when the projectile reaches its maximum
height.
h
x
k
k
(a)
(b)
How high will the projectile rise if the spring
is compressed three-fifth its previous value
before shooting?
Correct answer: 10.44.
Explanation:
Let : the direction be straight up
g = 9.8 m/s2 ,
m = 64 g ,
x1 = 30 cm ,
h1 = 49 cm , and
3
3
x2 = x1 = (30 cm)
5
5
= 18 cm .
The gun converts the elastic energy of the
compressed spring to kinetic energy of the
projectile, which is then converted to the projectile’s potential energy as it rises,
1
1
k x2 →
m v2 → m g H .
2
2
Note: The total height H should include the
spring’s compression H = h + x .
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
In the 18 cm shooting, the spring’s com3
pression is of the compression in the 30 cm
5
shooting,
1
k x21 = m g (h1 + x1 ) , and
2
1
k x22 = m g (h2 + x2 ) , or
2
2
3
3
1
k
x1 = m g h2 + x1
2
5
5
2 3
1
3
2
k x1 = m g h2 + x1
5
2
5
2
3
3
m g (h1 +x1 ) = m g h2 + x1
5
5
3
9
(h1 + x1 ) = h2 + x1 , so
25
5
3
9
(h1 + x1 ) − x1
h2 =
25
5
9
15
=
(h1 + x1 ) −
x1
25
25
9 − 15
9
h1 +
x1
=
25
25
6
9
h1 −
x1
=
25
25
9
=
(49 cm)
25
6
− (30 cm)
25
= 10.44 cm .
h2
Note: If the compression of the spring is not
included in the height h , we have
9
=
h1
25
9
(49 cm)
=
25
= 17.64 cm , percent error is
|h2 − h′2 |
∆h2
=
100
h2
h
2
9
h1 − 6 x1 − 9 h1 25
25
25 100
=
6
9
h1 −
x1
25
25
|−6 x1 |
100
=
9 h1 − 6 x1
h′2
15
6 (30 cm)
100
9 (49 cm) − 6 (30 cm)
(180 cm)
=
100
(441 cm) − (180 cm)
= 68.9655 % . ← big error
=
Question 26, chap 13, sect 4.
part 1 of 1
10 points
At a certain instant the position of a stone
in a sling is given by x = 1.7 mı̂. The linear
momentum ~p of the stone is 14 kg m/s ̂.
Calculate the magnitude of its angular mo~ = ~r × ~p
mentum: L
Correct answer: 23.8.
Explanation:
Basic Concepts:
~ = ~r × ~p
L
~ = ~r × ~p
L
= (1.7 mı̂) × (14 kg m/s̂)
= 23.8 kg m2 /s k̂
Question 27, chap 7, sect 3.
part 1 of 1
10 points
A 4.38 kg block initially at rest is pulled to
the right along a horizontal, frictionless surface by a constant, horizontal force of 17.4 N.
Find the speed of the block after it has
moved 2.44 m.
Correct answer: 4.40299.
Explanation:
The weight of the block is balanced by
the normal force, and neither of these forces
does work since the displacement is horizontal. Since there is no friction, the resultant
external force is the 17.4 N force. The work
done by this force is
W = F s = (17.4 N) (2.44 m) = 42.456 J
Using the work-energy theorem and noting
that the initial kinetic energy is zero, we obtain
1
W = Kf − Ki = m v 2 − 0 .
2
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
Therefore,
s
r
2W
2 (42.456 J)
vf =
=
= 4.40299 m/s .
m
4.38 kg
The final momentum is
pf = (m1 + m2 ) vf .
7 kg 7 m/s
20◦
17 kg
8. 6
m/ s
What is the speed of the two objects after
the collision if they remain stuck together?
Correct answer: 8.04059.
Explanation:
Let : m1
m2
mf
v1
v2
p1
p2
p1 p2
= 7 kg ,
= 17 kg ,
= m1 + m2 = 24 kg ,
= 7 m/s ,
= 8.6 m/s ,
= m1 v1 = 49 kg m/s ,
= m2 v2 = 146.2 kg m/s ,
= m1 v1 m2 v2
= 14327.6 kg2 m2 /s2 ,
px = p1 + p2 cos θ
= 186.383 kg m/s ,
py = p2 sin θ
= 50.0033 kg m/s ,
θ = 20◦ , and
π − θ = 160◦ .
vf
mf
m1
v1
φ
θ
v2
m2
(1)
Momentum is conserved
Question 28, chap 11, sect 4.
part 1 of 1
10 points
A(n) 7 kg object moving with a speed of
7 m/s collides with a(n) 17 kg object moving
with a velocity of 8.6 m/s in a direction 20 ◦
from the initial direction of motion of the 7 kg
object.
16
~p1 + ~p2 = ~pf .
(2)
Using the law of cosines, we have
p21 + p22 − 2 p1 p2 cos(π − θ) = p2f .
Solving for vf , we have
q
p21 + p22 − 2 p1 p2 cos(π − θ)
vf =
m1 + m2
1
=
(7 kg) + (17 kg)
h
× (2401 kg2 m2 /s2 )2
+ (21374.4 kg2 m2 /s2 )2
− (14327.6 kg2 m2 /s2 ) cos(160◦ )
= 8.04059 m/s .
i1/2
Since the direction of the total momentum
before and after the collision does not change,
the initial and final momentum direction is
py
φ = arctan
p
x
50.0033 kg m/s
= arctan
186.383 kg m/s
◦
= 15.0178 .
Alternate Solution: Since
p2f = p2x + p2y ,
we have
vf =
q
p2x + p2y
m + m2
1
1
=
(7 kg) + (17 kg)
h
× (34738.6 kg2 m2 /s2 )
i1/2
2 2 2
+ (2500.33 kg m /s )
= 8.04058 m/s .
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
Question 29, chap 8, sect 5.
part 1 of 1
10 points
17
T1
1
A single constant force 19.1 N acts on a
particle of mass 6.74 kg. The particle starts
at rest at t = 0.
What is the instantaneous power delivered
by the force at t = 2.87 s?
Correct answer: 155.342.
T3
T3
W1
T2
T2
2
W2
Explanation:
The instantaneous power delivered by a
force F is
At pulley 2,
P = F v = F at
F
F2 t
=F
t=
m
m
2
(19.1 N) (2.87 s)
=
6.74 kg
= 155.342 W
2 T2 = W2
W2
T2 =
.
2
At the weight W1 ,
T3 = T2 + W1
W2
+ W1 .
=
2
Question 30, chap 5, sect 5.
part 1 of 1
10 points
The system is in equilibrium, and the pulleys are weightless and frictionless.
The acceleration of gravity is 9.8 m/s2 .
T
T3
T3
12 N
T2
At pulley 1,
T1 = 2 T3
= W2 + 2 W1
= (14 N) + 2 (12 N)
= 38 N .
T2
14 N
Find the tension T .
Correct answer: 38.
Explanation:
Let : W1 = 12 N and
W2 = 14 N .
Question 31, chap 12, sect 5.
part 1 of 1
10 points
Three identical thin rods of length 18 m
and mass 59.7 kg are placed perpendicular to
each other with their centers intersecting each
other, as shown below.
This setup is rotated about an axis “parallel to the z-axis” that passes through the end
of one rod “along the y-axis” and is perpendicular to another “along the x-axis”.
final 01 – JYOTHINDRAN, VISHNU – Due: May 14 2007, 8:00 pm
x
z
18 m
y
Determine the moment of inertia of this
arrangement.
Correct answer: 17730.9.
Explanation:
Let :
m = 59.7 kg
L = 18 m .
and
Since r is much smaller than L, we can use
the thin rod approximation. Let the junction
be at the center of our coordinate system.
The axis of rotation is parallel to the z-axis
and the y-axis is along the rod touching the
axis of rotation. For the rod along the z-axis
the parallel axis theorem gives:
m r2
Iz =
+m
2
2
L
≈m
2
1
= m L2 .
4
2
L
2
For the rod along the y-axis, from the table
Iy =
L/2
L2 x =2
+
L
3
4
0
3
3
L
L
2m
+
=
L
24
8
1
= m L2 .
3
m x3
59.7 kg per rod
f
so
i
x
a
on
at i
t
o
r
18
1
m L2 .
3
In the rod along the x-axis, the bit
ofmaterial
m
d x and
between x and x + d x has mass
L
s
2
L
is at a distance r = x2 +
from the
2
axis of rotation, so the total moment of inertia
for this rod is
Z L/2 L2 m 2
dx
Ix = 2
x +
4
L
0
Hence, the overall moment of inertia of the
three-rod system is
I = Iz + Iy + Ix
1 1 1
=
m L2
+ +
4 3 3
11
m L2
=
12
11
=
(59.7 kg) (18 m)2
12
(1)
= 17730.9 kg m2 .
Alternate Solution: The moment of intertia of a thin rod about its center-of-mass is
1
mL2 . Therefore the center-of-mass
Icm =
12
moment-of-inertia of the three perpendicular
rods about an axis concentric to one of the
rods is
Icm = Iz + Iy + Ix
1
1
m L2 +
m L2
=0+
12
12
1
2
= mL .
6
Using the “parallel axis theorm”, we have
2
L
I = Icm + 3 m
2
3
1
= m L2 + m L2
6
4
11
2
mL ,
=
12
which is the same as Eq. 1.