Download Problem 1: Second Law and projectile motion

Problem 1 Earth and a Geostationary Satellite (10 points)
The earth is spinning about its axis with a period of 23 hours 56 minutes and 4 seconds.
The equatorial radius of the earth is 6.38 × 106 m . The latitude of Cambridge, Mass is
42° 22′ .
a) Find the magnitude of the velocity and the centripetal acceleration with respect to
the earth’s axis of a person at MIT as they undergo circular motion about the
earth’s axis of rotation.
A geostationary satellite goes around the earth once every 23 hours 56 minutes and 4
seconds, (a sidereal day, shorter than the noon-to-noon solar day of 24 hours) so that its
position appears stationary with respect to a ground station. The mass of the earth is
me = 5.98 × 1024 kg . The mean radius of the earth is Re = 6.37 × 106 m . The universal
constant of gravitation is G = 6.67 × 10−11 N ⋅ m 2 ⋅ kg −2 . Your goal is to find the radius of
the orbit of a geostationary satellite.
b) Describe what motion models this problem.
c) What is the radius of the orbit of a geostationary satellite? Approximately how
many earth radii is this distance?
Solution:
For recent values of the physical constants used in this problem, see
http://pdg.lbl.gov/2007/reviews/contents_sports.html - constantsetc .
a) The rotational period of the earth is given by
Tearth, 2005 = (23 hr) ( 3600 s ⋅ hr −1 ) + (56 min) ( 60 s ⋅ min −1 ) + 4s = 86164 s . (1.1)
Note that this period is less than 24 hr . Twenty-four hours is one solar day (noon to
noon), while the above period is one sidereal day; sidereal means with respect to the
“fixed” stars, and you should be able to see why the two are different.
A person at MIT undergoes circular motion about the axis of the earth. The radius of this
circle is R = Re sin θ where θ is the angle between MIT and the axis of rotation. Since
the latitude λ is measured form the equator, sin θ = sin(π / 2 − λ ) = cos λ (the angle θ is
sometimes called the “colatitude”). Hence the radius of the orbit of a person at MIT is
R = Re sin θ = Re cos λ = (6.38 × 106 m) ( cos 42.36° ) = 4.71× 106 m .
(1.2)
Since the motion is uniform, during one period of rotation the person travels a distance
s = 2π R = vT
(1.3)
where v is the magnitude of the velocity and s = 2π R is the circumference.
The magnitude of the velocity (the speed) is then given by
v=
2π R
T
(1.4)
and so for a person at MIT, the magnitude of the velocity is
6
2π R (2π ) ( 4.71× 10 m )
v=
=
= 3.44 × 102 m ⋅ s −1 .
T
86164 s
(1.5)
The magnitude ar of the centripetal acceleration is given by
−1
2
v 2 ( 3.44 × 10 m ⋅ s )
=
= 2.51× 10−2 m ⋅ s −2 .
ar =
6
4.71×10 m
R
2
(1.6)
b) The satellite’s motion can be modeled as uniform circular motion. The gravitational
force between the earth and the satellite keeps the satellite moving in a circle. The
acceleration of the satellite is directed towards the center of the circle, that is, along the
radially inward direction.
The figure below is close to a scale drawing.
c) Choose the origin at the center of the earth, and the unit vector r̂ along the radial
direction. This choice of coordinates makes sense in this problem since the direction of
acceleration is along the radial direction.
G
G
Let r be the position vector of the satellite. The magnitude of r (we denote it as rs ) is the
distance of the satellite from the center of the earth, and hence the radius of its circular
orbit. Let ω be the angular velocity of the satellite, and the period is T = 2π / ω . The
acceleration is directed inward, with magnitude rsω 2 ; in vector form,
G
a = − rsω 2ˆr .
(1.7)
Apply Newton’s Second Law to the satellite for the radial component. The only force in
this direction is the gravitational force due to the Earth,
G
Fgrav = − msω 2 rs rˆ .
(1.8)
The inward radial force on the satellite is the gravitational attraction of the earth,
−G
ms me
rˆ = − msω 2 rs rˆ .
2
rs
(1.9)
ms me
= msω 2 rs .
2
rs
(1.10)
Equating the r̂ components,
G
Solving for the radius of orbit of the satellite rs ,
1/ 3
⎛Gm ⎞
rs = ⎜ 2 e ⎟ .
⎝ ω ⎠
(1.11)
The period T of the satellite’s orbit in seconds is 86164 s (see Equation (1.1)) and so
the angular speed is
2π
2π
=
= 7.2921×10−5 s −1 .
(1.12)
ω=
86164 s
T
Using the values of ω , G and me in Equation (1.11), we determine rs ;
rs = 4.22 × 107 m = 6.62 Re .
(1.13)
Problem 2 (10 points):
a) A bicycle tire rolls without slipping along the ground with its center of mass moving
with speed v0 . You may neglect any rolling resistance. What is the speed with respect to
the ground of each of the four marked points on the tire shown in the figure below?
Explain each of your answers.
b) A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which
diagram correctly shows the forces acting on her? Explain your reasoning.
Answers to (a): As an intermediate calculation (introducing an “auxiliary parameter”),
let the wheel radius be Rwheel . The angular speed of the wheel about its center is then
ω 0 = v0 / Rwheel if the wheel is to roll without slipping. The speed of any point on the rim
of the wheel with respect to the center is then v = ω 0 Rwheel = v0 .
This should be obvious, and such a detailed calculation may seem pointless. However,
for more complicated systems, or, if we needed to consider points other than those on the
rim, we have to include this intermediate step.
Speed of point a: This point is moving parallel to the velocity of the center of the wheel,
so the speed is the sum v0 + v0 = 2v0 , a perhaps apparent result.
Speed of point b: This point has a horizontal velocity component v0 and a vertical
velocity component, also v0 , and so the speed is 2 v0 .
Speed of point c: For this point, the velocity with respect to the center of the wheel is
directed opposite to the velocity of the center of the wheel, and so the speed with respect
to the ground is v0 − v0 = 0 . Or, we could say quite correctly that if the wheel is rolling
without slipping, the contact point, being in contact with the ground, cannot be moving
with respect to the ground.
Speed of point d: This is the same speed as that of point b, 2 v0 . The vertical
component of velocity is of course down, but the magnitude is the same.
Answer to (b): 1. Consider the vertical and horizontal forces on the rider separately. For
the rider not to slip in the barrel, there must be no net vertical force acting on her. The
two vertical forces acting on her are her weight down and a frictional force between her
and the wall acting up. These forces must be equal in magnitude. Then there must also be
a force acting on the rider that always points toward the center of the barrel to provide the
rider with centripetal acceleration to keep her going in a circle. Where does that force
come from? The normal component of the contact force exerted on her by the wall
supplies that centripetal acceleration. You may wonder how the wall can exert a normal
force on her if there is no other force acting to press her into the wall. Ordinarily we think
of contact forces arising because some other force -- typically gravity -- is pushing an
object against another object, and the contact force arises in response to that push. If you
think carefully about the motion of the rider, her velocity is always directed tangent to the
vertical wall of the barrel of fun, so the rider in a sense is constantly colliding with the
part of the wall that is right behind her as it travels in a circle. So the wall behind her
pushes out on her because otherwise her inertia would mean that she would go straight
through the wall, tangent to her velocity at any moment. She exerts a contact force on the
wall due to her inertia. As your reading remarked, one thing to be careful about is to
realize that “centripetal force” is not an independent force that somehow arises because
something is going in a circle. Rather, if an object is going in a circle, there must be a
force acting on it from its environment -- a string tied to it or a contact force from an
adjacent surface, for example -- which acts in the appropriate direction to be a
centripetal force.
So, in the case of the barrel of fun, there are three forces: gravity, the upward frictional
force and the inward normal force, and the correct answer is diagram 1.
Note that it would be possible to combine the friction force and the normal force into a
single contact force. This contact force would have upward and inward components,
hence directed up and to the left in the diagram. None of the five choices given matches
this model (although a case could be made for “other,” but such this would need to be
explained).
Problem 3 Object Undergoing Circular Motion on the Inside of a Cone (10 points)
A body of mass m slides without friction on the inside of a cone. The axis of the cone is
vertical and gravity is directed downwards. The apex half-angle of the cone is θ as
shown in the figure below. The path of the object happens to be a circle in a horizontal
plane. The speed of the particle is v0 . Find the radius of the circular path and the time it
takes to complete one circular orbit in terms of the given quantities and g . Include all
relevant free-body diagrams in your answer.
Solution: The free body diagram of the forces acting on the object is shown in the
figure below.
We choose cylindrical coordinates with unit vectors r̂ pointing in the radial outward
direction and k̂ pointing upwards. The two forces acting on the object are the normal
G
G
force N of the wall on the object and the gravitational force mg (we are given that there
K
K
is no friction). The two vector components of Newton’s Second Law F = m a become:
−m v0 2
,
r
kˆ : N sin θ − m g = 0 .
rˆ : − N cos θ =
(3.1)
(3.2)
These equations can be re-expressed as
v0 2
,
r
N sin θ = m g .
N cos θ = m
(3.3)
(3.4)
We can divide these two equations,
N sin θ
mg
,
=
v0 2
N cos θ
m
r
(3.5)
yielding
rg
.
v0 2
(3.6)
v0 2
tan θ .
g
(3.7)
tan θ =
Equation (3.6) can be solved for the radius,
r=
Note that the radius is independent of the mass because the component of the normal
force in the vertical direction must balance the gravitational force, and so the normal
force is proportional to the mass. Therefore the radially inward component of the normal
force is also proportional to mass. Since this force is causing the object to accelerate
inward, by Newton’s Second Law, the mass will cancel and the acceleration is
independent of the mass. Since the radial component of the acceleration is inversely
proportional to the radius, ar = −v0 2 / r , therefore the radius is also independent of
the mass.
The period for the orbit is given by
T=
2π r 2π v0 tan θ
=
.
v0
g
(3.8)
Problem 4: Circular Motion (10 points)
A governor to control the rotational speed of a steam engine was invented by James Watt.
Two spheres were attached to a rotating shaft by rigid arms that were free to rotate up and
down about a pivot where they attached to the shaft, as in the diagram above. As the arms
pivoted up and down they actuated a mechanism to control the throttle of the steam
engine. Assume the rigid arms have length l and no mass. All of the mass is then
concentrated in the two spheres at the end of the arms, each having mass m.
a) Describe the acceleration of the spheres. Explaining and quantifying your knowledge
of the acceleration will help you model the problem. Show all relevant free
body diagrams.
b) Show that there is a minimum angular velocity ω min below which the governor will not
function as intended.
c) Derive an expression for the radius r of the circular path followed by the spheres.
Express your answer only in terms of as few of the quantities m, ω , l, and g (the
acceleration due to gravity) as you can. (Do not use the angle φ in your answer).
Solutions:
a) Each sphere moves in a horizontal circle of radius r = l sin φ with period T = 2π / ω ,
and hence an inward radial acceleration of magnitude
ar = ω 2 r = ω 2l sin φ .
(4.1)
Thus, there must be a net inward force of magnitude Fnet = m ar = mω 2l sin φ . This force
G
G
is the vector sum of the gravitational force mg and the force Frod that the rod applies to
G
the ball (since we call the period “ T ,” we don’t want to use “ T ” for a “tension force”).
G
The gravitational force has no horizontal component, so the vertical component of Frod
G
G
must balance the weight mg and the horizontal component of Frod must have magnitude
Fnet = m ar = mω 2l sin φ .
A free body diagram for the right-hand sphere is shown below (the diagram for the lefthand sphere is of course just a reflection and won’t be shown here).
b) Expressing the answer to part a) mathematically, the two components of Newton’s
Second Law are:
kˆ : Frod cos φ − mg = 0
rˆ : Frod sin φ = mω 2l sin φ .
(4.2)
Eliminating sin φ (why can we exclude the case sin φ = 0 ?) from the second expression
in (4.2) yields Frod = mω 2l and substitution into the first and rearranging yields
cos φ =
g
lω 2
.
(4.3)
We must have cos φ < 1 , or ω 2 > g / l , so ω min = g / l .
c) Using r = l sin φ = l 1 − cos 2 φ from part a) and cos φ = g / lω 2 from part b) above, we
can eliminate the angle φ with the result
r = l 1 − g 2 / ( lω ) = l 2 − g 2 / ω 4 .
2
(4.4)
Note that this is consistent with part b); for a positive argument of the square root, we
must have ω > ω min .
Extra: It’s probably worth noting the differences between this problem and Problem 3.
Except for notation, the free body diagrams are identical. For instance, in Problem 3 the
speed was given and the angle was fixed. In Problem 4, as ω varies, the speed, radius
and angle φ vary (which is the point of a governor).
Problem 5: Whirling Objects, U-Control Model Airplane (10 points)
A U-control airplane of mass M is attached by wires of length L and negligible mass to
the “pilot” who controls the lift provided by the wing. (The wires control the plane’s
elevator.) The plane’s engine keeps it moving at constant speed v .
a) Briefly describe how you intend to model the motion of the object. What
directions are you choosing for analyzing the components of your forces? Give
your reasons.
b) Find the total tension T in the wires when the plane is flown overhead in a circle
so that the wires make an angle θ with the ground. Remember that the wings can
provide lift only in the direction perpendicular to their area, that is, in a direction
perpendicular to the wires. Think carefully before selecting the directions of the
axes of your coordinate system.
c) The plane will go out of control and crash if the tension is not maintained. Given
a particular speed v of the plane, is there some angle θ crit which you would advise
the pilot not to exceed? If possible, exhibit a speed vsafe at which the plane would
be safe at any angle.
Solutions:
(See the problem set for the original figure, which will not be reproduced here.)
a) The “object” (the airplane) is moving in a horizontal circle. The radius of the circle
and the hence the period, for a given speed, will depend on the angle that the wires make
with respect to the horizontal. Anticipating parts b) and c), the radius of the circle will be
r = L cosθ and hence the period of the circular motion will be 2π r / v = 2π L cosθ / v .
One choice of coordinates is that shown in the free body diagram on the left below, with
k̂ directed vertically upward and r̂ directed radially outward. Two advantages to this
choice are that the acceleration and the gravitational force are along the coordinate
directions, and that this is the “standard,” already used in Problems 3 and 4 above.
Another choice would be that shown on the right below. The advantage to this system is
that the unknown forces, including the tension to be determined in part b), are along the
G
coordinate axes, and need not be decomposed. However, the weight mg and the net
force must be decomposed along the given axes.
b) We’ll do this using both of the coordinate systems, starting with the “tilted” system on
the right below. Note that the î - and ĵ -directions are not the horizontal and vertical
directions.
Applying Newton’s Second Law, with a = ar as the magnitude of the centripetal
acceleration, for the two coordinates, we have
ˆi : T + mg sin θ = ma cosθ
ˆj : N − mg cosθ = ma sin θ .
(5.1)
Now we go back to the question asked: We want the find the tension but not the normal
force. We know that a = v 2 / r = v 2 / L cosθ , so the first expression in (5.1) leads almost
immediately to
T = m ( v 2 / L − g sin θ ) ,
(5.2)
a very tidy result.
If we used the coordinate system on the left above, Newton’s Second Law becomes
kˆ : N cosθ − T sin θ = mg
rˆ : N sin θ + T cosθ = ma.
(5.3)
The expressions in (5.3) are two equations in the two unknowns N and T and may be
solved by standard methods. For our immediate purpose, however, eliminate N by
multiplying the first by sin θ and the second by cosθ and then subtract the first from the
second, leading to Equation (5.2).
Decide for yourself which coordinate system is “better” or “easier,” but keep in mind that
either will work.
c) Either way, we are now set to find the critical angle θ crit at which the tension as found
in Equation (5.2) would go to zero,
⎛ v2 ⎞
(5.4)
θ crit = sin −1 ⎜ ⎟ .
⎝ gL ⎠
We see that if v 2 > gL , the critical angle is never reached, and all angles would be safe,
so we have
vsafe = gL .
(5.5)
Notice that if θ = 0 the rope lies in the plane of the airplane’s circular orbit, sin θ = 0 ,
G
and the tension T = mv 2 / L > 0 for all velocities (assuming the normal force N , acting
vertically upward, is sufficient to keep the plane flying). The other extreme value occurs
when θ → π / 2 . This corresponds to the radius of the orbit r → 0 , cosθ → 1 and the
tension T → ( mv 2 / L ) − mg . In order for the tension to stay positive, v > gL (which is
consistent with θ crit → π / 2 ).
G
Extra: This problem did not ask for the normal force N (often called the “lift”), but it’s
not hard to find
N = mg ( v 2 / L ) tan θ + cosθ .
(5.6)
(
)