midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 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
⊥
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 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
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
3
Question 1, chap 9, sect 5.
part 1 of 1
10 points
The gravitation force of the Moon obeys
Newton’s law of general gravitation
4.
Moon
Earth
5.
Moon
Earth
~ m,e = −G Mm Me R̂m,e
F
2
Rm,e
and is responsible for tides on the oceans of
Earth.
Select the diagram which best describes the
effect of the tides on the oceans on Earth.
The diagrams show an exaggerated amount
of water on the surface of the Earth (shown
in gray) and make the assumption that the
whole Earth is covered with water. Assume
also that the only relevant force causing the
tides is the gravitational force.
1.
Moon
Earth
2.
Moon
Earth
correct
3.
Moon
Explanation:
The side of the Earth closest to the Moon
will feel a stronger attraction to the Moon
since it is closer to the Moon and vice versa;
the side farthest from the Moon will feel a
weaker attraction from the Moon. Therefore
the shape of the water on the Earth’s surface
will be an oblate ellipsoid.
The center of mass of the whole Earth
is balanced between gravitational attractions
and centripetal forces. The center of mass of
the Earth and the water on the Earth will be
at the same point (to first order), since the
gravitational attraction of the water by the
Earth is very much stronger than the gravitational attraction of the water (on the Earth’s
surface) by the Moon, Sun, planets, etc.. That
is, the spherical shape of the Earth and the
oblate ellipsoidal shape of the water have the
same center of mass.
This results in the ocean sticking out on
either side of the Earth.
A small higher order effect results in the
tide being very slightly higher on the side of
the Earth facing the Moon than on the side
opposite the Moon.
Earth
Question 2, chap 7, sect 4.
part 1 of 1
10 points
Energy is required to move a 793 kg mass
from the Earth’s surface to an altitude 0.922
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
times the Earth’s radius RE . (RE =6.37 ×
106 m)
What amount of energy is required to accomplish this move?
Hint: G and ME can be eliminated in favor
of parameter of the gravitational acceleration
at the surface of the earth; i.e., the acceleration of gravity on Earth is 9.8 m/s2 .
Correct answer: 2.37474 × 1010 .
Explanation:
4
Consider a frictionless roller coaster such as
depicted below.
The acceleration of gravity is 9.8 m/s2 .
A
C
hA
16 m
Let : RE = 6.37 × 106 m ,
h RE = 0.922 RE , so
Rmass = (1 + h) RE , and
m = 793 kg .
Using the formulae
U = −G
and
g=G
m ME
R
ME
,
2
RE
(where ME is the mass of the Earth and g is
the gravitational acceleration at the Earth’s
surface), we obtain that the change, ∆U , in
the potential energy of the mass-Earth system
is
∆U = (Ufinal − Uinitial )
1
1
−
= G m ME
Rinitial Rfinal
1
1
= G m ME
−
RE
RE (1 + h)
1
= m g RE 1 −
1+h
h
= m g RE
1+h
= 793 kg × 9.8 m/s2 × 6.37 × 106 m
0.922
×
1 + 0.922
= 2.37474 × 1010 J .
Question 3, chap 8, sect 1.
part 1 of 1
10 points
B
Passenger cars start at point A with zero
initial speed, accelerate as they go down to
point B, swing around the circular vertical
loop B → C → B of radius 16 m, then go
on towards further adventures (not shown).
When a car goes through the top of the loop
(point C), the passengers feel weightless (for
just a moment).
What is the height hA of the starting
point A above the loop’s bottom B?
Correct answer: 40.
Explanation:
Let :
R = 16 m and
g = 9.8 m/s2 .
A passenger feels weightless when his acceleration ~a is precisely equal to the freefall acceleration ~g . At the top of the loop
(point C), passengers have no tangential acceleration while their normal acceleration is
directed straight down and has magnitude
aN =
v2
.
R
Weightlessness happens when this acceleration equals g, therefore the cars should go
through point C at speed
vC =
p
Rg.
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
5
Correct answer: 8.41634.
In the absence of friction, the mechanical
energy of a car is conserved as it follows the
coaster’s route, thus
Explanation:
KA + UA = KC + UC
=⇒
Let : g = 9.8 m/s2 ,
m = 0.448 kg ,
k = 12000 N/m ,
v = vx ,
d = 0.055 m , and
ℓ = 1.3 m .
m 2
m 2
vA + m g hA =
v + m g 2 R.
2
2 C
Therefore, in light of the initial condition
vA = 0, we have
m 2
v
2 C
m
Rg
= mg2R +
2
5
= mgR
2
5
hA = R
2
5
= (16 m)
2
= 40 m .
m g hA = m g 2 R +
b b b b
b b
2.3 m
5.5 cm
1.3 m
µ=0.4
b
ℓ
b
b
b
since vi = 0 m/s.
1
1
Since K = m v 2 , Us = k d2 , and W =
2
2
µmgℓ,
1
1
m vx2 = k d2 − µ m g ℓ
2
2
k d2
− 2µgℓ
vx2 =
m
Thus
r
k d2
vx =
−2µgℓ
m
(12000 N/m) (0.055 m)2
=
0.448 kg
2
− 2 (0.4) (9.8 m/s ) (1.3 m)
vx
b b b b
b b
b
∆x
Applying Conservation of Mechanical Energy,
Ui = Uf + Kf + W
A block is pushed against a spring with
spring constant 12 kN/m (located on the lefthand side of the track) and compresses the
spring a distance 5.5 cm from its equilibrium
position (as shown in the figure below).
The block starts at rest, is accelerated by
the compressed spring, and slides across a
frictionless track except for a small rough area
on a horizontal section of the track (as shown
in the figure below).
It leaves the track horizontally, flies through
the air, and subsequently strikes the ground.
The acceleration of gravity is 9.8 m/s2 .
448 g
h
d
Question 4, chap 8, sect 1.
part 1 of 2
10 points
12 kN/m
vx
k
b
b
= 8.41634 m/s .
b
b
b
∆x
What is the speed v of the block when it
leaves the track?
b
1/2
Question 5, chap 8, sect 1.
part 2 of 2
10 points
What is the horizontal distance x the block
travels in the air?
b
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
6
Correct answer: 5.76619.
Explanation:
Choosing the point where the block leaves
the track as the origin of the coordinate system,
∆x = vx ∆t
Earth
1
∆y = − g ∆t2
2
13 R
2
R
Satellite
since axi = 0 m/s2 and vyi = 0 m/s.
At ∆y = h (below the jump off height),
1
h = − g t2
s2
−2 h
.
t=
g
(8)
(9)
Since x = vx t , we have
x = vx t
s
r
k d2
−2 h
=
−2µgℓ
m
g
s 2h
k d2
−2µgℓ
= −
m
g
(12000 N/m) (0.055 m)2
= −
(0.448 kg)
2
− 2 (0.4) (9.8 m/s ) (1.3 m)
2 (−2.3 m)
×
(9.8 m/s2 )
= 5.76619 m .
1/2
Question 6, chap 9, sect 5.
part 1 of 2
10 points
The radius of the satellite around the center
13
of the Earth is r =
R , where R is the
2
radius of the Earth.
Hint: You may find it useful to take into
account that the gravitational force is a conservative force.
Hint: The universal gravitational force law
is
~ = G M m r̂ .
F
r2
Caution: Neglect the rotational kinetic energy due to the Earth’s rotation.
Find the energy required to launch a satellite from Earth into the circular orbit at the
13
R.
specified radius r =
2
13 G M m
1. E =
12
R
13 G M m
2. E =
10
R
12 G M m
3. E =
correct
13
R
4. None of these
10 G M m
13
R
11 G M m
6. E =
10
R
11 G M m
7. E =
13
R
14 G M m
8. E =
13
R
15 G M m
9. E =
13
R
11 G M m
10. E =
12
R
Explanation:
Basic Concepts: Force of gravity between
two masses m1 and m2 at a distance r
5. E =
m1 m2
and
r2
m1 m2
Ug = G
,
r
Fg = G
where G is the gravitational constant.
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
Centripetal acceleration for circular motion
is
v2
ar =
,
r
for the radial acceleration, where v is the
tangential speed.
Solution: Since the gravitational force supplies the centripetal force,
m ar = m
v2
mM
=G 2
r
r
so that
m v2 = G
mM
mM
=G
.
13
r
R
2
Then the kinetic energy of the satellite in
orbit is found to be
Kr =
1
m v2 = G
2
mM
.
13
R
2
2
Conservation of energy gives us
UR + ∆E = Ur + Kr
∆E = Ur − UR + Kr
GM m GM m GM m
=−
+
+
13
R
13 R
R
2 2
13
1
GM m
− +
+
=
R
13 13 13
=
12 G M m
.
13
R
Question 7, chap 9, sect 5.
part 2 of 2
10 points
Consider the case where the rocket is fired
vertically to rise to a maximum height h =
11
R , the same height as in Part 1; i.e., r =
2
13
R+h=
R.
2
Assume: All the rocket fuel is expended at
the initial instant of firing; e.g., as in the case
of a rail gun.
Caution: Here you must take into account
that the gravitational force is being reduced
as the rocket moves upward into outer space.
7
GM
Given: g =
.
R2
Find the initial velocity, immediately after
it is fired.
r
5
1. vR =
gh
12
r
5
gh
2. vR =
11
r
3
gh
3. vR =
14
4. None of these
r
1
5. vR =
gh
3
r
2
gh
6. vR =
7
r
4
g h correct
7. vR =
13
r
1
gh
8. vR =
5
r
3
gh
9. vR =
13
r
5
10. vR =
gh
13
Explanation:
Using conservation of energy, we have
KR + UR = Kr + Ur = Ur , or
KR = Ur − UR
GM m GM m
=−
+
13
R
R
2
11 G M m
, and
=
13
R
1
11 G M m
m vR 2 =
, so
2
13
R
r
11 G M
vR = 2
r 13 R
22 G M
R
=
13 R2
v
u 22 G M h
=u
t
13 R2 11
2
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
=
r
4
gh ,
13
h
GM
.
and R =
2
11
R
2
Question 8, chap 12, sect 3.
part 1 of 1
10 points
since g =
Given: A rotating bicycle wheel has an
angular speed of 71 ◦ /s at t1 = 3.9 s and a
constant angular acceleration of 22◦ /s2 .
Given: With the center of the wheel at the
origin, the valve stem is on the positive x-axis
(horizontal) at t0 = 1.1 s.
Using 0◦ ≤ θ ≤ 360◦ , what angle θ does the
valve stem make with its original direction at
t2 = 6.9 s?
Correct answer: 64.56.
Explanation:
8
= (9.4◦ /s) [5.8 s]
1
+ (22◦ /s2 ) [5.8 s]2
2
= 424.56◦
= (424.56◦ ) − 1 (360◦)
= 64.56◦ .
Question 9, chap 7, sect 3.
part 1 of 1
10 points
A projectile of mass 0.363 kg is shot from
a cannon, at height 6.3 m, as shown in the
figure, with an initial velocity vi having a
horizontal component of 6.8 m/s.
The projectile rises to a maximum height of
∆y above the end of the cannon’s barrel and
strikes the ground a horizontal distance ∆x
past the end of the cannon’s barrel.
The acceleration of gravity is 9.8 m/s2 .
vi
t0
t1
t2
ω0
ω1
θ0
θ2
= 1.1 s
= 3.9 s
= 6.9 s
=?
= 71◦ /s
= 0◦
=?
First, find the angular velocity ω0 where the
angle θ0 = 0◦ is know. The angular velocity
at t0 = 1.1 s is
ω0 = ω1 + α [t0 − t1 ]
= (71◦ /s) + (22◦ /s2 ) [(1.1 s) − (3.9 s)]
= (71◦ /s) + (22◦ /s2 ) [−2.8 s]
= 9.4◦ /s .
Now the final angle can be determined. The
angle θ2 at t2 = 6.9 s is
1
θ2 = ω0 [t2 − t0 ] + α [t2 − t0 ]2
2
h
i
◦
= (9.4 /s) (6.9 s) − (1.1 s)
h
i2
1
◦ 2
+ (22 /s ) (6.9 s) − (1.1 s)
2
∆y
6.3 m
Given :
◦
41
∆x
Find the magnitude of the final velocity
vector when the projectile hits the ground.
Correct answer: 14.306.
Explanation:
Let : vxi
xi
yi
θ
= 6.8 m/s ,
= 0 m,
= 6.3 m , and
= 41◦ .
For the horizontal component of vector vi ,
we have vxi = vi cos θ; consequently
vxi
cos θ
(6.8 m/s)
=
cos 41◦
= 9.01009 m/s .
vi =
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
9
q
The work done by gravity depends only
= 2 (9.8 m/s2 ) [(6.3 m) + (1.78274 m)]
on the vertical distance yi , since the work
= 12.5866 m/s , so
involving ∆y adds and subtracts; i.e., cancels.
q
The work done by gravity is
vf = vy2f + vx2i
q
W = m g yi .
= (12.5866 m/s)2 + (6.8 m/s)2
1
1
From ∆K = m vf2 − m vi2
2
2
= 14.306 m/s .
and
the work-energy theorem ∆K = W , the final
velocity is
Question 10, chap 8, sect 5.
part 1 of 1
10 points
1
1
m vf2 = W + m vi2
2
r 2
2W
+ vi2
vf =
m
r
2 m g yi
+ vi2
=
m
q
= 2 g yi + vi2
h
= 2 (9.8 m/s2 ) (6.3 m)
i1/2
+ (9.01009 m/s)2
A(n) 547 kg elevator starts from rest. It
moves upward for 4.26 s with a constant acceleration until it reaches its cruising speed of
1.98 m/s.
The acceleration of gravity is 9.8 m/s2 .
Find the average power delivered by the
elevator motor during this period.
Correct answer: 5.55869.
Explanation:
The height y is
y = vave t
1
= vt
2
1
= (1.98 m/s) (4.26 s)
2
= 4.2174 m .
= 14.306 m/s .
Alternate Solution: The initial velocity
in the vertical direction is
vyi = vxi tan θ
= (6.8 m/s) tan 41◦
= 5.91115 m/s , and
the vertical velocity vytop = 0 at the top, therefore we have
vy2i = vy2top + 2 g ∆y , so
s
vy2i
∆y =
2g
s
(5.91115 m/s)2
=
2 (9.8 m/s2 )
= 1.78274 m .
The magnitude of the final vertical velocity
vyf is
p
|vyf | = 2 g [yi + ∆y]
Since the elevator starts from rest, the power
P supplied by the motor is totally transferred
into Kinetic Energy K and Potential Energy
U , where E = K + U , is
E
U +K
=
t
t
1
m g y + 2 m v2
=
t
1
m g 2 v t + 21 m v 2
=
t
mv
(g t + v)
=
2t
(547 kg) (1.98 m/s)
=
2 (4.26 s)
× [(9.8 m/s2 ) (4.26 s) + 1.98 m/s]
= 5558.69 W
= 5.55869 kW .
P =
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
µ = 0.2 , and
k = 5 N/m .
Question 11, chap 8, sect 2.
part 1 of 1
10 points
The two blocks are connected by a light
string that passes over a frictionless pulley
with a negligible mass. The 2 kg block lies
on a rough horizontal surface with a constant
coefficient of kinetic friction 0.2. This block
is connected to a spring with spring constant
5 N/m. The second block has a mass of 7 kg.
The system is released from rest when the
spring is unstretched, and the 7 kg block falls
a distance h before it reaches the lowest point.
The acceleration of gravity is 9.8 m/s2 .
Note: When the 7 kg block is at the lowest
point, its velocity is zero.
h
5 N/m
2 kg
2 kg
µ = 0.2
Solution:
ext
WA→B
= (KB − KA ) + (UBg − UAg )
dis
+ (UBsp − UAsp ) + WA→B
.
For the present case, the external work
ext
= 0, A corresponds to the initial state
WA→B
and B the state where m2 has descended by a
distance s. The sum of the kinetic energy of
m1 plus that of m2 at B is given by
K = KB
dis
= (UAg − UBg ) + (UAsp − UBsp ) − WA→B
1
= m2 g s − k s2 − µ m1 g s .
(1)
2
Based on the Eq. 1 at s = h, KB = 0, we
have
7 kg
µ m1 g h = m2 g h −
h
7 kg
Explanation:
Basic Concepts:
Work-Energy Theorem
Spring Potential Energy
Frictional Force according to the WorkEnergy Theorem
h
k
m1
m1
µ
m2
h
Given :
m1 = 2 kg ,
m2 = 7 kg ,
1
k h2 .
2
In turn,
Calculate the mechanical energy removed
by friction durning the time when the 7 kg
mass falls a distance h .
Correct answer: 101.418.
m2
10
2 g [m2 − µ m1 ]
(2)
k
2 (9.8 m/s2 ) [(7 kg) − (0.2) (2 kg)]
=
(5 N/m)
= 25.872 m .
h=
We know that Einitial = Ef inal + Eµ , where
Eµ is the mechanical energy removed by friction. In order to solve the second part of the
problem we need to calculate the initial and final energies. Let y1 be the vertical position of
m1 and y2 the vertical position of m2 , where
say y = 0 is the initial vertical position of m2 .
The total energy of the system is E = U + K,
where U = Ugrav1 + Ugrav2 + Uspring . Initially, Ugrav1,in = m1 g y1 , Ugrav2,in = 0,
Uspringi n = 0 (the spring is unstretched) and
Kin = 0 . In the final situation the masses
have a null velocity, and so we have once
again Kf in = 0 . The potential energies are
Ugrav1 = m1 g y1 , Ugrav2,f in = m2 g (y1 − h)
1
and Uspring = k h2 .
2
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
Finally, letting y1 = 0 m and using Eq. 2
for h, we get
Eµ = Einitial − Ef inal
1
= −m2 g (−h) − k h2
2
1
= m2 g h − k h2
2
2
2g
m2 (m2 − µ m1 ) − (m2 − µ m1 )2
=
k
2 µ m1 g 2
=
[m2 − µ m1 ]
k
2 (0.2) (2 kg) (9.8 m/s2 )2
=
(5 N/m)
× [(7 kg) − (0.2) (2 kg)]
= 101.418 J .
11
A block is at rest on the incline shown
in the figure. The coefficients of static and
kinetic friction are µs = 0.4 and µk = 0.34,
respectively.
The acceleration of gravity is 9.8 m/s2 .
17 k
g
18◦
µ
What is the frictional force acting on the
17 kg mass?
Correct answer: 51.4823.
Explanation:
N
Alternate Solution:
Ff
Eµ = µ m1 g h
= (0.2) (2 kg) (9.8 m/s2 ) (25.872 m)
18◦
= 101.418 J .
Question 12, chap 7, sect 99.
part 1 of 1
10 points
An advertisement claims that a(n) 1700 kg
car can accelerate from rest to 23 m/s in 7.6 s.
Given: Unit conversion: 746 W/hp.
What average power must the motor produce to cause this acceleration, if we ignore
friction and air resistance?
Correct answer: 79.3089.
Explanation:
1
W = P t = Ef − Ei = E = m v 2
2
m v2
2t
(1700 kg) (23 m/s)2
1
=
2 (7.6 s)
746 W/hp
P =
= 79.3089 hp .
Question 13, chap 6, sect 1.
part 1 of 3
10 points
mg
The forces acting on the block are shown
in the figure. Since the block is at rest, the
magnitude of the friction force should be equal
to the component of the weight on the plane
of the incline
Ff = M g sin θ
= (17 kg) (9.8 m/s2 ) sin 18◦
= 51.4823 N .
Question 14, chap 6, sect 1.
part 2 of 3
10 points
What is the largest angle which the incline
can have so that the mass does not slide down
the incline?
Correct answer: 21.8014.
Explanation:
The largest possible value the static friction
force can have is Ff,max = µs N , where the
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
normal force is N = M g cos θ. Thus, since
Ff = M g sin θ,
M g sin θm = µs M g cos θm
tan θm = µs
θm = tan−1 (µs )
= tan−1 (0.4)
= 21.8014◦ .
12
The test mass will
1. oscillate back and forth about the center
of the shell.
2. move in a circular trajectory around the
center of the shell.
3. move to the surface of the shell farthest
from the test mass (to the left).
Question 15, chap 6, sect 1.
part 3 of 3
10 points
What is the acceleration of the block down
the incline if the angle of the incline is 25◦ ?
Correct answer: 1.12184.
Explanation:
When θ exceeds the value found in part 2,
the block starts moving and the friction force
is the kinetic friction
Fk = µk N = µk M g cos θ.
Newton’s equation for the block then becomes
M a = M g sin θ − Ff
= M g sin θ − µk M g cos θ
and
a = g [sin θ − µk cos θ]
= (9.8 m/s2 ) [sin 25◦ − (0.34) cos 25◦ ]
= 1.12184 m/s2 .
4. move to the surface of the shell closest to
the test mass (to the right).
5. remain stationary. correct
Explanation:
Consider a thin spherical shell. The gravitational force on a test mass anywhere inside
the gravitational shell can be expressed by
considering a solid angle subtended in opposite directions. The shell’s mass will be proportional to the area within the solid angle.
y
R
A ∝ r2
m
x
r
R
sh
el
l
A test mass m is placed at rest inside a
thin spherical shell with uniform areal mass
density.
y
sh
el
l
Question 16, chap 15, sect 1.
part 1 of 1
10 points
m
x
As can be seen the distance from the test
mass m to the surface scales as r and the surface area scales as r 2 . Since the gravitational
1
force scales as 2 and the mass scales as the
r
surface area of the sphere which scales as r 2 ,
the force to the left is equal and opposite the
force to the right.
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
Therefore, no net gravitational force on the
test mass m and no net gravitational field
within the shell.
Since there is no net force the mass remains
stationary.
Question 17, chap 6, sect 3.
part 1 of 2
10 points
The coefficient of static friction between the
person and the wall is 0.8 . The radius of the
cylinder is 6.97 m .
The acceleration of gravity is 9.8 m/s2 .
An amusement park ride consists of a large
vertical cylinder that spins about its axis fast
enough that any person inside is held up
against the wall when the floor drops away.
6.97 m
support the person vertically, the maximal
friction force must be larger than the force of
gravity m g, so that the actual force, which
is equal to or less than the maximum µ N , is
allowed to take on the value m g in the positive vertical direction. In other words, the
“ceiling” µ N on the frictional force has to be
raised high enough to allow for the value m g.
The normal force supplies the centripetal acv2
celeration
on the person, so from Newton’s
R
second law,
mv 2
N =
.
R
Since
fmax = µ N =
2
µ m vmin
= m g , or
R
s
gR
.
vmin =
µ
From this we immediately find the angular
speed
vmin
rR
g
=
µR
s
9.8 m/s2
=
(0.8) (6.97 m)
ωmin ≡
Explanation:
Let :
µ m v2
≥ mg,
R
the minimum speed required to keep the person supported is at the limit of this inequality,
which is
ω
What is the minimum angular velocity
ωmin needed to keep the person from slipping downward?
Correct answer: 1.32572.
13
R = 6.97 m and
µ = 0.8 .
= 1.32572 rad/s .
Basic Concepts: Centripetal force
F =
m v2
.
r
Frictional force
f ≤ µ N = fmax .
Solution: The maximum force due to
static friction is fmax = µ N , where N is
the inward directed normal force exerted by
the wall of the cylinder on the person. To
Question 18, chap 6, sect 3.
part 2 of 2
10 points
Suppose the person, whose mass is m, is
being held up against the wall with an angular
velocity of ω ′ = 2 ωmin .
The magnitude of the frictional force between the person and the wall is
1. F =
1
mg.
4
2. F
3. F
4. F
5. F
6. F
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
i
+ (2.9 N/m) y (̂ · ı̂) dx
= 4mg.
Z 2.4 m h
i
=
(2.6 N/m) x (ı̂ · ı̂) dx
1
= mg.
0
2.4 m
2
1 2
1
= (2.6 N/m) x 
= mg.
2 0
5
= (2.6 N/m) (2.88 m2 )
= 2mg.
= 7.488 J .
1
= mg.
3
14
7. F = 3 m g .
Question 20, chap 8, sect 5.
part 1 of 1
10 points
8. F = 5 m g .
9. F = m g . correct
Explanation:
As discussed above,
f ≤ fmax = µ N
.
Once the angular velocity has increased past
the minimum angular velocity ωmin required
to keep the person pinned against the wall,
there is no “incentive” for the force of friction
to increase any more. Therefore, regardless of
the maximum frictional force allowed by the
angular speed, f stays at its value m g.
Question 19, chap 7, sect 2.
part 1 of 1
10 points
~ = a x ı̂+b y ̂ , where a = 2.6 N/m
A force F
and b = 2.9 N/m, acts on an object as the
object moves in the ı̂ direction along the xaxis from the origin to c = 2.4 m.
Find the work done on the object by the
force.
Correct answer: 7.488.
Explanation:
The work is found through the expression
Z sf
~ · d~s .
F
W =
si
For the force given,
Z 2.4 m h
W =
(2.6 N/m) x (ı̂ · ı̂)
0
A particle of mass m moves along the x
axis. Its position varies with time according
to x = (5 m/s3 ) t3 − (8 m/s2 ) t2 .
What is the power delivered to the particle
at any time t?
h
1. P = 18 m t (25 m2 /s6 ) t2
i
−(15 m2 /s5 ) t + (2 m2 /s4 )
h
2. P = 98 m t (9 m2 /s6 ) t2
i
−(9 m2 /s5 ) t + (2 m2 /s4 )
h
3. P = 2 m t (729 m2 /s6 ) t2
i
2 5
2 4
−(324 m /s ) t + (32 m /s )
h
4. P = 36 m t (32 m2 /s6 ) t2
i
−(12 m2 /s5 ) t + (1 m2 /s4 )
h
5. P = 2 m t (225 m2 /s6 ) t2
i
−(315 m2 /s5 ) t + (98 m2 /s4 )
h
6. P = 4 m t (72 m2 /s6 ) t2
i
−(126 m2 /s5 ) t + (49 m2 /s4 )
h
7. P = 8 m t (81 m2 /s6 ) t2
i
−(27 m2 /s5 ) t + (2 m2 /s4 )
h
8. P = 8 m t (9 m2 /s6 ) t2
i
2 5
2 4
−(9 m /s ) t + (2 m /s )
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
y
h
9. P = 2 m t (729 m2 /s6 ) t2
−(162 m2 /s5 ) t + (8 m2 /s4 )
h
10. P = 2 m t (225 m2 /s6 ) t2
2
5
15
2
4
i
F
i
−(360 m /s ) t + (128 m /s ) correct
Explanation:
The velocity of the particle is
dx
dt
d (5 m/s3 ) t3 − (8 m/s2 ) t2
=
dt
= (15 m/s3 ) t2 − (16 m/s2 ) t .
r
x
m
planet
v=
Determine tne the effective force constant k
of the harmonic motion; i.e., use Hook’s Law
F = −k x .
Correct answer: 0.00289927.
Explanation:
The acceleration of the particle is
The power delivered to the particle at any
time t is
Mp = 7.67 × 1024 kg ,
and
G = 6.67259 × 10−11 N m2 /kg2 .
When the object is in the tunnel, the gravitational force exerted on the object acts towards the Earth’s center and is given by the
equation
GmM
Fg = −
r,
R3
where r is chosen to lie on the r-axis.
y
Rp
P =Fv
= mav
= m (30 m/s3 ) t − (16 m/s2 )
× (15 m/s3 ) t2 − (16 m/s2 ) t
h
= 2 m t (225 m2 /s6 ) t2
i
2 5
2 4
−(360 m /s ) t + (128 m /s ) .
Let : m = 114 kg ,
Rp = 2.72 × 106 m ,
r
dv
a=
dt
d =
(15 m/s3 ) t2 − (16 m/s2 ) t
dt
= (30 m/s3 ) t − (16 m/s2 ) .
m
x
Question 21, chap 15, sect 1.
part 1 of 1
10 points
Given: G = 6.67259 × 10−11 N m2 /kg2
An object of mass 114 kg moves in a smooth
straight tunnel dug through the center of a
planet of mass 7.67×1024 kg and radius 2.72×
106 m as shown in the figure.
Note: If we consider only the mass outside
the Gaussian surface (a spherical shell), this
mass produces no gravitational field inside the
Gaussian surface (inside this spherical shell).
midterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 7 2007, 11:00 pm
A ∝ r2
R
sh
el
l
Only mass inside the Gaussian surface contributes to the gravitation field.
Note: Another way of obtaining the same
result is to consider a thin spherical shell. The
gravitational force on a test mass anywhere
inside the gravitational shell can be expressed
by considering a solid angle subtended in opposite directions. The shell’s mass will be
proportional to the area within the solid angle.
y
m
x
r
As can be seen the distance from the test
mass m to the surface scales as r and the surface area scales as r 2 . Since the gravitational
1
force scales as 2 , the force to the left is equal
r
and opposite the force to the right. Therefore
there is no net gravitational force on the test
mass m and there is no net gravitational field
within the shell.
The mass Mr inside the Gaussian surface
(dashed circle indicating a spherical surface)
is
Mr = Mp
Vr
Vplanet
4
π r3
= Mp 3
, so
4
π Rp3
3
G m Mp
r.
Fr = −
R3
Applying Newton’s second law to the motion
along r and Hook’s Law, we have
Fr = −G
m Mp
r = −k r ,
Rp3
16
therefore the effective force constant k is equal
to
k=G
m Mp
Rp3
= (6.67259 × 10−11 N m2 /kg2 )
(114 kg)(7.67 × 1024 kg)
×
(2.72 × 106 m)3
= 0.00289927 N/m .