Download Solution Key

Homework 1 Solutions
Andre Goran
February 3, 2017
g = 9.8 sm2 (value of g used throughout these solutions, note a negative indicating downward
is not included in the value and must be provided where needed)
1
Ch1: 40, 46, 49
1.40) A cannon shoots a ball at angle θ above the horizontal ground. A) Neglecting air
resistance, use Newton’s second law to find the ball’s position as a function of time:
d2 x
d2 y
ax = 2 = 0
ay = 2 = −g
dt
dt
define (0,0) as the cannon’s location then you get equations of motion
1
y = v0 sin(θ)t − gt2
x = v0 cos(θ)
2
B) Let r(t) denote the ball’s distance from the cannon. What is the largest possible value
of θ if r (t) is to increase throughout the ball’s flight?
Clearly, if the ball is fired nearly horizontally, its distance from the cannon increases until
it lands, and if it is fired vertically, it returns to the cannon after achieving some height.
What is the threshold θ at which the ball’s distance increases for the entire trajectory, but
at any greater θ this no longer holds?
Just after launch the ball’s distance from the cannon must increase, regardless of θ, and
thus dr
> 0. If at any point the ball ceases to gain distance, the derivative of r will no
dt
longer be positive, so let’s examine the conditions for the derivative remaining positive for
the duration of the shot:
g 2 t4
4
> 0 and
r2 = x2 + y 2 = v02 t2 − v0 gsin(θ)t3 +
Since r is a real positive quantity, we may take
statements.
1
dr
dt
d 2
r
dt
> 0 as equivalent
d 2
r = 2v02 t − 3v0 gsin(θ)t2 + g 2 t3
dt
If the ball fails to continue gaining distance from the cannon, then
0 = 2v02 t − 3v0 gsin(θ)t2 + g 2 t3
has a solution during the trajectory, since the ball cannot begin returning without a
moment of zero change in distance (not necessarily a moment of zero net velocity though!).
Eliminate the potential solution of t=0 as it is not physical, and you are left:
0 = 2v02 − 3v0 gsin(θ)t + g 2 t2
This is a quadratic equation in t, with standard ax2 + bx + c = 0 coefficients of a = g 2 ,
b = −3v0 gsin(θ), and c = 2v02 . The number of real solutions t which satisfy is determined
by the discriminant, b2 − 4ac, as appears under the square root in the quadratic formula.
b2 − 4ac = 9g 2 v02 sin2 (θ) − 8v02 g 2
No real solution to the quadratic exists if the value is negative, equivalently if
8v02 g 2 > 9g 2 v02 sin2 (θ)
sin2 (θ) <
8
9
3
22
3
π
Since physically θ ∈ [0, 2 ] , we can conclude that θ < 1.23
sin(θ) <
1.46) Consider the experiment of Problem 1.27, in which a frictionless puck is slid straight
across a rotating turntable through the center O. A) Write down the polar coordinates r, φ
of the puck as functions of time, as measured in the inertial reference frame of an observer
on the ground. (Assume that the puck was launched along the axis φ = 0 at t = 0)
Assuming the puck starts at linear speed v, distance r0 out from the origin along φ = 0 .
Then for the observer in the inertial ground frame, the puck’s coordinates may be expressed as
2
r = |r0 − vt|
φ=0
which holds until the puck passes through the origin at t =
Azimuthal component becomes
r0
,
v
after which the puck’s
φ=π
B) For an observer sitting at the center of the rotating frictionless table, no difference
in r behavior would be observed, as at any given instant the observers would agree on the
puck’s absolute distance from the center of the circle. However, the rotation of S 0 is seen in
their observation of the puck’s φ value:
r0 = |r0 − vt|
φ0 = −ωt
The observer S 0 sees the puck spiral into the center of the circle, and spiral out. Though
the puck feels no net force, it’s instantaneous velocity in S 0 is not constant, so this frame is
non-inertial.
1.49) Imagine two concentric cylinders, centered on the vertical z axis, with radii R ,
where is very small. A small frictionless puck of thickness 2 is inserted between the two
cylinders, so that it can be considered a point mass that can move freely at a fixed distance
from the vertical axis. If we use cylindrical polar coordinates (ρ, φ, z) for its position
(Problem 1.47), then ρ is fixed at ρ = R. while φ and z can vary at will. Write down and
solve Newton’s second law for the general motion of the puck, including the effects of gravity.
Describe the puck’s motion
The geometry of the problem lends itself to being solved using cylindrical polar coordinates
ρ, φ, z. First, let us express the unit vectors of a cylindrical system in familiar terms. We
may keep the unit vector for the z direction, k̂ as we know it from Cartesian. Radius ρ
and the azimuthal angle φ cannot be so simply stated in Cartesian terms, as the directions
indicated by φ̂ and ρ̂ are position dependent. In the case of ρ̂, this is a vector of length 1 in
the direction from the origin to the point where ρ̂ is being calculated. We may say:
ρ̂ =
ρ~
xî + y ĵ
ρ cos φî + ρ sin φĵ
=
=
= cos φî + sin φĵ
ρ
ρ
ρ
3
We may easily see that this is a unit vector (length 1) for any point in the plane, due to
the relationship cos2 φ + sin2 φ = 1. The definition of a right handed coordinate system then
demands that ẑ × ρ̂ = φ̂, which allows us to finish the set of unit vectors
φ̂ = k̂ × (cos φî + sin φĵ) = − sin φî + cos φĵ
The position vector in cylindrical coordinates is given by the sum of the two coordinates
with dimensions of length
~r = ρρ̂ + z ẑ
While the velocity must explicitly include a term for the tangential motion
~v = ρ̇ρ̂ + ż ẑ + (ρφ̇)φ̂
and a derivative gives acceleration as normal
˙
~a = ρ̈ρ̂ + ρ̇ρ̂˙ + z̈ ẑ + ż ẑ˙ + (ρφ̇)φ̂ + φ̂(ρφ̈ + φ̇ρ̇)
And we simplify this expression using the relationships derived above:
~a = (ρ̈ − ρφ̇2 )ρ̂ + (ρφ̈ + 2ρ̇φ̇)φ̂ + ẑ z̈
The constraints of the problem require that ρ̇ and all higher derivatives be 0, so dropping
those terms, we have:
~a = (−ρφ̇2 )ρ̂ + (ρφ̈)φ̂ + ẑ z̈
The first term is in the −ρ direction, or always directed toward the center of the circle.
Tangential speed in polar coordinates is of course ρφ̇, so we see that our first acceleration
2
term is the familiar vr , or a centripetal acceleration under uniform circular motion. There is
not torque about the central axis causing the mass to experience changing angular velocity,
so we also observe that φ̈ must be 0, and we finally have
~a = (−ρφ̇2 )ρ̂ + ẑ z̈
Which we may solve by inspection to give equations of motion
ρ=R
φ = φ0 + ωt
1
z = z0 + v0z t − gt2
2
Where ω is the angular velocity of the constrained object about the cylindrical axis at t = 0.
As time passes, the mass will move down with free fall acceleration while tracing out a circle
at constant angular velocity in the plane perpendicular to gravity.
2
CH2: 3,10,23,47,53
2.3) A)The quadratic and linear drag forces on a moving sphere in a fluid are given by (2.84)
and (2.82) (Problems 2.4 and 2.2). Show that the ratio of these two kinds of drag force can
f
R
be written as fquad
= 48
, where R is the dimensionless Reynolds number.
lin
4
fquad = κρAv 2
flin = 3πηDv
fquad
κρAv 2
=
flin
3πηDv
Flow-facing area of a sphere is just the area of its cross section, which we can relate to
diameter:
2
κρπ D4 v 2
fquad
κρDv
κ
=
=
= R
flin
3πηDv
12η
12
For a sphere,κ = 14 , so we end with
fquad
flin
=
R
48
B) Find the Reynolds number for a steel ball bearing (diameter 2 mm) moving at 5 cm/s
through glycerin (density 1.3 g/cm3 and viscosity 12 Ns/m2 at STP).
R=
kg
(.002m)(.05 ms )(1300 m
Dvρ
3)
=
= .0108
Ns
η
12 m2
2.10) For a steel ball bearing (diameter 2 mm and density 7.8 g/cm 3) dropped in glycerin
(density 1.3 g/cm3 and viscosity 12 Ns/m2 at STP), the dominant drag force is the linear
drag given by (2.82) of Problem 2.2. A) Find the characteristic time τ and the terminal
speed vter . [In finding the latter, you should include the buoyant force of Archimedes. This
just adds a third force on the right side of Equation (2.25).] How long after it is dropped
from rest will the ball bearing have reached 95terminal speed?
Fdrag = 3πηDv
0 = Mball × g − ρg × Vball × g − 3πηDvter
5
cm
ρg × Vball × g − Mball × g
= .12
−3πηD
s
m
m
=
= 1.44 × 10−4 s
τ=
b
3πηD
vter =
Allowing that the ball starts at some speed v0 and must eventually converge to vter , we
obtain
t
v = vter + (v0 − vter )e− τ
So we can find that 95% terminal of .0011
m
s
is reached when:
t
.0011 = .0012(1 − e− .000144 )
log(
t
.1
)=−
1.2
.000144
(3)
which gives t = 3.6 × 10−4 or about 2.5 times the characteristic time τ
f
B) fquad
= Dvρ
= 5.42 × 10−6 This is a very low Reynolds number situation, meaning it was
48η
lin
appropriate to ignore quadratic drag.
2.23) Find the terminal speeds in air of (a) a steel ball bearing of diameter 3 mm, (b) a
16-pound steel shot, and (c) a 200-pound parachutist in free fall in the fetal position. In all
three cases, you can safely assume the drag force is purely quadratic. The density of steel is
about 8 g/cm3 and you can treat the parachutist as a sphere of density 1 g/cm3 .
Getting all necessary values into compatible units, Qair = 1.225kg/m3 , Qstee = 8000kg/m3 , Qperson =
Qwater = 1000kg/m3 . The mass of the parachutist is 90.7kg, which, given the density, gives
a volume of .09m3 and thus a radius of .407m. Using
r
mg
v=
κQA
Ballbearing converges to 22 m/s, steel shot to 140 m/s, balled-up free-falling human to 107
m/s
6
2.47) For each pair find z+w, z-w, zw, and z/w
A) z=6+8i and w=3-4i
z + w = 9 + 4i
z/w =
z − w = 3 + 12i
zw = 18 + 24i − 24i + 32 = 50
6 + 8i
6 + 8i 3 + 4i
−14 + 48i
=
×
=
= −.56 + 1.92i
3 − 4i
3 − 4i 3 + 4i
25
B)
z = 8eiπ/3 = 8cos(π/3) + 8isin(π/3) = 4 + 6.93i
√
w = 4eiπ/6 = 4cos(π/6) + 4isin(π/6) = 2 3 + 2i
z + w = 7.464 + 8.93i
z − w = 5.36 + 4.93i
z/w = 2eiπ/6 =
zw = 32
√
3+i
2.53) A charged particle of mass m and positive charge q moves in uniform electric and
magnetic fields, E and B, both pointing in the z direction. The net force on the particle is
→
−
→
− − →
−
F = q( E + →
v × B ). Write down the equation of motion and resolve it into components.
Solve and describe the motion.
→
−
→
− − →
−
F = q( E + →
v × B)
→
−
F = qE k̂ + q(vy B î − vx B ĵ)
→
−
F = (qvy B, −qvx B, qE))
ẍ =
q ẏB
m
ÿ = −
7
q ẋB
m
z̈ =
qE
m
m
qBt
m
qBt
1 qE 2
t
x = x0 −
cos(
)
y = y0 +
sin(
)
2m
qB
m
qB
m
This object accelerates in the z direction as it traces out a circle at constant angular velocity
ω = qB
in the x,y plane.
m
z = z0 + v0z t +
8