Newton’s Laws
1
Introduction
Newton’s laws are the basic objects of study in classical mechanics. Here, we’ll briefly review Newton’s three laws
and point out a few features of using them to solve problems. Finally, we’ll review accelerating reference frames. In
these frames, Newton’s laws don’t hold in their standard form, but we can still do physics in accelerating frames by
modifying Newton’s laws.
2
Newton’s First Law
Newton’s first law says that if there’s no net external force on an object, it moves at constant velocity (i.e. at
constant speed in a straight line.) The converse is also true - if an object is moving in a straight line at constant
speed, there must be no net force on it.
This doesn’t mean that such an object has no forces at all on it, only that the forces on it sum to zero. For example,
a car driving down a straight road at constant speed has numerous forces on it: gravitational force from the Earth,
normal force from the road, friction from the road pushing the tires forward, rolling resistance, air drag, lift or
downforce from the air, and even obscure forces such as a force from radiation pressure due to sunlight shining on
the car. If the car drives at constant velocity, we can conclude that all these forces sum to zero.
No car drives at truly constant velocity; there will be little speed-ups and slow-downs all the time. In this sense,
Newton’s first law, like most laws of physics, can be thought of as, at best, approximately applying to any given real
system. If the car’s velocity is very nearly constant over time, the net force on it is very nearly zero.
Not all parts of a car move with the same velocity. The tires rotate, for example. The pistons in the engage move up
and down relative to the engine itself, and the parent in the front seat turns around and yells, “I swear I’ll turn this
car around if you don’t all be quiet!” to the kids in the back. Although the car’s motion is complicated, we can apply
Newton’s first law to its center of mass. This subject will be explored further in a dedicated section.
Additionally, a car driving on the surface of the Earth is following a curved path, even if it’s on a straight road. This
is due to Earth’s rotation. We will usually analyze the motion of the car not from the point of view of an observer in
an inertial reference frame in outer space, but from our own point of view standing on Earth. Doing so requires the
introduction of fictitious centrifugal and Coriolis forces, to which we also dedicate a small section. The Coriolis and
centrifugal forces are small enough that they are simply ignored in most physics problems, and the Earth reference
frame is considered inertial.
Newton’s first law can be viewed in combination with the principle of relativity and statics. According to the
principle of relativity, you can analyze a physical situation from any inertial reference frame and the basic equations
used will be the same. In Newtonian mechanics, this means that forces are the same in all inertial reference
frames, and so are accelerations.
Suppose that an object moves with constant velocity ~v in some frame S. Then we can adopt a new reference frame
S 0 so that any given point in S 0 moves at speed ~v in S. In the frame S 0 , the object will have velocity zero (at all
times). When we studied statics, we learned that an object that always has velocity zero is in static equilibrium and
has zero net force on it in frame S 0 . The principle of relativity says that the net force on the object is zero in frame S
as well. This is Newton’s first law.
Problem 2.1.
In deep space, an astronaut throws a baseball. The baseball continues away from the astronaut indefinitely with
constant velocity. After the astronaut (or their spacesuit) is no longer touching the ball, what force makes the
ball continue going forward?
(a) The force of the astronaut’s throw, which the baseball carries forward with it.
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Newton’s Laws
(b) The force of the baseball itself, which never dies out in outer space.
(c) The force of the spin the astronaut put on the ball.
(d) The force of whatever the ball is going to crash into eventually.
(e) The gravitational force; the ball cannot keep going without gravity.
(f) There is no force on the baseball after the astronaut throws it.
Solution:
The baseball is traveling with constant velocity. By Newton’s first law, there is no net force on the baseball. In fact,
there are no forces on it at all. No force is necessary to make the ball continue going forward, so the answer is
(f) .
Many students think of force as “traveling with an object”. They might think that an object thrown upward near the
surface of Earth has a force of the throw which gradually gets used up by gravity, at which point the ball stops and
changes directions. These ideas are much closer to the concept of momentum than force. Momentum is a
conserved quantity which can pass from object to object and can be carried by an object. Force, in contrast,
describes the rate that momentum flows into an object. The baseball is carrying momentum with it as it travels
through deep space, and the momentum is constant in time. This means no momentum is flowing into or out of
the baseball. There are no forces on it.
Problem 2.2.
Suppose the baseball in the previous problem is spinning. Spinning is a circular motion that implies acceleration.
How is this possible if there is no force on the baseball?
Solution:
If we take the entire baseball as a system, then Newton’s first law says that because there is no net force on the
baseball, its center of mass moves with constant velocity. So although the baseball can spin, it will spin in such a
way that its center of mass moves in a straight line. Suppose there is an insect on the ball, crawling around and
redistributing mass. (Let’s ignore the mechanism by which the insect stays alive.) Then the ball could shake back
and forth a bit as it flies through space. The rotational motion could include wobbling as well. But however the ball
shakes or wobbles in its flight, it will do so in such a way that the center of mass of the insect-ball system will
continue moving in a straight line at constant speed.
On the other hand, suppose we take as our system an individual seam on the edge of the ball. This seam, relative to
the ball, is moving in a circle as the ball spins. The seam is undergoing centripetal acceleration, and there must be
a net force on it. There is such a force. It is a force from the rest of the baseball on the seam. Perhaps the string
from which the seam is made will be under tension, or the cowhide covering through which the seam runs will exert
friction forces on the seam. Whatever the mechanism, there is a force from the rest of the baseball on the
seam.
There is also a force from the seam on the rest of the baseball. By Newton’s third law, these forces are equal in
magnitude and point in opposite directions. When we consider the entire baseball as a system, the net force on
that system is still zero because the sum of the force from the seam on the rest of the baseball and the force from
the rest of the baseball on the seam is zero.
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Newton’s Laws
Problem 2.3.
A planet orbits a star at constant speed in a circle. Is there a force on the planet?
(a) Yes, the force of gravity pushing the planet forward in its orbit.
(b) Yes, the force of gravity pulling the planet towards the star.
(c) No. The speed is constant, so there is no force on the planet.
Solution:
Newton’s first law only applies to situations of constant velocity. Constant speed is not enough. If the direction of
motion is changing, the velocity is changing, and by Newton’s first law, there must be a net force on the object. In
this case, that means there is a net force on the planet because as time goes on, its velocity vector rotates in a
circle.
Newton’s first law alone is not enough to determine the direction in which the force points, but we can use other
knowledge to determine the direction. Forces that point in the direction of motion of an object speed it up without
changing its direction. Forces that point opposite the direction of motion slow the object down without changing
its direction. Forces that point perpendicular to the motion do not change the object’s speed. In this case, the
speed of the planet is constant, so the force on it must be perpendicular to its motion. The correct choice is (b) ,
there is a force pointing toward the star.
Problem 2.4.
A propeller-drive airplane of mass 105 kg flies at constant velocity. The density of air is 0.3 kg/m3 . The plane’s
speed is 240 m/s, and its drag coefficient is CD = 0.02. The lift-to-drag ratio of the plane is 12. What is the total
force from air on the plane?
Solution:
The plane experiences forces from two sources: Earth (the gravitational force) and air. The forces from the air can
be broken down into lift and drag, but this is not necessary to solve the problem. Because the plane flies at
constant velocity, by Newton’s first law, the net force on it is zero. The gravitational force is downward and has a
magnitude of 106 N. The total force from air on the plane is therefore upward and has magnitude 106 N .
Exercises
2.1. You are sitting in your tree house. You have a bucket on a long rope. You lower it down to the ground at
constant velocity, then pull it back up, again at constant velocity. The mass of the rope can be neglected. Which of
the following is true of the constant-velocity periods while you are raising / lowering the bucket?
(a) You must exert upward force on the rope in each case, but more upward force when raising the bucket.
(b) You must exert the same upward force on the rope in each case.
(c) You must exert a downward force on the rope to lower it and an upward force on the rope to raise it.
(d) You must exert an upward force on the rope to lower it and a downward force on the rope to raise it.
(e) You only need to exert a force on the rope to raise the bucket. On the way down, you don’t need to exert any
force on the rope.
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Newton’s Laws
2.2. A bumblebee can fly vertically upward with a maximum speed v1 , and downward with a speed v2 . Assuming
that the bumblebee’s thrust F does not depend on the direction of flight and the air resistance is proportional to the
bumblebee’s speed, determine the bumblebee’s maximum speed with it flies at an angle α with the
horizontal.
(Boris Korsunsky, Quantum Magazine, V3N3 pp 25)
3
Newton’s Second Law
Newton’s second law says
F~net = m~a.
(1)
Objects accelerate in the direction of the net force on them. This is not necessarily the same as the direction of
movement, so objects do not necessarily move in the direction of the force on them.
Like the first law, Newton’s second law only applies in inertial reference frames (unless we make special
accommodations to adapt it, as we’ll see in the final sections of this handout). The second law tells us the direction
of the acceleration of the center of mass, a point we’ll return to in the next section.
As the second law is at the core of Newtonian mechanics, we’ll do several example problems.
Problem 3.1.
A mass m1 is suspended from the ceiling by a string. A second mass m2 is suspended from the first mass by
a spring of spring constant k. The string is cut. What is the acceleration of the mass m1 immediately after the
string is cut?
m1
m2
Solution:
The only physical insight required is that the tension in the spring remains the same momentarily when the string is
cut, since the length of the spring does not immediately change.
From here we’ll show three approaches. (All three are equivalent, of course. Which do you find most
intuitive?)
1. Since the lower mass is at rest, the net force on it is zero. This net force is the sum of the tension in the spring
upwards and its weight m2 g downwards. Hence the tension in the spring is T = m2 g. After the string is cut, the
upper mass experiences two forces, both downwards: the tension T and its weight m1 g. From Newton’s second
law
Fnet = ma
we have
m2 g + m1 g = m1 a
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Newton’s Laws
a=
(m1 + m2 )g
.
m1
2. The tension in the string before it is cut is equal in magnitude to the weight of the entire spring-and-mass
system, (m1 + m2 )g, by similar reasoning as before. Since the net force on the upper mass is zero at this time, the
other forces on the mass must total (m1 + m2 )g downward. After the string is cut only these other forces remain,
so again from Newton’s second law
(m1 + m2 )g = m1 a
and we can solve to get
a=
(m1 + m2 )g
.
m1
3. (A preview of material we’ll explore further below.) Consider the spring-and-mass system as a whole. After the
string is cut the only external force on this system is its weight (m1 + m2 )g, so its momentum must change at a
rate
dp
= Fnet = (m1 + m2 )g
dt
But the forces on the lower mass do not change at the moment the string is cut, so its acceleration remains zero.
Any change in the momentum of the system must be due to acceleration of the upper mass:
dp
= m1 a
dt
Combining these, once again
(m1 + m2 )g = m1 a
and
a=
(m1 + m2 )g
.
m1
Problem 3.2.
Prove the work-energy theorem for a point-like object subjected to a constant force. Assume the object moves
in one dimension.
Solution:
We imagine that the object has some initial velocity v0 and its mass is m.
Suppose a force F acts on the object for a time ∆t. Then by Newton’s second law, the object experiences a
constant acceleration, a. If its initial velocity is v0 , its final velocity is
(2)
vf = v0 + a∆t
Solving for ∆t,
∆t =
vf − v0
a
(3)
Because the acceleration is constant, the particle’s average velocity is halfway between its initial and final
velocities, so its distance traveled is
vf + v0
d=
∆t
2
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(4)
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Newton’s Laws
Plugging ∆t from equation (3) into equation (4), we have
vf + v0 vf − v0
d=
2
a
(5)
If we multiply both sides by ma and expand the right hand side, we get
mad =
1
m(vf2 − v02 ).
2
(6)
Substituting F for ma in the previous equation,
Fd = ∆
1
mv 2
2
(7)
We interpret the right hand side of this equation as the change in kinetic energy, ∆T , and the left hand side as the
work, W . Then the equation can be written concisely as
W = ∆T.
Problem 3.3.
Suppose a particle moving in one dimension begins at x = 0, moving with velocity v0 . The particle is subject to
a retarding force
F = −kv n
(8)
Find the total distance traveled by the particle.
Solution:
We present a partial solution and leave the remainder of the solution as an exercise.
The force on the particle is
(9)
F = −kv n
Using Newton’s Second Law to replace F with ma, then using the definition a =
dv
= −kv n
dt
(10)
dv
k
= − dt
n
v
m
(11)
m
Separating variables,
dv
, we have
dt
Integrating,
v 1−n
k
v 1−n
=− t+ 0
1−n
m
1−n
where we have chosen the constant of integration so that v(0) = v0 .
Then solving for v,
v=
v01−n +
(12)
1/(1−n)
(n − 1)k
t
m
(13)
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Newton’s Laws
The total distance traveled is
∞
Z
(14)
vdt
d=
0
Carrying out the integral, we find
v02−n m
2−n k
(15)
However, we caution that this result does not hold for all n. (See exercise 3.2)
It’s often useful to see another way to do a calculation, so below is an alternative route.
Newton’s second law tells us
m
But
dv
dt
=
dv dx
dx dt
by the chain rule, and
dx
dt
dv
= −kv n
dt
is just v, so
dv
= −kv n
dx
dv
k
= − v n−1
dx
m
mv
Separating variables,
Z
0
v0
m dv
−
=
k v n−1
Z
d
dx
0
and the same result follows.
Exercises
3.1. Return to problem 3.2 deriving the work energy theorem. Prove the work-energy theorem in one dimension
even when the force is non-constant,
Z
x2
∆T =
F (x)dx
(16)
F~ · d~x
(17)
x1
If you dare, prove the three-dimensional version,
Z
∆T =
C
where
R
C
denotes a line integral over an arbitrary curve.
3.2. Return to problem 3.3 on the particle experiencing a drag force F = −kv n . For which n is the result derived in
that problem true? What happens for other n?
3.3. A worker W is pushing a packing crate of mass m1 = 4.2 kg. In front of the crate is a second crate of mass
m2 = 1.4 kg. Both crates slide across the floor without friction. The worker pushes on crate 1 with a force
P1W = 3.0 N. Find the accelerations of the crates and the force exerted by crate 1 on crate 2.
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Newton’s Laws
(Robert Resnick, David Halliday, Kenneth S. Krane. Physics, Volume 1, 5th Edition (Page 55).)
3.4.
British Olympiad Round 1, 2018 Question 3
3.5.
Australian Exam 2016 q13
4
The Center of Mass
Many physics problems treat objects as points. In these cases, all the forces on an object act at the same point,
simply because the object itself is modeled as a point.
There are several ways in which the situation can be more complicated.
• If a problem involves rigid bodies, force can be applied to different parts of an object, causing it both to
accelerate and rotate. We studied rigid bodies last week in every problem involving torques.
• If a problem involves deformable bodies, they can additionally change their shape. We studied these objects
last week when learning about the Young’s modulus and stress and strain.
• If a system consists of multiple sub-parts, each can move and accelerate independently.
In many cases involving these extended systems, we can simplify the analysis by considering the system’s center
of mass. As we mentioned in sections (2) and (3), Newton’s first and second laws apply to the center of mass of an
extended system. (So does the third law, covered in section 6.)
For example, suppose a pinata is flying through the air, following some parabolic trajectory, when the pinata
explodes mid-flight The pinata pieces and candy will go in all directions, but their center of mass continues to move
along the same parabola. (Ignoring air resistance and other effects other than a uniform gravitational field).
Or, suppose there’s a ruler on your (frictionless) desk which you push around with your finger. You can push on the
center of the ruler, or on some off-center point. If you push on the center of the ruler, you’ll simply make it
accelerate. If you push on the off-center point, you’ll make the ruler spin as well as accelerate, but in each case
you’ll give the center of mass of the ruler the same acceleration.
To define the center of mass, we take an average location of all the mass in a system. For example, if there were
x0
two particles, each of mass m, one at x = 0 and one at x = x0 , the center of mass would be at
.
2
If there were three particles, each of mass m and at locations x1 , x2 , and x3 , the center of mass would be at
1
xcm = (x1 + x2 + x3 )) .
3
What if there were two particles, one of mass m at x1 and one of mass 2m at x2 ? This is the same as having three
1
particles with the second two at location x2 , so xcm = (x1 + 2x2 ). We can rewrite that as
3
1
xcm =
(x1 m + (2m)x2 ) . What we see is that in the sum, each particle’s position is weighted according to its
3m
mass, and we divide by the total mass to get an average position.
Turning to continuous systems, but retaining the same idea, we can define the position of the center of mass
by
Z
1
~xcm =
~xdm,
M
where M is the total mass of the system.
We are often given the mass density instead of an infinitesimal mass element dm. In one dimension, we might
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Newton’s Laws
dm
. Then the center of mass is given by
dx
Z b
1
λ(x)xdx,
=
M a
write the mass density, mass per unit length, as λ =
xcm
(18)
where a and b are the limits of the system.
M is found via
Z
b
M=
(19)
λ(x)dx.
a
In three dimensions,
~xcom
1
=
M
Z
(20)
ρ(~x)~xdV.
V
Here, the integral is taken over the entire volume of the system, and ρ(~x) is the mass per unit volume at location
~x.
Intuitively, the center of mass of a rigid body is a point where, if you push on it, it accelerates without rotating. If you
have a cardboard cutout of your state, you could balance it on the head of a pin if that pin were under the center of
mass of the state.
The center of mass is not a point so that half the mass is on either side. For example, imagine a dumbbell
consisting of two points masses, M and m. Suppose M > m. The point masses are connected by massless rod.
The center of mass of this system is somewhere in the middle of the rod, closer to M than to m. The entire mass
M is on one side of the center of mass and the entire mass m on the other side. So it is not half the mass on either
side of the center of mass.
Here we will focus on applying the center of mass concept to dynamics. For practice locating the center of mass of
a given mass distribution, see the references. Aside from brute-forcing integrals, there are several nice tricks in the
Feynman Lectures.
Problem 4.1.
A star of mass M is instantaneously stationary at the origin in some given reference frame. A planet of mass
m M is orbiting the star at a distance r in a circular orbit. Find the approximate location of the star after a
time t, which is very long compared to the orbital period.
M
m
Solution:
The basic idea is that there are no external forces on the system, so its center of mass obeys Newton’s first law
and moves in a straight line at constant velocity. Because the star is much more massive than the planet, the
center of mass position is nearly the same as the position of the star, so if we find how the center of mass moves,
we find approximately how the star moves on average over time. In more detail, the star moves in a small orbit
around the center of mass of the system, but this small motion is regular and cyclic. After many periods of rotation,
it has only a small and ignorable contribution to the location of the star.
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Newton’s Laws
We would like to find the initial velocity of the center of mass, since this velocity is constant. To find the velocity, we
can find the location of the center of mass and differentiate. The location is
~rcom =
m~rm + M~rM
.
m+M
(21)
The velocity of the center of mass is the time derivative of equation (21),
~vcom =
m~vm + M~vM
.
m+M
(22)
This equation is worth examining briefly. Multiplying both sides by the total mass of the system,
(23)
(m + M )~vcom = m~vm + M~vM .
The right hand side is the total momentum of the system. The left hand side is the total mass times the velocity of
the center of mass. This is a general result for any system: the total momentum is equal to the total mass times
the velocity of the center of mass.
We know that the original velocity of the star is zero, ~vM = 0.
To find the original velocity of the planet, we note that the force on it comes from Newton’s gravitational law,
Fm =
GM m
r2
(24)
The planet can be thought of as moving in a circle of radius r around the sun (although this is not exact, because in
fact the planet orbits around the center of mass of the system, so we will find an approximate answer).
Then the acceleration of the planet comes approximately from the formula for acceleration in uniform circular
motion
v2
am ≈ m
(25)
r
According to Newton’s second law, if we multiply equation 25 by m, we have Fm , the force on the planet. This
means we can use equation (25) to eliminate Fm from equation (24). The result is
2
mvm
GM m
≈
.
r
r2
Solving for vm , we find
r
vm ≈
(26)
GM
.
r
(27)
Using vm = |~vm | and returning to equation (22), we have
vcom
m
≈
m+M
r
GM
,
r
(28)
or more simply
r
vcom ≈ m
G
.
Mr
(29)
r
So after a long time t, the location of the star is approximately tm
G
away from its starting location in the
Mr
same direction as the original velocity of the planet.
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Newton’s Laws
See the exercises for an approach to this problem when the mass of the planet is not much smaller than that of the
star.
Problem 4.2. A uniform rod stands vertically on the ground. The rod is given a negligibly-small push sideways
and falls. The friction with the ground prevents the rod from sliding. What angle θC does the rod make with the
ground at the moment the rod loses contact with the ground?
θ
θC
Solution:
The rod’s center of mass accelerates according to Newton’s second law,
~acom =
F~net
m
(30)
We are concerned with when the rod lifts off the table. We note that the table can only provide upward normal
forces, not downward ones. One way to find the point where the rod lifts off the table is to assume that the rod
does not lift off the table and find the normal force from the table on the rod. If the normal force is negative (i.e. if it
points downward), our solution does not make physical sense and our assumption that the rod has not lifted off
the table must be wrong. So the rod lifts off the table just at the moment that the normal force goes from positive
to negative. That is, at the moment the normal force equals zero.
At that moment, the only force on the rod in the vertical direction is the gravitational force, so the y-component of
the rod’s center-of-mass acceleration is
ay; com = −g.
(31)
In order to find when the rod lifts off the table, we will need to find when, assuming it stays on the table, the vertical
component of its acceleration is −g. To do that, we need to understand the acceleration of the center of mass of
the rod in more detail, assuming that the rod is touching the table.
The center of mass of the rod moves on a circle, so it has centripetal, or radial acceleration pointing toward the
point of contact with the table. The rod also speeds up as it falls, so the center of mass has tangential acceleration
as well.
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Newton’s Laws
at
ar
θ
Taking the vertical components of these accelerations,
(32)
ay;com = −ar sin θ − at cos θ.
The radial acceleration is
ar =
1 2
`ω
2
(33)
at =
1
`ω̇.
2
(34)
and the tangential acceleration is
From plugging equations (33) and (34) into equation (32), the acceleration in the y direction is
1 2
1
acom;y = −
`ω sin θ + `ω̇ cos θ .
2
2
(35)
Setting this acceleration to −g gives us an equation for when the end of the rod will leave the table:
g=
1 2
1
`ω sin θc + `ω̇ cos θc .
2
2
(36)
In order to use these equations, we need both ω and ω̇. We can find the acceleration of the center of mass of the
rod using conservation of energy. The total energy of the rod at the moment when it’s standing upright is
gravitational potential energy,
1
E = `mg
(37)
2
where ` is the length of the rod and m is its mass. (Because the initial push is very small, we can ignore any initial
1
kinetic energy.) The fraction comes from the fact that the center of mass of the rod is a distance `/2 above the
2
ground when the rod is stood on end.
When the rod is falling, it has both gravitational potential energy and kinetic energy. The gravitational potential
energy can be found from the height of the center of mass multiplied by the weight, and the kinetic energy can be
1
found from the equation T = Iω 2 .
2
1
11 2 2
E = `mg sin θ +
m` ω
(38)
2
23
1
where we have used I = m`2 .
3
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Newton’s Laws
Putting equation (38) into equation (37) to eliminate E, we have
1
Solving for ω 2 ` sin θ, we have
2
1
11 2 2 1
lmg =
m` ω + m`g sin θ.
2
23
2
(39)
1 2
3
ω ` sin θ = g(1 − sin θ) sin θ.
2
2
(40)
This equation will give us half of the right hand side of equation (39) in terms of only variables in the problem. For
the other half of the right hand side, we need ω̇.
To find ω̇, we can use the rotational form of Newton’s second law, τ = I ω̇. We take the end of the rod as a pivot
point, so the torque is from the gravitational force. We have
`
Solving for ω̇ cos θ, we have
2
1
1
mg` cos θ = m`2 ω̇.
2
3
(41)
`
3
ω̇ cos θ = g cos2 θ.
2
4
(42)
Plugging the results from equations (41) and (42) into equation (39), we have
g=
3
3
g(1 − sin θ) sin θC + g cos2 θC .
2
4
(43)
Using cos2 θC = 1 − sin2 θC , this is a quadratic equation in sin θC , equivalent to
3 sin2 θC − 2 sin θC +
The solution is
sin θC =
1
= 0.
3
(44)
1
3
(45)
or
(46)
θC ≈ 0.34 .
Exercises
4.1. Revisit problem (4.1). In that problem, we approximated the radius of the circular path followed by the planet.
We also worked in a frame where the path of the planet is not exactly a circle. Rework the problem eliminating
M
these approximations. Note that the factor
is the factor by which mass is reduced in the reduced mass,
M +m
which we will study further when we work on orbits in lesson 4.
4.2. A tube of glycerol (a very viscous liquid) sits on a scale. A metal ball of mass m is falling through the glycerol
at constant speed v. How does the reading on the scale depend on the height of the ball? What is the momentum
of the glycerol if the density of the ball is ρb and the density of the glycerol is ρg ?
4.3. A “pizza slice” is a section of a circular disk with angle θ and radius r. The center of mass of the pizza slice is
on the symmetry line of the pizza slice, some distance rcm from the tip. Find rcm as a function of r and θ. You may
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Newton’s Laws
2
the way along a median. Try to approach this exercise as a
3
dynamics problem instead of an integral found via the definition of the center of mass. (We will revisit this problem
in the homework, so you could delay working it until after class.)
use the fact that for a triangle, the center of mass lies
5
Momentum and Force
The discussion in this section closely follows that of Morin’s Introduction to Classical Mechanics, Appendix
C.
In section (2) we mentioned that the net force on a system can be thought of as the rate of flow of momentum into
the system. This picture replicates Newton’s second law in most cases.
Suppose we define the net force on a system to be equal to its rate of change of momentum. Then
d
F~net = (mvcom ),
dt
(47)
where we have used the result, mentioned in section (4), that the momentum of a system is equal to its mass times
the velocity of its center of mass.
If the mass of the system is constant, then
d~vcom
F~net = m
= m~acom .
dt
(48)
This is Newton’s second law. Our conclusion is that, so long as the mass of a system is constant (so ṁ = 0), the
rate of change of momentum of the system is equal to the external force on it. For this reason, we can think of
force as a flow of momentum.
For example, you take a book, drop it from a high height, and it lands on a table. At the moment you drop the book,
it has no momentum. There is a downward gravitational force on the book, which means downward momentum
flows into the book. This means the book picks up some velocity and falls. The book falls faster and faster because
it is continually gaining more downward momentum from the gravitational force on it. Then the book hits the table.
By observation, the book stops very suddenly. It very quickly loses all its downward momentum, which means there
was a huge flow of downward momentum into the table - the book exerted a very large downward force on the
table. The table, in turn, exerted a very large upward force on the book. This resulted in a sudden flow of upward
momentum into the book, leaving it with zero momentum total. Finally, the book rests stationary on the table.
Downward momentum continues to flow into it via the gravitational force, but upward momentum flows into it at
the same rate due to the normal force from the table, and the book remains in a state of zero momentum.
In situations where the mass of the system can change, the situation is more subtle. For example, consider a
situation where sand is continually being poured into a train, and you are pushing the train forward.
µ
F
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Newton’s Laws
Suppose the train is moving at constant velocity. Then the acceleration of the train is zero, and the net force on the
train is zero. However, we cannot conclude that you do not have to push on the train. That’s because it’s the net
force on the train that’s zero, and there is a force on the train from the sand.
A useful trick to solve problems such as this one is to consider the system consisting of the train and all the sand
in it. This system’s mass changes, but despite the change in mass, we can still apply the equation
d
F~net = p~total
dt
(49)
to the system.
Problem 5.1.
Suppose you are pushing the train at speed v. Sand enters the cart at mass per unit time ṁ. Find the force you
must exert on the cart.
Solution:
The train-sand system gains momentum ṁv per unit time, so the force you must exert is ṁv .
Exercises
5.1. I hear there’s a class called PhysicsPOOT where they try to teach you physics, but often get it all backwards.
The PhysicsPOOT instructors solved problem (5.1) as an example in their class, and here’s the solution they
gave:
Work in the reference frame of the train. In this frame the train and the sand in it always have zero
velocity, so their total momentum is always zero. Hence the force exerted is zero.
Who’s right - PhysicsWOOT or PhysicsPOOT? What’s the mistake made by whoever got it wrong?
5.2.
6
BPhO 2017, Round 1, question (e)
Newton’s Third Law
Robert Goddard was an early proponent of rocketry, and published his ideas about how rockets could travel into
outer space, where there is no atmosphere. On January 13, 1920, the New York Times published an editorial which
said,
That professor Goddard, with his “chair” in Clark College and the countenancing of the Smithsonian
Institution, does not know the relation of action to reaction, and of the need to have something better
than a vacuum against which to react - to say that would be absurd. Of course he only seems to lack the
knowledge ladled out daily in high schools.
Newton’s third law might seem to tell us that if there is a force on a rocket from something, then there is also a
force from the rocket back onto the something! And because in space, there is nothing around to play the role of
“something”, rockets can’t propel themselves in space.
To see why this is wrong, let’s return to the astronaut and the baseball, first considered in section (2). If we think of
the astronaut and the baseball together as a single system, then by Newton’s first law, the center of mass of this
system has constant velocity. Because there are no external forces on this system, this is true even if the astronaut
throws the baseball. If the entire system were initially stationary and the astronaut threw the baseball to the right,
for example, the astronaut would drift slowly left so that the center of mass of the system remained stationary.
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Newton’s Laws
(And if the astronaut and ball were originally drifting, the center of mass of the system would remain drifting with
the same velocity, even if the astronaut threw the ball.)
We can think of the rocket in space as analogous to the astronaut. By expelling fuel at high speed behind it, the
rocket can be pushed forward. In other words, the “something” a rocket pushes against is its own fuel.
Because the momentum of a system is the total mass times the velocity of the center of mass, the momentum of
the astronaut+ball system or the rocket+fuel is conserved - the mass doesn’t change, and neither does the velocity
of the center of mass.
The New York Times did eventually admit Goddard was right.
Further investigation and experimentation have confirmed the findings of Isaac Newton in the 17th
century, and it is now definitely established that a rocket can function in a vacuum as well as in an
atmosphere. The Times regrets the error.
They published that retraction on July 17, 1969, the day after the launch of Apollo 11. So do not worry if you catch a
mistake in PhysicsWOOT. Just let us know, and we’ll try to fix it in a bit less than 49.5 years.
Now let us think about the astronaut and the ball as two separate systems. If the astronaut throws the ball to the
right, they put rightward momentum into the ball. By momentum conservation, the ball puts leftward momentum of
equal magnitude into the astronaut. Further, if the astronaut is in the process of throwing the ball and putting
rightward momentum into it at some rate, the ball is simultaneously putting leftward momentum into the astronaut
at the same rate.
We saw in section (4) that the rate of change of momentum of a system is the net external force on it. So the rate
that the astronaut is putting rightward momentum into the ball is also the force that the astronaut exerts on the
ball. The ball exerts a force on the astronaut of the same magnitude but in the opposite direction. This is Newton’s
third law,
F~a→b = −F~b→a
(50)
The force from the astronaut on the ball and the force from the ball on the astronaut are called a third-law
pair.
We have seen Newton’s third law as a consequence of momentum conservation, but angular momentum is
conserved as well. The rate that angular momentum flows into a system is the net external torque on the system.
For forces to conserve angular momentum, therefore, they need to obey another condition on top of Newton’s third
law, which is that both forces in any third-law pair have the same moment arm about any axis.
Many students have some difficulty identifying third-law pairs at first. It is helpful to remember that any two forces
acting on the same object cannot be a third-law pair because third law pairs are always of the form of a force from
a to b and another force from b to a. For example, when you stand stationary on the ground, there is a gravitational
force downward on you and a normal force upward from the ground. These forces are equal in magnitude and
opposite in direction. They are applied at the same location (wherever your feet touch the ground), so they have the
same moment arm about any axis. Still, they are not a third-law pair. They cannot be, because they are both forces
on you. These forces are in fact not always equal in magnitude. For example, if you jump, then while you are
accelerating upward, the normal force on you from the ground (the force that pushes you upward while you jump)
is greater than the gravitational force.
The third-law pair for the gravitational force on you is the gravitational force that you exert on the Earth. The third
law pair for the normal force that the ground exerts on you is the normal force that you exert on the ground.
The centrifugal and Coriolis forces we will study in section (8) do not have third law pairs. These forces only appear
in accelerating reference frames, where Newton’s laws require modification.
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Newton’s Laws
Problem 6.1. A child catches some insects and keeps them in a jar. They place the jar on a scale. When is the
reading on the scale higher - when the insects are sitting on the bottom, or when they are flying around?
Solution:
The reading we get when placing the jar on a scale is the normal force between the jar and the scale.
Whether the insects are sitting on the bottom or flying around, the center of mass of the insects+jar+(air in jar)
system does not accelerate, so the net force on it is zero. In the vertical direction, the only forces are a gravitational
force and a normal force. The gravitational force is not affected by whether the insects are flying around, so neither
is the normal force (because it has the same magnitude as the gravitational force because the net force in the
vertical direction is nearly zero). So the scale reading is the same.
It is possible that the scale reading will fluctuate a little. The scale reading depends on the acceleration of the
center of mass of the jar+insects+air system. The center of mass could accelerate up or down for a little while as
the insects fly. If the reading were consistently lower or higher than baseline, though, it would indicate that the
center of mass of the jar were consistently accelerating up or down. Because the jar is not flying up into the air or
burrowing down into the table, over time, the average acceleration of the center of mass must be close to zero, and
the scale reading must be the same, on average, as if the insects were sitting on the bottom of the jar.
To see this result with Newton’s third law, we consider an insect hovering in the middle of the jar and compare this
to the “usual” situation in which no insect is in the jar at all.
The insect experiences a downward gravitational force. In order to remain in place, it must experience an upward
force from the air equal to its own weight. By Newton’s third law, the insect exerts a downward force equal to its
own weight on the air. Then, because the air on net is not accelerating, the bottom of the jar must exert an upward
force on the air larger than usual by the weight of the insect. By Newton’s third law, this means the air exerts a
downward force on the jar larger than usual by the weight of the insect. Then, because the jar is not accelerating,
the normal force from the scale on the jar must be larger than usual by the weight of the insect. By Newton’s third
law, this means the normal force from the jar on the scale is larger than usual by the weight of the insect. This
means the reading on the scale is larger than usual by the weight of the insect.
This is just the same result we get when the insect sits on the bottom of the jar - the reading on the scale goes up
by the weight of the insect.
Problem 6.2.
Ignore gravity in this problem.
A spring is connected to a U -shaped tube of uniform cross-section A with water of density ρ sitting stationary
in it. Next, the water is pushed through the tube at speed v. Find the deflection of the spring in the steady state.
~v
−~v
Solution:
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Newton’s Laws
To find the deflection of the spring, we need to find the force from the tube on the spring. By Newton’s third law, this
is the same as the force from the spring on the tube.
The tube itself isn’t accelerating, so the force from the spring on the tube is equal in magnitude and opposite in
direction from the force from the water on the tube (although these forces are not a third law pair).
By Newton’s third law, the force from the water on the tube is equal in magnitude and opposite in direction to the
force from the tube on the water. So to find the deflection of the spring, we need to find the force from the tube on
the water.
The force from the tube on the water is the external force accelerating the water as it changes directions in the
U -tube. To find this force, we can calculate the rate of change of momentum of the water.
A small bit of water with mass dm approaches the spring with velocity ~v , then turns around and has velocity −~v .
Then change in its velocity was −2~v , so the change in its momentum was
(51)
d~
p = −2~v dm.
~v is a constant because we are considering the steady state. (The velocity of an individual piece of water changes,
but that is not ~v . Instead ~v has been defined as the velocity of whatever water is approaching the spring.) This
means that if we take the time derivative of equation (51), we find
dm
dp
= −2~v
.
dt
dt
(52)
This quantity is the force on the water, and therefore also tells us the magnitude of the force on the spring.
Fspring = 2v
dm
.
dt
(53)
Here, Fspring is the magnitude of the force on the spring.
Using Hooke’s law, Fspring = k∆x for the spring, we have
k∆x = 2v
dm
.
dt
(54)
The rate that mass flows through the tube, dm/dt, is vρA, so
k∆x = 2v 2 ρA.
(55)
2v 2 ρA
.
k
(56)
or
∆x =
Exercises
6.1.
If a can of compressed air is punctured and the escaping air blows to the right, the can will move to the left in a
rocket-like fashion. (Top picture). Now consider a vacuum can that is punctured in the air. The air blows in to the
left as it enters the can. After the vacuum is filled, what direction (if any) does the can move?
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Newton’s Laws
(Epstein, Lewis Carroll. Thinking physics)
6.2. In a classic problem, a horse is attached to a buggy. The horse wants to pull the buggy forward, so it exerts a
force on the buggy. The buggy, by Newton’s third law, exerts a backward force of the same magnitude on the horse.
These forces sum to zero, so how can the horse and buggy ever go anywhere?
6.3. A rope of mass m and length ` is held vertically above a scale, with the bottom of the rope touching the scale.
The rope is released and falls onto the scale. Find the reading of the scale over time.
As usual in physics problems, this is a magical scale which can read out the force on it in real time with no delay.
Assume that the rope cannot support any compression forces, and that it collides perfectly inelastically with the
scale.
7
Uniformly-Accelerating Reference Frames
We examined uniformly-accelerating reference frames last class, where we solved some problems with constant
acceleration by using uniformly-accelerating frames. This let us turn dynamics problems into statics
problems.
In the statics lesson, we stated that when we transform from an inertial frame to a frame accelerating at ~a, every
object of mass m experiences a force −m~a in the accelerating frame.
To see this, let’s suppose that the position vector for an object in some inertial frame is ~r, and in a frame uniformly
accelerating with acceleration ~a, the position vector is r~0 .
Consider a particle p at the origin in the inertial frame, so ~rp = 0. In the accelerating frame, this particle appears to
be accelerating with acceleration −~a, so
1
r~0 p = − ~at2 .
(57)
2
Next consider a particle at a time-dependent position ~r. The displacement between this particle and the particle p
is the same in all frames, so
~r − ~rp = r~0 − r~0 p .
(58)
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Newton’s Laws
Using ~rp = 0, this becomes
~r = r~0 − r~0 p
(59)
We can use equation (59) to eliminate r~0 p from equation (57), obtaining
Solving for r~0 , we find
1
~r = r~0 + ~at2 .
2
(60)
1
r~0 = ~r − ~at2 .
2
(61)
This gives us the position of the particle in the accelerating frame. If we take two time derivatives and multiply by
m, the mass of the particle, we get
d2~r
d2 r~0
(62)
m 2 = m 2 − m~a.
dt
dt
d2~r
is the net force on the particle in an inertial frame, F~net; inertial . We can use the
dt2
left hand side of the equation as the definition of the net force on the particle in the accelerating reference frame,
F~net; accel . So we have
F~net; accel = F~net; inertial − m~a.
(63)
On the right hand side, the term m
This extra force −m~a on the particle is called an “inertial force” or a “fictitious force”. The idea is that, if we include
this fictitious force and all the physical forces (gravitational forces, friction forces, etc.), then Newton’s second law,
F~net; accel = m~a0
(64)
is obeyed in the accelerating frame, where ~a is the acceleration of the particle in the accelerating frame.
0
Some people are very worried about whether the the force −m~a “is real” or is “just a trick to make Newton’s laws
work”, etc. etc. Fortunately, no physics contest problems has ever required students to write a philosophical essay
about the ontological status of this force. We can instead just use it to solve problems.
We reiterate the note from part (6) that the force −m~a in the accelerating frame does not obey Newton’s third law.
There is no reciprocal force m~a on some other object. Consequently, momentum is not conserved in accelerating
reference frames.
The inertial force is exactly analogous to a gravitational force from a uniform gravitational field. In particular, it can
be considered to act at the center of mass of extended bodies.
Problem 7.1.
A certain rope is pulled from the right-hand side. The rope has uniform density, but gets thinner towards the left
in such a way that the stress in the rope is uniform when the rope is accelerated to the right. Find the thickness
of the rope as a function of distance from the right-hand side of the rope.
Solution:
We go into a frame where the rope is stationary. This frame accelerates to the right with acceleration ~a, so in this
frame, all pieces of the rope of mass dm experience a force −dm a to the right (i.e. a force dm a to the left).
Consider a thin slice of the rope of thickness dx, extending from x to x + dx. This piece of rope is stationary and in
equilibrium. It experiences three forces: tension on the right, tension on the left, and an inertial force dma to the
left.
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adm
TL
TR
(The inertial force acts at the center of mass of this slice, but we’ve drawn it higher to avoid overlapping with the
tension force.)
The sum of these forces is zero, so relating just their magnitudes,
(65)
adm = TR − TL
The right hand side of this equation is a force to the right due to tension (the rightward tension is slightly higher than
the leftward tension) and the left hand side of the equation is a force to the left, which is the inertial force.
The stress σ in the rope is assumed to be constant. The tension is
(66)
T = Aσ
where A is the cross-sectional area of the rope.
TR − TL is the change in tension dT across the infinitesimal section of rope, so we have
(67)
TR − TL = dT = σdA.
Combining equation (65) with equation (67) to eliminate TR − TL , we have
adm = σdA.
(68)
aρAdx = σdA.
(69)
aρ
dA
dx =
.
σ
A
(70)
aρ
x + c = ln(A).
σ
(71)
The mass dm is ρAdx, so
Separating variables,
Integrating both sides,
Solving for A,
aρ
x
A = A0 e σ .
(72)
It might superficially appear that this result says that the shape of the rope depends on the acceleration a, so that a
rope like this one must be finely-tuned to a particular acceleration, and if we pulled with a different acceleration, the
stress would not be uniform throughout the rope.
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This is not so. If we were to pull the rope with twice the acceleration, the stress σ would also be doubled, and the
a
ratio would be unchanged.
σ
There is no unique solution to the problem; the profile
(73)
A = A0 ex/L
works for any L, and the stress in the rope is given by
(74)
σ = Laρ.
Problem 7.2.
A train accelerates horizontally at acceleration a. A point mass m hangs from a massless rope of length ` from
the ceiling of the train. What is the frequency of small-amplitude oscillations of the mass?
Solution:
In a frame accelerating along with the train, the mass experiences an inertial force of magnitude ma in the
horizontal direction. This can be thought of as indistinguishable from a gravitational force of the same magnitude
in the horizontal direction, to combine with ordinary gravitational force mg in the vertical direction. By the
Pythagorean theorem, the combined effective gravitational force on the mass is
p
(75)
mgeff = m g 2 + a2 .
The frequency of a simple pendulum with small oscillations is
r
1
geff
f=
.
2π
`
(76)
The frequency of oscillations is therefore
1
f=
2π
sp
g 2 + a2
.
`
(77)
Exercises
7.1. Reconsider the derivation of the inertial force −m~a added to free body diagrams in accelerating frames. Does
the derivation of this force depend on the acceleration ~a being constant in time? If the acceleration were a varying
function of time, ~a = ~a(t), could we add the force −m~a(t) to an object and find it obeying Newton’s laws in the
time-varying accelerating reference frame?
7.2. Consider objects in free fall near the surface of the Earth. In what reference frame do they move in straight
lines? Briefly explain why objects appear to be weightless on the ”Vomit Comet”, an airplane which flies in parabolic
trajectories with acceleration ~g .
7.3. A block of mass m sits on a frictionless wedge of mass M. The wedge sits on the flat frictionless ground.
The wedge makes an angle θ with the horizontal. Find the acceleration of the wedge.
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8
Uniformly Rotating Reference Frames
In section (2), we mentioned that the surface of Earth is not an inertial reference frame because the Earth is
spinning. The reference frame in which Earth is stationary is a uniformly rotating reference frame. In such frames,
there is more to the inertial force than in the uniformly-accelerating frames we studied in the previous
section.
Last week, in statics, we introduced the centrifugal force. It applies to all objects in a rotating frame and pushes
them away from the axis of rotation with a force of magnitude
(78)
Fcentrifugal = mω 2 r.
But consider an object that is stationary in an inertial frame. There are no forces on it in that frame. Now, we will
view this object in a frame rotating at angular frequency ω around an axis of rotation through the origin. In this
frame, the object (and all other objects) experience a centrifugal force of magnitude mω 2 r directed away from the
axis of rotation. Additionally, the object moves in a circle at angular frequency ω.. This means that the object
accelerates towards the origin , while the centrifugal force throws it out away from the origin!”
Given the circular path of the object in this reference frame, the net force on the object in the rotating reference
frame must be mω 2 r towards the origin (from the formula for acceleration of an object moving in uniform circular
motion) and, as discussed above, the centrifugal force has magnitude mω 2 r in the direction away from the origin.
So there must be another inertial force pushing the object toward the origin, and its magnitude must be 2mω 2 r.
This force is the Coriolis force.
The Coriolis force has magnitude
(79)
FCoriolis = 2mωv sin θ
where v is the speed of the particle and θ is the angle between the velocity vector of the particle and the axis of
rotation.
The Coriolis force points in the direction perpendicular to both the velocity and the axis of rotation. Which of the
two perpendicular directions it points depends on the direction of spin of the rotating coordinate system. We saw
in the previous example that for a particle that is stationary in an inertial frame, the Coriolis force pushes it toward
the origin in a rotating frame.
For those familiar with vector cross products, the Coriolis force and its direction can be written
F~Coriolis = −2m~
ω × ~v .
(80)
This equation can be used to find the direction of the Coriolis force via the “right hand rule” for cross products. For
a basic reference on cross products, see this pdf from MIT OpenCourseWare.
The Coriolis force is important to planet-sized motions, such as the circulation of ocean currents, winds, and
convection in the mantle of the Earth. (This last problem is extremely complicated because the Earth is conductive
and has a magnetic field, and there are large temperature gradients, so mechanics, electromagnetism, and
thermodynamics all play important roles.)
The physics of rotating reference frames is only partially included in the IPhO syllabus, which lists
Inertial and non-inertial frames of reference: inertial force, centrifugal force, potential energy in a
rotating frame.
The list does not include the Coriolis force, but some problems do test the phenomena it describes. We won’t go
over the origin of the Coriolis force in great detail, but a good and short reference is section 19-4 of the Feynman
Lectures.
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Problem 8.1. A child in a circular, rotating space station tosses a ball in such a way that once the station has
rotated through one half rotation, the child catches the ball. From the child’s point of view, which plot shows the
trajectory of the ball? The child is at the bottom of the space station in the diagram below, but only the initial
location of the ball is shown.
(F=ma 2018, Exam B Q12)
Solution:
First, let’s figure out how it’s possible that the child both throws and catches the ball, imagining we are observers
watching from an inertial reference frame. In this reference frame, the spaceship spins. The child throws the ball at
the bottom and catches it at the top.
catch
throw
There are no forces on the ball between the toss and the catch, so by Newton’s first law it must have moved in a
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Newton’s Laws
straight line between between them.
catch
throw
Next, let’s plot some locations of the child and the ball during their path.
catch
5
5
4
4
3
3
2
2
1
1
throw
Consider when the child and ball are both at location 1. Let’s draw in the line from the child to the center of the
station (dashed black line). This is what the child considers “straight ahead” and corresponds to the vertical axis
through the center of the image in the answers.
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Newton’s Laws
catch
5
5
4
4
3
3
2
2
1
1
throw
The dashed red line is from the child to the location of the ball. Evidently, the child sees the ball as being off to the
left.
We can draw the same lines for other positions.
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catch
5
5
4
4
3
3
2
2
1
1
throw
At position 3, when the child is halfway around the space station, the ball reaches the center of the station. Then it
appears to move to the right as it approaches.
This heuristic approach narrows the choices down to (C) and (D), and if we were more careful about it, we could
see that the answer is (C). For example, consider the first moments when the child throws the ball. In (D), the ball
would have to move directly horizontally from the child’s point of view, which is not the case.
This heuristic method works, and indeed can be used to find an exact equation for the path of the ball as seen by
the child (by finding the coordinates of the ball in the inertial frame, then transforming those coordinates into the
coordinates in the rotating frame).
Another method is to analyze the dynamics in the rotating frame directly. In this frame, the ball experiences both a
centrifugal force, which pushes it out away from the origin, and a Coriolis force, which pushes at a right angle to the
motion of the ball. We can think about the answer choices in the problem based on these forces.
Choice (A) and choice (E) do not show the ball returning to the child, so we don’t need to consider them.
Choice (B) shows the ball moving purely radially in the rotating frame. But in that case, there would be a Coriolis
force on the ball pushing sideways, and we don’t see that happening, so that answer can’t be right.
The ball moves in a circle in (D), which is the motion when a constant force is always perpendicular to the motion,
the way the Coriolis force is. Choice (D) shows a path that obeys only the Coriolis force, and no centrifugal
force.
Only choice (C) remains. The centrifugal and Coriolis forces combine to form this shape.
Incidentally, the shape was not drawn perfectly; a more-accurate image for the trajectory of the ball is
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27
Newton’s Laws
catch
throw
Exercises
8.1.
Consider the following passage from Bad Astronomy, a book by Philip Plait.
Let’s say you are driving a car north at 100 kph (60 mph) in Wiscasset, Maine. The Coriolis effect
deflects you by the teeny amount of 3 millimeters (0.1 inch) per second. After a solid hour of driving,
that amounts to a deflection of only 10 meters (33 feet). You couldn’t possible notice this.
How do you think the author came up with the figures he quoted? What did he do wrong?
8.2.
Show that the angle between a freely-hanging rope and the direction to Earth’s center is approximately
Rω 2 sin(2θ)
tan α =
, where ω is Earth’s rotational angular velocity, θ is the latitude, R the radius of Earth, and g
2g
gravitational acceleration near Earth’s surface. Assume that shape of the Earth is a perfect sphere (Taylor
9.13).
8.3. The rotation of hurricanes is caused by the Coriolis force. Explain which direction hurricanes rotate in each
hemisphere.
8.4. A river near the North pole flows south, directly away from the pole, at speed v. The width of the river is w.
Which bank of the river is higher? By how much?
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Newton’s Laws
9
Further Reading and Problems
The sources in this section are optional and contain more complete explanations of the material here, additional
practice problems, or both.
9.1
Newton’s Laws
Halliday, Resnick, Krane Physics, 5th ed. 2002. Ch. 3,4,5
Feynman, Richard. Lectures on Physics Chapter 9
Morin, David. Problems and Solutions in Introductory Mechanics, 2014. Ch. 4 for practice problems
Morin, David. Introduction to classical mechanics: with problems and solutions. Cambridge University Press, 2008.
Ch 3.
IsaacPhysics has more practice questions.
Kalda, Jaan. Mechanics Sections 2 and 4 contain many challenging problems
9.2
Center of Mass
Halliday, Resnick, Krane Physics, 5th ed. 2002. Section 7-4.
Kowalski, Stan. Momentum, Systems of Particles, and Conservation of Momentum Section 10.5 and on
Feynman, Richard. Lectures on Physics Lecture 18 and Lecture 19. Especially lecture 19 contains useful
tricks.
9.3
Momentum and Force
Morin, David. Introduction to classical mechanics: with problems and solutions. Cambridge University Press, 2008.
Appendix C.
9.4
Accelerated and Rotating Reference Frames
Morin, Problems and Solutions in Introductory Mechanics, 2014. Ch. 12
Morin, David. Introduction to classical mechanics: with problems and solutions. Cambridge University Press, 2008.
Ch 10.
Feynman, Richard. Lectures on Physics. Chapter 19-4 gives a view of Coriolis forces based on energy.
Kalda, Jaan. Mechanics Appendix 4 contains a brief derivation using vector notation; several problems are
scattered throughout the document
Taylor, John R. Classical mechanics. University Science Books, 2005. Ch 9. (advanced)
Vandiver, J. Kim, Fictitious Forces and Rotating Mass, MIT Opencourseware. (video, 72 min)
9.5
Cross Product
Cross Product, MIT OpenCourseWare
Right Hand Rule, Wikipedia. (Most Wikipedia articles are good basic references in physics.)
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Newton’s Laws
10
exercise answers
2.1: (b)
2.2: v =
1
2
q
(v2 − v1 )2 sin2 α + 4v1 v2 − (v2 − v1 ) sin α .
3.3: a = .536 m/s2 for each crate. F1→2 = 0.75 N.
r
G
4.1: tm
in the direction of the original motion of the planet
(m + M )r
4.2: The scale reading is constant and does not depend on v. The momentum of the glycerol is m
ρg
v.
ρb
4 sin(θ/2)
r
.
3
θ
5.1: PhysicsPOOT is wrong. Those chuckleheads!
4.3: rcm =
6.1: The can is stationary after the vacuum is filled.
6.2: The buggy is pushed forward by the force from the horse. The force that the buggy exerts on the horse is
irrelevant to the buggy’s motion because that is not a force exerted on the buggy. The horse is pushed force by a
friction force from the ground. This force must be larger than the force from the buggy on the horse in order for the
horse to accelerate.
6.3:
F =
3 g 2 `t2
,
2 m
2l
g
2l
t>
g
0<t<
F = mg,
7.1: No, the derivation does not rely on ~a being constant in time. Yes, we can add the force −m~a(t) even when ~a
varies in time.
7.2: Objects in free fall move in straight lines in any frame accelerating toward Earth’s center at ~g .


 sin θ cos θ 
7.3: a = −g 
.
M
sin2 θ +
m
8.4: The west bank of the river is higher by about w
2ωv
2π
, where ω =
.
g
24 h
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