This formally proves that my guess of = I as being the correct
form of Newtons Second Law for this simple rotating rigid body holds
up pretty well no matter where we apply the force(s) that make up the
torque, as long as we dene the torque:
= rFFt = rFF
millimeter is still absurdly large on an atomic scale. We could make
the cube 1 micron (1106 meter, a thousandth of a millimeter) and
because atoms have a generic size around one Angstrom 11010
meters we would expect it to contain around (106/1010)3 = 101
physics! As a rule in this course, if you are a math, physics, or
engineering major I expect you to go the extra mile and nish o the
algebra, but if you are a life science major who came into the course
terried of anything involving algebra, well, Im prou
will need to do just a teensy bit of work to show that this is true and
extract any essential conceptual insight to be found along the way.
Week 5: Torque and Rotation in One Dimension 243
5.3: The Moment of Inertia
We begin with a specic example to help
symmetric across the axis of rotation. Now each of the two masses has
a torque acting on it due to the rod connecting it with the origin, each
mass has a vector angular momentum that at right angles to both ~r
and ~v, but the components of ~ L in the x-y
c) Compute the average force being exerted on the stream of beads by
the pan over a second (assuming that N 1, so that many beads
strike the pan per second). d) Use Newtons Third Law to deduce the
average force exerted by the beads on the pan, and from th
This problem will help you learn required concepts such as:
Center of Mass of Continuous Mass Distributions Integrating Over
Distributions
so please review them before you begin.
In the gure above a rod of total mass M and length L is portrayed
that has
(block at rest)
This problem will help you learn required concepts such as:
Momentum Conservation The Impact Approximation Elastic
versus Inelastic Collisions The Non-conservative Work-Mechanical
Energy Theorem
so please review them before you begin.
In
In the gure above, two identical billiard balls of mass m are sitting in
a zero gravity vacuum (so that we can neglect drag forces and gravity).
The one on the left is given a push in the x-direction so that it
elastically strikes the one on the right (wh
where = d dt = d2 dt2 is the angular accleration of the particle.
Although the magnitude of vr = 0, we note well that the direction of
~vt is constantly changing and we know that ar = v2/r = r2 which
we derived in the rst couple of weeks and by now have u
This now reveals the only point where we have to do real work in this
frame (or any other) nding If around the center of mass! Lots of
opportunities to make mistakes, a need to use Our Friend, the Parallel
Axis Theorem, alarums and excursions galore. Howe
Im not going to quite nish this one for you, as there are a lot of
things one can ask and it is a homework problem. But I do want you to
get a good start.
The spool in gure 67 is wrapped many times around with string. It is
sitting on a level, rough table
mass right after the collision is exactly the same as that of the free rod
found above.
Week 6: Vector Torque and Angular Momentum 295
m v0
m
M
M
f
L
pivot
Figure 87: A blob of putty of mass m travelling at speed v0 strikes the
rod of mass M and length L
pivot r is a constant but the angle can vary in time as external
forces act on the system.
The very rst things we need to do are to bring to mind the set of
rotational coordinates that
107In particular, we solved elastic collisions in the center of mass
f
~ L = ~rp~ = ~rm~v (623) which points up and to the left at the
instant shown in gure 88. Note well that ~ L is perpendicular to both
the plane containing ~r and ~v, and that as the mass moves around in
a circle, so does ~ L! In fact the vector ~ L sweeps
vo
o
fv
This problem will help you learn required concepts such as:
Newtons Third Law Momentum Conservation Fully Inelastic
Collisions
so please review them before you begin.
In the gure above, a large, heavy truck with mass M speeds through
a red light
This is displayed in gure 61. A massless rod as long or longer than
both rm and rF is pivoted at one end so it can swing freely (no
friction). The mass m is attached to the rod at the position rm. A force
~ F is applied to the rod at the position ~rF (on
1
2
2
Figure 70: Atwoods Machine, but this time with a massive pulley of
mass M and radius R. The massless unstretchable string connecting
the two masses rolls without slipping on the pulley, exerting a torque
on the pulley as the masses accelerate to mat
32
max (509)
or:
ax =
23
g sin() (510)
We can then substitute this back into the equation for fs above to get:
fs =
12
max =
13
mg sin() (511)
In order to roll without slipping, we know that fs sN or 1 3 mg
sin() smg cos() (512)
or
s
13
tan() (513)
If s
The four places where the tires are in contact with the pavement, of
course. Those points arent sliding on the pavement, they are rolling,
and rolling means that they are coming down at rest onto the
pavement and then lifting up again as the tire rolls on
before, the limit of this sum as N is by denition the integral, and
in this limit the sum exactly represents the moment of inertia of the
rod.
We can easily evaluate this. To do so, we chant our ritual expression:
The mass of the chunk is the mass per uni
the lungs, where it is oxygenated and returned to the heart by means
of pulmonary veins. These two distinct circulations do not mix and
together, form a closed double circulation loop. The heart is the
pump that serves both systemic and pulmonary circulat
and to keep the pulley accelerating up with the string, the string has to
exert a torque on the pulley due to the unequal forces.
Week 5: Torque and Rotation in One Dimension 253
One last thing to note. We are being rather cavalier about the normal
force
d dt
c) The angular acceleration as a function of time,
(t) =
d dt
=
d2 dt2 Hopefully the analogy between these one dimensional
angular coordinates and their one dimensional linear motion
counterparts is obvious. Forces applied to a rigid object
perpendic
we have already introduced for doing kinematics of a rotating object.
Since r is xed, the position of the particle is uniquely determined by
the positive angle (t), measured by convention as a counterclockwise
rotation about the z-axis from the +x-axis as
through the symmetric center of mass of the object, because we can
use the parallel axis theorem and the perpendicular axis theorem to
nd the moments of inertia around at least some alternative axes.
Shape Icm
Rod from L/2 to L/2
1 12
ML2
Ring MR2
Disk
12
6.5.1: More General Collisions
As you can imagine, problems can get to be somewhat more dicult
than the previous two examples in several ways. For one, instead of
collisions between point masses that stick and rods (pivoted or not) one
can have collisions
F
r2
1r
mm12
y
rF
Figure 62: A single torque = rFF sin() is applied to a rod with two
masses, m1 at r1 and m2 at r2.
The WKE theorem for this picture is now (note that v1 and v2 are both
necessarily tangential):
dW = rFF sin() d = d = dK = m1v1 dv1 + m2v2
which works because is the same for all chunks dm in the blob and
is hence a constant that can be taken out of the integral, leaving us
with the integral for I.
If we combine this with the theorem proved at the end of the last
chapter we at last can preci
Once tiny vortices start to appear and reach a certain size, they rapidly
grow with the velocity gradient and cause a change in the character of
the drag force coupling across the uid. There is a dimensionless scale
factor called the Reynolds Number Re162