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Review:
Uniform Circular Motion
In Part 2 we looked at uniform circular motion:
r = constant (def. of circular motion)
q = t (q changes with time at uniform rate
where  = constant)
vr = 0 (radius doesn’t change)
vq = r (where  = Dq/Dt)
ar = 2r (v changes direction toward center)
aq = 0 (angular speed doesn’t change)
Angular Acceleration
Here we generalize circular motion to include
the case where the angular speed can
change. We define angular acceleration
as: a = D/Dt .
[Recall arclength = s, and q=s/r, so s=rq, vq=r]
Since aq = Dvq/Dt = D(r)/Dt = r(D/Dt) = ar.
We still have all the equations we had before,
except aq is no longer zero but is instead ar.
NON-uniform Circular Motion
Rectangular viewpoint
Circular motion is defined by: r = constant.
We still have dq/dt = , but  is no longer
held constant. This means that d/dt = a
rather than zero.
Converting polar to rectangular, we have:
x = r cos(q) = r cos[q(t)]
y = r sin(q) = r sin[q(t)] .
Non-uniform Circular Motion
Rectangular viewpoint
x = r cos[q(t)]
y = r sin[q(t)]
Now we use: vx = dx/dt and vy = dy/dt:
vx = dx(t)/dt = dx(q)/dq * dq/dt =
r (- sin(q) ) *  = - r sin(q)
vy = dy(t)/dt = dy(q)/dq * dq/dt =
r (+ cos(q) ) *  =  r cos(q)
Non-uniform Circular Motion
Rectangular viewpoint
x = r cos(q)
y = r sin(q)
vx = - r sin(q)
vy =  r cos(q)
This v is the same as for
uniform circular motion
Now we use: ax = dvx/dt and ay = dvy/dt. Since
 is not constant, we need to use the product rule.
Acceleration
vx = - r sin(q)
vy =  r cos(q)
Now we use: ax = dvx/dt and ay = dvy/dt. Since
 is not constant, we need to use the product rule:
ax = dvx/dt = d[- r sin(q) ]/dt
= r {d[-]/dt} sin(q) + - r d[sin(q)]/dt
= r {d[-]/dt} sin(q) + - r [dsin(q)]/dq]*dq/dt
= -a r sin(q) + -2 r cos(q) .
In a similar way, we get:
ay = +a r cos(q) + -2 r sin(q) .
Non-uniform Circular Motion
Back to Polar
To convert back to polar for position, we use
the inverse transformation equations:
x = r cos(q)
y = r sin(q)
r = [x2 + y2]1/2 = [r2 cos2(q) + r2 sin2(q)]1/2
= r[cos2(q) + sin2(q)]1/2 = r
q = inv tan[y/x] = inv tan[ r sin(q) / r cos(q)]
= inv tan[tan(q)] = q .
Non-uniform Circular Motion
Back to Polar
To convert back to polar for velocity, we use the inverse
transformation equations:
vx = - r sin(q)
vy =  r cos(q)
v = [vx2 + vy2]1/2 = [2r2 sin2(q) + 2r2 cos2(q)]1/2
= r[sin2(q) + cos2(q)]1/2 = r
qv = inv tan[vy/vx] = inv tan[r cos(q) / -r sin(q)]
Note that cos(q) = sin(q+90o) and –sin(q) = cos(q+90o) so
qv = q  90o . Note that the direction of the velocity,
qv, is perpendicular to the direction of the position
(the radius), which means the velocity is tangent to
the circle.
Non-uniform Circular Motion
Back to Polar
qv = q  90o
Note that the direction of the velocity, qv, is
perpendicular to the direction of the
position (the radius), which means the
velocity is tangent to the circle.
Non-uniform Circular Motion
Back to Polar
To convert acceleration:
ax = - a r sin(q) + -2 r cos(q)
ay = +a r cos(q) + -2 r sin(q) .
a = [ax2+ay2]1/2 = [a2r2 sin2(q) + 2a2r2sin(q)cos(q)
+ 4r2 cos2(q) + a2r2 cos2(q) - 2a2r2sin(q)cos(q) +
4r2 sin2(q)]1/2 = [(ar)2 + (2r)2]1/2 = [at2+ar2]1/2.
We can identify (2r) as the radial acceleration due to the turning of the velocity, and we now
identify (ar) as the tangential acceleration - due
to the changing speed of the velocity.
Non-uniform Circular Motion
Back to Polar
ax = - a r sin(q) + -2 r cos(q)
ay = +a r cos(q) + -2 r sin(q) .
a = [at2+ar2]1/2 where at = ar and ar = 2r.
If we look at just the ar terms, then the direction of these
terms is: qat = inv tan[ cos(q) / -sin(q) ] = q+/- 90o
(just like for qv which is tangent to the circle),
And if we look just at the 2r terms, then the direction of
these terms is: qar = inv tan[ -sin(q) / -cos(q) ] q+/- 180o
(which is towards the center).
Constant Angular Acceleration
In the special case of constant angular
acceleration (a = constant), we have equations
analogous to those we had for constant
(regular) acceleration:
d/dt = a = constant, becomes o d = t=0t a dt, or
 = o + at like vx = vxo + axt .
 dq/dt =  becomes qoq dq = t=0t  dt = t=0t (o+at) dt , or
q = qo + ot + (1/2)at2 like x = xo+vxot+(1/2)axt2.
Constant Angular Acceleration
If a car rolls on wheels that do not slip, we can
relate the motion of the car to the spinning of
the wheels:
s = distance of car = arclength of wheels,
v = speed of car = vq of wheels,
a = acceleration of car = aq of wheels;
q = angle of wheel,
 = angular speed of wheel, and
a = angular acceleration of wheel.
Constant Angular Acceleration
For circular motion:
s = rq,
r = constant (circular)
v = r = vq
vr=0
a = ra = aq
ar = 2r (due to turning)
For constant acceleration:
s = so + vot + (1/2)at2
v = vo + at
q = qo + ot + (1/2)at2
 = o + at .
This gives many relations so that, by knowing just a few things, we
can solve for many other things about the motion.
Example
Constant Angular Acceleration
Consider a car that accelerates from rest to a speed of 20
m/s (45 mph) in 10 seconds on wheels (tires) of
outside diameter 66 cm. Assume that the tires roll
nicely and do not slip on the road. Through what
angle (θ) do the tires rotate during the 10 seconds?
What is the final angular speed (ω) at the 10 second
mark? What is the angular acceleration (α) of the
wheels during the 10 seconds?
Example
We can use the equations for constant angular
acceleration, or we can use the relations
between distance and angle.
t = 10 seconds
x = ? v = 20 m/s
t=0
xo= 0 m vo = 0 m/s
a=?
α=?
θo = 0 rad ωo = 0 rad/sec
θ=? ω=?
d = 66 cm, r = d/2 = 33 cm
Example
Since the car is rolling (and not slipping), we
recognize that x = s = rθ, and that
v = vq = r .
Using the equations for constant angular
acceleration along with r = vq :
q = qo + ot + (1/2)at2
 = o + at
v = 20 m/s, t = 10 sec., qo = 0 rad, r = .33 m,
vo = 0 m/s, so o = 0 rad/sec.
Example
(1) r = vq becomes: .33 m *  = 20 m/s
(2) q = qo + ot + (1/2)at2 becomes
q = 0 + 0 + (1/2)*a*(10 sec)2
(3)  = o + at becomes:  = 0 + a*(10 sec)
From Eq. 1:  = (20 m/s) / .33 m = 60.6 rad/s
From Eq. 3: a = /t = (60.6 rad/s) / 10 sec = 6.06 rad/s2.
From Eq. 2: q = (1/2)*a*(10 sec)2
= (1/2)*(6.06 rad/s2)*(10 sec)2
= 303 rad * (1 cycle/2p rad) = 48.2 revolutions
Example – another way
Instead of using the constant angular acceleration
formulas, we could solve the regular motion
and use the relations between regular quantities
and angular quantities:
s = rq,
r = constant (circular)
v = r = vq
vr = 0
a = ra = aq
ar = 2r (due to turning)
Example another way
s = rq,
r = constant (circular)
v = r = vq
vr = 0
a = ra = aq
ar = 2r (due to turning)
s = so + vot + (1/2)at2
v = vo + at
20 m/s = 0 + a*(10 sec),
so a = (20 m/s) / (10 s) = 2 m/s2 , so
a = ar, or a = a/r = (2 m/s2) / (.33 m) = 6.06 rad/s2 .
s = 0 + 0 + (1/2)*a*(10 sec)2 =
(1/2)*(6.06 rad/s2)*(10 sec)2 = 303 radians.
v = r, or  = v/r = (20 m/s) / (.33 m) = 60.6 rad/s.
Rotational Kinematics
The computer homework program,
Rotational Kinematics, Volume 2 #7,
contains explanations and problems
involving Rotational Kinematics using the
equations in the previous slide.
Rotational Force (Torque)
Forces cause change in the motion, but in Part 2
we only considered motion that changes the
position of the object.
What about changing the “spin” or rotation of an
object?
To get a nice introduction to the idea of torque,
see the computer homework program on
Introduction to torque (Vol.2 #5), and
continue with the next slide.
Torque
There are two important quantities in torque:
Force and where you apply the force (called
radius).
t = r x F = r F sin(qrF)
To get a large torque, we need to use a large
radius, a large force, and apply the force
perpendicular to the radius!
Statics and Torque
Just as S F = 0 when the object is static, so
also S t = 0 when the object is not spinning
(or spinning at a constant rate).
In static cases, there is no obvious center to
measure the radius from, so we are free to
choose any point. However, some points
may be simpler to use than others.
Your elbow
Let’s consider as an example of torque how
your muscles, bones and joints work.
Consider holding up a ball of weight 5 lb.
How does this work? (We will choose to measure
distances from the elbow.)
First we draw a diagram:
triceps
= elbow
biceps
weight
rw
rb
Your elbow
In terms of forces and distances, the diagram
looks like this:
Estimate the distance
from your elbow joint
to your biceps connect
point, rb; and to
your hand, rw.
Fc
Fb
rc=0 rb
rw
W
Your elbow
If the ball weights 5 lb, how much force does
your biceps pull up with? How much force
of contact does your upper arm push down
with on your lower arm at the elbow?
What is the basic principle to use? Statics:
S F = 0 and S t = 0.
Your elbow
From S F = 0 we have:
-Fc + Fb - W = 0
And from S t = 0 and measuring from the
elbow gives: Fc*rc + Fb*rb - W*rw = 0 .
We have two equations and we have two
unknowns (Fc and Fb).
Your elbow
We can use the torque equation first, since
rc=0 eliminates one of the unknowns, Fc.
Fc*rc + Fb*rb - W*rw = 0 or Fb = W*rw/rb .
Then we can use the force equation to find Fc :Fc + Fb - W = 0, or Fc = Fb - W.
Your elbow
By putting in reasonable values for rb and rw, you
can see that the biceps have to exert a large
force to hold up a relatively light weight!
What advantage does this give? Note how far the
biceps have to contract in order to move the
weight! This is the advantage of the elbow setup!
In practice, we use clubs and rackets to make this
difference even greater!
Car going around a turn:
Sliding off the road or flipping over
For a car going around a turn, if the car goes too fast the
car will either slide off the road or flip over. What
determines which will occur, and how fast can the car
go around the turn without either happening?
Since we now introduce “flipping over”, we have to
consider rotational motion, and hence consider torque.
This means we need to consider not only what the
forces are, but where they act.
Sliding or Flipping
We’ll consider a car making a right turn on an
unbanked (horizontal) turn.
We have the weight of the car (W=mg), which acts
at the center of gravity.
We have the contact force, which acts on each of the
left and right wheels at the road: FcL and FcR.
We have the friction force, which also acts on each
of the left and right wheels at the road: FfL and
FfR, where Ff  mFc
We also have an acceleration to the right: ac = v2/r .
Sliding or Flipping
The weight, W=mg, acts at the center of gravity,
which is at a height, h, above the road.
The contact forces, Fc, act at the tires which are
on the road. The friction forces, Ff, also
act at the road.
ac = v2/r
The tires are a distance,
w, apart.
h
w
Sliding or Flipping
SFx = FfL + FfR = m v2/r
SFy = FcL + FcR - mg = 0
Let’s choose to consider the center of gravity as the
point for rotations. The distances for the torques
need to be distances that are  to
2/r
a
=
v
c
the direction of the forces.
h
We will also choose clockwise
as positive for torques.
w
Sliding or Flipping
SFx = FfL + FfR = m v2/r
SFy = FcL + FcR - mg = 0
St =
-FfL*h - FfR*h + FcL*(w/2) - FcR*(w/2) + mg*0 = 0.
Since the FcL is the only force causing
a + torque, it must increase and
FcR must decrease if the St = 0.
This can happen until FcR = 0 and
FcL = mg. When this happens, FfR= 0 .
ac = v2/r
h
w
Sliding or Flipping
SFx = FfL + 0 = m v2/r, or FfL = m v2/r
SFy = FcL + 0 - mg = 0, or FcL = mg
St = -FfL*h - 0*h + FcL*(w/2) - 0*(w/2) + mg*0 = 0,
or FfL*h = FcL*(w/2) .
If friction is maximum (limit for sliding) we have:
FfL = m v2/r with Ff = mFc = mmg,
so mmg = m v2/r, or vmax = [mgr]1/2 .
If flipping is to be prevented we have FfL = mv2/r,
FcL=mg, or mv2/r * h = mg*(w/2), or
vmax = [grw/2h]1/2 .
Sliding or Flipping
vmax = [mgr]1/2 applies when sliding is a concern;
vmax = [grw/2h]1/2 applies when flipping over is a
concern.
Which maximum speed is lower - that tells us the
true maximum speed to continue around the turn:
If m < w/2h, then sliding is the main concern;
if m > w/2h, then flipping is the main concern.
For big trucks, h is large, and unless the street is very
slick, flipping over will be bigger; for sports cars,
h is small and sliding is more of a concern.
Statics and Torque
The computer homework program, Statics
and Torque, Volume 2 #6, contain
explanations and problems dealing with this
topic.
Torque and Rotations
Forces (when not balanced) cause changes in
motion. Torques (when not balanced) cause
changes in rotational motion. Forces are
related to acceleration by Newton’s Second
Law: S F = ma. How are t and a related?
Rotational Dynamics
Consider: S F = ma and t = r F sin(qrF) .
If we multiply both sides of S F = ma by
r sin(qrF), we get: S t = m a r sin(qrF) ;
but a sin(qra) = aq = ar, so S t = mr2a .
If the mass is distributed at various radii, then
the mr2 becomes S miri2.
We define the moment of inertia be be
I = S miri2 so that we have S t = Ia .
Moment of Inertia
Note that the moment of inertia relates the
torque to the angular acceleration just
like the mass relates the force to the regular
acceleration.
Note that the moment of inertia depends on
the mass of the object and also on the
shape of the object.
Moment of Inertia
Objects that have most of their mass in the center
have smaller moments of inertia than objects
that have most of their mass on the outside.
Using the calculus, we can replace the summation
of small mass pieces with an integral:
I = S miri2 becomes I = r2 dm.
Each small dm has its own radius, or distance
from the axis of rotation, so r is really a
function of which dm we are using: r(m).
Moment of Inertia
I = r2 dm
It is awkward to express r(m). We can, however,
use density: r = mass/Volume = dm/dV to
convert dm into r dV. Then the integral is
really a Volume integral (triple integral).
I = r2r dV
Using this, we can derive nice formulas for the moment
of inertias of certain shapes: Iring = MR2
Isphere = (2/5)MR2;
Icylinder = (1/2)MR2 .
Sphere versus Cylinder
The blue object in the figures is the side
view of a cylinder with the same radius
and volume (mass) as the orange object
(a sphere).
[The top view would like the same: circles.]
Note that most of the mass is in the same
place, but the top and bottom of the
sphere must be moved away to provide
the corners for the cylinder. Hence,
the Moment of Inertia of the cylinder
is slightly larger than that of the sphere:
Isphere = (2/5)MR2 ; Icylinder = (1/2)MR2 .
axis
R
Rotational Energy
Now that we have the equivalent of Newton’s
Second Law for rotations: S t = Ia , let’s see
if we can get a rotational energy relation.
Recall that Work = F  ds. If we exert a force
to spin an object, the ds becomes a small arc
length that is perpendicular to the radius.
Recall the definition of an angle in radians:
q = s/r, so that ds = r dq. But F times r is
torque, so we get: Work = t  dq .
Rotational Power
Work = t  dq
The definition of power is: P = dWork/dt,
and so for rotations we have:
P = dW/dt = d[t  dq]/dt = t  .
This formula for rotational power is similar to
that for regular power: P = F  v
P=t.
Rotational Kinetic Energy
If something is spinning, it can still not be moving so
that it’s regular kinetic energy is zero. However,
parts of the object are moving, so there is kinetic
energy associated with its spinning. We use three
relations we have already developed:
KE = ½mv2, vq = r, and I = Smiri2
KEtotal = S ½mivi2 = S ½mi(ri)2 = ½2 S miri2
= ½I2 .
This energy is used in the Conservation of Energy
relation just as any other energy term is used.
Angular Momentum
Another important tool in solving problems is
Conservation of Momentum. Is there a
similar tool for rotations?
We already have: p = mv and SF = dp/dt.
If we multiply both sides of Newton’s Second
law by the radius we have (with v = r):
St = r SF = r  dp/dt = d(r  p)/dt = dL/dt
(where we have used the fact that dr/dt  p = v  mv = 0)
where we have defined L = r  p,
where L is called the Angular Momentum.
Conservation
of Angular Momentum
St = dL/dt
Just like SF = dp/dt leads to Conservation of
momentum if no external forces are present, so
St = dL/dt leads to Conservation of Angular
momentum if no external torques are present.
Note: p = mv, and L = r  p =
r  mv = r m vq = r m r = mr2 = I .
Angular Momentum
In class we will observe several demonstrations
about Angular Momentum and talk about its
uses in controlling rockets and spinning
athletes (divers and gymnasts in particular).
We will also show how angular momentum
relates to riding bicycles, and how this relates
to the spinning earth!
Review of Rotational Equations
Basically we replace F with t, m with I, v
with , a with a and p with L (where L is
the angular momentum): I = Smiri2
S F = ma
S t = Ia
Work = = F  ds
Work = t  dq
Power = F  v
Power = t 
KE = (1/2)mv2
KErotation = (1/2)I2
p = mv
L = I
S F = Dp/Dt
S t = DL/Dt .
Rotational Motion
The computer homework program, Volume
2 #8 on Rotational Dynamics gives an
overview and several problems involving
rotational motion.
Example: rolling ball
How fast will a ball be going at the base of a
ramp if it is released at the top of the ramp,
where the length of the ramp is 2 meters and
the high end is 50 cm above the floor (base of
the ramp)? We assume the ball rolls without
slipping. Does the mass or radius of the ball
affect the speed? If a cylinder of the same
mass and radius were also released, would the
ball or cylinder win the race down the ramp?
Rolling ball
Ball:
vi = 0, i=0
mass = m
radiius = r
hi = 50 cm
s = 2 meters
vf = ?
f = ?
Rolling ball
We could use Newton’s Laws of Motion:
SF = ma, and St = Ia along with a=ar and
the equations for constant acceleration, or
we could use Conservation of Energy
(KEregular + KErotational + PEgravity)initial =
(KEregular + KErotational + PEgravity)initial + Elost
along with v=r.
Rolling ball
Since we’re not interested in direction or time,
we’ll use Conservation of Energy.
We know that for a ball (sphere), I = (2/5)mr2.
We know that KEregular = (1/2)mv2 .
We know that KErotational = (1/2)I2.
We know that PEgravity = mgh .
We know that without slipping, even though
friction causes the ball to roll, there is no
Elost because the ball does not slide.
Rolling ball
Conservation of Energy
(KEregular + KErotational + PEgravity)initial =
(KEregular + KErotational + PEgravity)initial + Elost
0 + 0 + mgh = (1/2)mv2 + (1/2)I2 + 0 + 0.
Substituting I=(2/5)mr2 and =v/r gives:
mgh = (1/2)mv2 + (1/2)[(2/5)mr2][v2/r2] , or
mgh = (1/2)mv2 + (1/5)mv2 = (7/10)mv2 .
Rolling ball
mgh = (7/10)mv2 which becomes when
solving for v: v = [(10/7)gh]1/2 =
[(10/7)*9.8m/s2 * .50m] = 2.645 m/s.
Note: both the mass and the radius cancel out
of the solution, and so the mass and radius
don’t matter.
Rolling ball
Note: the angle of the ramp doesn’t make
any difference either, since only the height
and not the length is involved. The steeper
the ramp, the faster the ball will reach its
final speed, but the final speed will not be
affected!
Sphere versus Cylinder
Would a cylinder or a sphere win a race down
the ramp?
What difference does the shape make in our
equation? The difference comes in the KE
for rotations in the moment of inertia, I. For
a sphere (ball), I = (2/5)mr2, while for a
cylinder it would be I = (1/2)mr2. This
ultimately changes the (10/7) factor into
(4/3), the value from 2.65 m/s to 2.55 m/s.
Sphere versus Cylinder
We can understand the slightly lower speed
for the cylinder by considering that the
moment of inertia of the cylinder is larger.
This will make the kinetic energy larger for
the same speed. Hence there will be less
kinetic energy for the regular speed if we
have the same potential energy to begin
with!