174 Week 3: Work and Energy
Problem 7.
H
v
m
R
R
A block of mass M sits at rest at the top of a frictionless hill of height
H leading to a circular frictionless loop-the-loop of radius R.
a) Find the minimum height Hmin for which the block barely goes
aro

= g
bm
v2 = b mv2 mg b (188)much as before. Also as before, we divide
all of the stu with a v in it to the left, multiply the dt to the right,
integrate, solve for v(t), set the constant of integration, and answer the
questions.
Ill do the rst few steps i

gasoline to drive the same distance at 100 kph ( 62 mph) than they
do at 50 kph, it isnt four times as much, or even twice as much. Why
not?
Things to think about include gears, engine eciency, fuel wasted
idly, friction, streamlining (dropping to Fd = bv

Eventually, the drag force will balance the gravitational force and the
object will no longer accelerate. It will fall instead at a constant speed.
This speed is called the terminal velocity.
It is extremely easy to compute the terminal velocity for a fal

=
1 MtotX i
mi
d2~xi dt2
=
1 MtotX i
mi
d~vi dt
(348)
188 Week 4: Systems of Particles, Momentum and Collisions
and can integrate twice on both sides (as usual, but we only do the
integrals formally). The rst integral is: d ~ X dt = ~ V = 1 MtotX i
mi~vi

given and discussed in this textbook (or any other), but do it yourself
and without looking after studying.
Problem 3.
L
m
k
D?
at rest
A block of mass m slides down a smooth (frictionless) incline of length
L that makes an angle with the horizontal as sh

Example 3.5.1: Block Sliding Down a Rough Incline
Suppose a block of mass m slides down an incline of length L at an
incline with respect to the horizontal and with kinetic friction
(coecient k) acting against gravity. How fast is it going (released
from

m x
=
dm dx
= lim x0
m A
=
dm dA
= lim x0
m V
=
dm dV
In each of these expressions, m is the mass in a small chunk of the
material, one of length x, area A, or volume V . The mass
distribution of an object is in general a complicated
91Near the Earths s

out of an airplane, for example) might also be expected to behave
within reason like a single object independent of the motion of the
baseballs inside, or the motion of the atoms in the baseballs, or the
motion of the electrons and quarks in the atoms.
We

So any time a mass moves down a distance H, its change in potential
energy is mgH, and since total mechanical energy is conserved, its
change in kinetic energy is also mgH the other way. As one increases,
the other decreases, and vice versa!
This makes ki

x
Hmax
Figure 41: A spring compressed an initial distance x res a mass m
across a smooth (k 0) oor to rise up a rough k 6= 0) incline.
How far up the incline does it travel before coming to rest?
In gure 41 a mass m is released from rest from a position o

three dimensions. In two dimensional collisions we are a bit better o
we have three conservation equations (two momenta, one energy)
and four unknowns (four components of the nal velocity) and can
solve the collision if we know one more number, say the a

The energy in a given system is not, of course, usually constant in
time. Energy is added to a given mass, or taken away, at some rate. We
accelerate a car (adding to its mechanical energy). We brake a car
(turning its kinetic energy into heat). There are

m1
A block of mass m2 sits on a rough table. The coecients of friction
between the block and the table are s and k for static and kinetic
friction respectively. A much larger mass m1 (easily heavy enough to
overcome static friction) is suspended from a ma

v
m
Figure 40:
Here is a lovely problem so lovely that you will solve it ve or six
times, at least, in various forms throughout the semester, so be sure
that you get to where you understand it that requires you to use three
dierent principles weve learned

bushy tails or skin webs like those observed in the ying squirrel59
have a much lower terminal velocity than (say) humans and hence
have a much better chance of survival. One rather imagines that this
provided a direct evolutionary path to actual ight fo

c) Determine the speed v0 the ball must have at the bottom to arrive at
the top with this minimum speed. You may use either work or potential
energy for this part of the problem.
Week 3: Work and Energy 177
Problem 10.
H
m
vR
A block of mass M sits at the

if you like, to an actual chemical interaction between them) followed
by a hard core repulsion as the electron clouds are prevented from
interpenetrating by e.g. the Pauli exclusion principle. This second
potential energy well is often modelled by a Lenna

evaluating the algebraic expressions for the x and y components of the
center of mass separately:
xcm =
1 MtotX i
mixi =
18
(20 + 21 + 32 + 12) = 1.25 m (355)
ycm =
1 MtotX i
miyi =
18
(20 + 30 + 21 + 12) = 0.5 m (356)
Hence the center of mass of this sys

K = Kf Ki =
P2 tot 2Mtot p2 1,i 2m1
+
p2 2,i 2m2!
In a partially inelastic collision, the particles collide but dont quite
stick together. One has three (momentum) conservation equations and
needs six nal velocities, so one in general must be given three

basically slowly unstable) they are classically bound for a while
but eventually escape to innity.
However, in nature pairs of atoms in the metastable conguration
have a chance of giving up some energy (by, for example, giving up a
photon or phonon, where

The rst you should already be familiar with as the angular velocity,
the second is the angular acceleration. Recall that the tangential speed
vt = r; similarly the tangential acceleration is at = r as we shall see
below.
Work through the following exercis

dx (in one dimension) or xcm =Rx dm Rdm =Rx dA R dA = 1
MtotZ x dA and ycm =Ry dm Rdm =Ry dA R dA = 1 MtotZ y dA
(in two dimensions).
The Momentum of a particle is dened to be:
p~ = m~v
The momentum of a system of particles is the sum of the momenta of
t

87Again, more advanced math or physics students will note that these
should all be partial derivatives in correspondance with the force
being the gradient of the potential energy surface U(x,y,z), but even
then each component is the local slope along the

At the top (where we expect v to be at its minimum value, assuming it
stays on the circle) gravity points straight towards the center of the
circle of motion, so we get:
mg + N =
mv2 R
(312)
and in the limit that N 0 (barely looping the loop) we get the
c

Kids, dont try this at home! But if you ever do happen to fall out of an
airplane at a few thousand feet, isnt it nice that your physics class
helps you have the best possible chance at surviving?
Example 2.2.2: Solution to Equations of Motion for Stokes

=
dPi p~i dt
=
dp~tot dt
In this expression the internal forces directed along the lines between
particles of the system cancel (due to Newtons Third Law) and:
~ Ftot =X i
~F
ext i =
dp~tot dt
where the total force in this expression is the sum of only th

o
R
A ball of mass m is attached to a (massless, unstretchable) string and
is suspended from a pivot. It is moving in a vertical circle of radius R
such that it has speed v0 at the bottom as shown. The ball is in a
vacuum; neglect drag forces and friction

K < 0 forbidden
bE
Quadratic region
Classically forbidden domains (shaded)
K = E U > 0 allowed
Figure 44: The same potential energy curve, this time used to
illustrate turning points and classically allowed and forbidden regions.
Understanding the role of

below, one that I commend to all students when confronted by
problems of this sort.
Example 2.2.3: Dropping the Ram
The UNC ram, a wooly beast of mass Mram is carried by some
naughty (but intellectually curious) Duke students up in a helicopter
to a heigh

a number of further opportunities to solve equations of motion this
and next semester that are rst order, linear, inhomogeneous ordinary
dierential equations such as this one.
Given v(t) it isnt too dicult to integrate again and nd x(t), if we
care to, bu

To outline the solution, following a previous homework problem, we
write: F mg = ma (329)
or
a=
F mg m
(330)
We integrate twice to obtain (starting at y(0) = 0 and v(0) = 0):
v(t) = at =
F mg m
t (331)
y(t) =
12
at2 = 1 2
F mg m
t2 (332)
(333)
From this w

tanh1 (x/a) a (190) Now you know what those rarely used buttons
on your calculator are for. We substitute x > v, a mg/b, multiply
out thepmg/b and then take the hyperbolic tangent of both sides and
then multiply bypmg/b again to get the following result f

v0
L
m
L
H
Figure 39: Find the maximum angle through which the pendulum
swings from the initial conditions.
Heres the same problem, formulated a dierent way: A mass m is
hanging by a massless thread of length L and is given an initial speed
v0 to the righ

accelerations ~ai). From this, we construct a weighted average
acceleration of the system, in such a way that Newtons Second Law is
satised for the total mass.
internal equal and opposite force pairs between particles that help
hold the system together (s