Download Rotational Dynamics SL and Honors 2016 2017

Rotational
Dynamics
Torque,
Moment of
Inertia
(Option B1
for SL)
B.1 – Rigid bodies and rotational dynamics (CORE)
Essential idea: The basic laws of mechanics have an
extension when equivalent principles are applied to rotation.
Actual objects have dimensions and they require the
expansion of the point particle model to consider the
possibility of different points on an object having different
states of motion and/or different velocities.
Nature of science:
Modelling: The use of models has different purposes and has
allowed scientists to identify, simplify and analyze a problem
within a given context to tackle it successfully. The extension
of the point particle model to actually consider the dimensions
of an object led to many groundbreaking developments in
engineering.
on bodies
B.1 – Rigid bodies and rotational dynamics (CORE)
Understandings:
• Torque
• Moment of inertia
• Rotational and translational equilibrium
• Angular acceleration
• Equations of rotational motion for uniform angular acceleration
• Newton’s second law applied to angular motion
• Conservation of angular momentum
Applications and skills:
• Calculating torque for single forces and couples
• Solving problems involving moment of inertia, torque and angular
acceleration
• Solving problems in which objects are in both rotational and
translational equilibrium
• Solving problems using rotational quantities analogous to linear
quantities
• Sketching and interpreting graphs of rotational motion
• Solving problems involving rolling without slipping
B.1 – Rigid bodies and rotational dynamics (CORE)
Guidance:
• Analysis will be limited to basic geometric shapes
• The equation for the moment of inertia of a specific shape will be provided
when necessary
• Graphs will be limited to• angular
displacement–time,
The Γ represents
the torqueangular
causedvelocity–
by a force F applied
time and torque–time
at a displacement r from the center of a rotation.
• The θ represents the angle between the F and the r.
Data Booklet
• The I represents the rotational inertia of an extended
reference:
set of point masses.
• Γ = Fr sin θ
• I = Σmr2
• Γ = Iα
• ω = 2πf
• ωf = ω0 + α t
• ωf2 = ω02 + 2αθ
• θ = ω0t + (1/2 )αt 2
• L = Iw
• EKrot = ( 1/2 ) Iω 2
• The α represents the angular acceleration of a rotating
extended object.
• The ω represents the angular frequency and the f
represents the frequency of a rotating extended object.
• The ωf and ω0 represent the final and initial angular
speed of a rotating extended object.
• The θ represents the angular displacement of a rotating
object at a time t.
• The L represents the angular momentum of an
extended object.
• The EKrot represents the rotational kinetic energy of an
extended object (which may also have translational
kinetic energy EK = mv2/2).
B.1 – Rigid bodies and rotational dynamics (CORE)
Theory of knowledge:
• Models are always valid within a context and they are
modified, expanded or replaced when that context is altered
or considered differently. Are there examples of unchanging
models in the natural sciences or in any other areas of
knowledge?
Utilization:
• Structural design and civil engineering relies on the
knowledge of how objects can move in all situations
Aims:
• Aim 7: technology has allowed for computer simulations
that accurately model the complicated outcome of actions
Translational vs. Rotational Motion
In pure translational motion, all points on an
object travel on parallel paths.
The most general motion is a
combination of translation and
rotation.
Rotational Dynamics - Torque
• According to Newton’s second law, a
net force causes an object to have an
acceleration.
• What causes an object to have an
angular acceleration?
TORQUE
• The amount of torque depends on
where and in what direction the force is
applied, as well as the location of the
axis of rotation.
 These 3 scenarios to the right all have
a different torque for the same force
applied.
Rotational dynamics - Torque
Torque
• A torque  is just a force that can cause a rotation about
a pivot point.
• Consider a door as viewed from above:
WALL
WALL
r
F0
θ1θ2 θ3
r2
r1
F1
F2
• The location of the force and its size will determine the
ease with which the door opens.
• The torque is proportional to both the force F and the
moment arm (the distance), r. Thus  = Fr.
• But we note that the angle  between F and r also plays
a role. The closer to 90 the angle is, the more efficiently
the door is opened (or closed).
Rotational Dynamics - Torque
Torque
• In fact, the following equation describes the torque
completely.
 = Fr sin 
definition
where  is the angle between F and r
of torque
• Torque is a vector since it has a direction. For now we can
say clockwise (cw) or counterclockwise (ccw).
• The r is just the distance from the point the force is applied
to the pivot point.
FYI
• Note that torque has the units of a force (N) times a
distance (m) and is thus measured in Nm.
• Recall the work was also measured in Nm, which we called
Joules (J). Never express torque as a Joule though.
Rotational Dynamics - Torque
Lots of IB materials use the letter capital gamma Γ for torque instead of tau τ
torque = τ = F∙d = F∙l
• The lever arm l in a
simple case like
this is the distance
between the axis of
rotation and the
force.
• Note that the force
is perpendicular to
the lever arm here.
• Its not that easy if
its at an angle!!
Lever arm / Moment arm
F
Moment arm (lever arm)
F
F
• Consider a disk that is free to
rotate about its center.
B
A
C
• The application of the
F
r
identical forces to the disk’s
r
r
edge at points A, B, C, and D,
r

will produce very different
D 
outcomes:
• We define the moment arm
or the lever arm as that component
of r that is perpendicular to F.
• It turns out that that component is just r sin  and that
the force F times the lever arm r sin  is the torque.
 = force  moment arm definition of torque (alt.)
Torque – Method 1 for Lever Arm
METHOD # 1
• But what about this
case? Now the force
is acting at about a 45
degree angle to the
door.
• Draw a line from the
axis of rotation, normal
to the line of force.
• The distance from the
axis to the line of force
is the lever arm, l .
Method 1 : Lever Arm
Lots of IB materials use the letter capital gamma Γ for torque instead of tau τ
• So now
• torque = Force x distance
• τ = (F) (LsinФ) = F L sin Ф
• Extend the line of force
backwards with the
dotted black line.
• Draw a line from the
pivot point such that it is
perpendicular to the
extended line of force.
• The distance from the
axis of rotation, O, to
the dotted line of force
is d, the lever arm.
• d = L sin Ф in this figure.
Method 2: Lever Arm
L
•
•
•
•
So now
torque = Force x distance
τ = (FsinФ)(L) = F L sin Ф
OH LOOK! It comes out
exactly the same as
METHOD 1!
• Here’s what you can do
instead (it’s easier I think!)
• Leave the lever arm as the
whole length of the wrench, L.
• Now just take the component
of the force that is
perpendicular to the wrench,
FsinФ, and consider only that
force as being responsible for
creating torque.
• The parallel component has
zero contribution to the torque.
Torque – Lever
Arm Method 2
• METHOD 2
• Note how the force
vector in this figure
has been broken up
into its’ perpendicular and parallel components as
shown.
• This second method (which I like better) is to let the lever
arm be the distance from the axis to the force, shown
here as “r” and then just use the perpendicular
component of the force, shown by F
Rotational Dyamics - Torque
torque = τ = F∙d = F∙Lsinϴ
L
Ok, so looking at both figures now and considering Ф, explain
why method 1 and method 2 will always work out the same.
Depending on the nature of the problem, you may need to have
either of these methods in your “tool box” problem solving kit.
Lots of IB materials use the letter capital gamma Γ for torque instead of tau τ
Torque and lever arm
• In this case, the
line of force
passes right
through the axis
of rotation so
the distance is
zero and the
torque due to
this force is
zero.
Rotational dynamics - torque
Torque example
EXAMPLE: Suppose we apply a force of 80. N to a door
at a distance of 25 cm from the hinge, and at an angle
of 30° with respect to r. Find the torque.
SOLUTION:
• Use  = Fr sin . Then
 = Fr sin 
= (80. N)(0.25 m) sin 30
= 10. Nm.
• Never write the units for torque as J. Torque is not an
energy quantity.
Torque – direction + or Magnitude of Torque = (Magnitude of the force) x (Lever arm)
τ = F∙l
or
Γ = F∙l in IB letters
Direction:
• The torque is positive when the force produces
a counterclockwise rotation about the axis.
• The torque is negative when the force produces
a clockwise rotation about the axis.
SI Unit of Torque: newton x meter (N·m)
• Example 2 The Achilles Tendon
• The tendon exerts a force of
magnitude790 N.
• Determine the torque (magnitude and
direction) of this force about the ankle
joint.
• τ = FL
•
•
•
•
•
torque = force x lever arm
Method 1
cos 55 = L/(3.6E-2), so L=0.036cos55
τ = (790)(0.036cos55) = 16.3 N
Method 2
Angle between line of force and horiz.
is 35 degrees.
• Fperpindicular = (790)(sin35)
• τ = (790)(sin35)(0.036)= 16.3 N
790 N
Rigid Objects in Equilibrium
• If a rigid body is in equilibrium, neither its linear motion
nor its rotational motion changes.
linear
rotational
ax  a y  0
F
x
0
 0
F
y
0
  0
Translational and Rotational Equilibrium
• Recall that translational equilibrium was the state of a
system in which the sum of the forces was zero:
F = 0
condition for translational equilibrium
• Now we have an analogous condition for rotational
equilibrium – the state of a system in which the sum of
the torques is zero:
 = 0
condition for rotational equilibrium
FYI
• Note that the condition for translation equilibrium DOES
NOT imply that the system is not translating. As long as
it is not accelerating F = 0 is still true.
• Similarly, the condition for rotational equilibrium  = 0
DOES NOT imply that the system is not rotating.
Equilibrium
EQUILIBRIUM OF A RIGID BODY
A rigid body is in equilibrium if it has zero translational
acceleration and zero angular acceleration. In equilibrium,
the sum of the externally applied forces is zero, and the
sum of the externally applied torques is zero.
F
x
0
F
y
0
  0
Equilibrium Problems
Problem Solving Strategy
1. Select the object to which the equations for equilibrium
are to be applied.
2. Draw a free-body diagram that shows all of the external
forces acting on the object.
3. Choose a convenient set of x, y axes and resolve all
forces into components that lie along these axes.
4. Apply the equations that specify the balance of forces at
equilibrium. (Set the net force in the x and y directions
equal to zero.)
5. Select a convenient axis of rotation. Set the sum of the
torques about this axis equal to zero.
6. Solve the equations for the desired unknown quantities
Rotational equilibrium example
Translational and rotational equilibrium
• Suppose a uniform beam of mass m and length L is
placed on two scales, as shown.
• It is expected that each scale will read the same,
namely half the weight of the beam.
• Now we place a block of mass M on the beam,
closer to the left-hand scale.
M
x
• It is expected that the left scale will read higher than
the right one, because the block is closer to it.
Rotational equilibrium –
sum of the forces = 0
• To analyze an extended system we use what we will call
an extended free-body diagram. Suppose M, m and L
are known.
x
L/2
mg
Mg
N1
N2
• Find N1 and N2 in terms of x, L, m, M and g.
• From our balance of forces we have F = 0:
N1 + N2 – Mg – mg = 0
FYI
• We have one equation with two unknown normal forces.
• We will use  = 0 for our second equation.
Rotational equilibrium –
sum of the torques =0
• In order to use our balance of torques we need to choose a
pivot point. If a system is in static equilibrium you can use
ANY point! I have chosen N1’s location.
+
x
L/2
mg
Mg
N1
N2
• For bookkeeping purposes choose a torque direction.
• Note that Mg and mg want to rotate (+), and N2 (–).
• From our balance of torques we have  = 0:
N1  0 + Mg  x + mg  L / 2 – N2  L = 0
FYI
• Choosing the pivot (fulcrum) at the point of a force removes
that force from the torque equation! Why?
Solving equilibrium
problems
• We now resolve our system of equations:
N1 + N2 – Mg – mg = 0
N1  0 + Mg  x + mg  L / 2 – N2  L = 0
• Our second equation gives us N2:
N2 = ( Mx / L + m / 2)g.
• Our first equation gives us
N1 = (M + m)g – N2.
PRACTICE: If the 2.75-m long wood plank has a mass
of 45 kg, the box has a mass of 85 kg, and x = 0.50 m,
what do the two scales read?
SOLUTION: N2 = (850.50 / 2.75 + 45 / 2)10 = 380 kg.
N1 = (85 + 45)10 – 380 = 920 kg.
Equilibrium
Example 3
Example 3 A Diving Board
A woman whose weight is 530 N is
poised at the right end of a diving board
with length 3.90 m. The board has
negligible weight and is supported by
a fulcrum 1.40 m away from the left
end.
Find the forces that the bolt and the
fulcrum exert on the board.
Equilibrium Example 3:
Sum of the torques = 0
• Remember counterclockwise (+)
clockwise (-)
• F1 no torque, because it goes through
the axis of rotation of the diving board.
  F 
2 2
 W W  0
W W
F2 
2
F2

530 N 3.90 m 

 1480 N
1.40 m
Equilibrium Example 3:
Sum of the forces = 0
• Now still need to find F1
• Can use equilibrium of
forces to find it
• Can’t use torque because F1
contributed no torque.
F
y
 F1  F2  W  0
 F1  1480 N  530 N  0
F1  950 N
Equilibrium Example 4 –
“the classic” ladder problem
This is a classic problem that tends to show up on a bunch of
AP, IB and college exams. Make SURE you understand it!
Ladder
example
• The forces
shown
are
follows:
• Gy = normal force
from ground
• WL = weight ladder = 355 N, at center of mass 4.00 m up
• WF = weight firefighter = 875 N, stands 6.30 m up ladder
• P = normal force of wall pushing back on ladder (no frictional
force on smooth wall, so no y-component here)
• Ground has friction, therefore Gx and Gy present.
• Length of ladder = 8.0 m, angle 50.0 degrees to ground
• Axis of rotation is at the ground (picture if ladder fell
backwards)
Method #1
Ladder
example
50o
• Net force acting on
ladder is zero – it
is not moving.
• ΣFx = Gx – P = 0
• ΣFy = Gy –WL –WF = 0
• Solve 2nd equation: Gy = WL + WF = 355N + 875N = 1230N
• Now let’s tabulate the torques for each force
Force x
WL = 355N
Lever Arm =
lL = 4.00 cos 50
Torque
-WL (4.00 cos 50)
lP = 8.00 sin 50
P
WF = 875 N lF = 6.30 cos 50
+ P (8.00 sin 50)
-WF (6.30 cos 50)
Ladder
example
The torque for Gx and Gy is zero
because those forces are applied at
the axis of rotation, the ground.
Force
Lever Arm
Torque
WL = 355N
lL = 4.00 cos 50
-WL (4.00 cos 50)
P
lP = 8.00 sin 50
+ P (8.00 sin 50)
WF = 875 N
lF = 6.30 cos 50
-WF (6.30 cos 50)
• Στ = -WL l L - WF l F + P l P = 0 (axis of rotation is at the ground)
• P = (WL l L + WF l F ) / l P
• P = (355)(4.00 cos 50) + (875)(6.30cos 50) / (8.00)(sin 50) = 727N
• Now back to 1st eq in x dir: Gx –P =0, so Gx = P = 727 N
Ladder
example
•
•
•
•
To summarize, you had to calculate
ΣFx = 0
ΣFy = 0
Στ = 0
in order to solve that problem.
• This will frequently be the case. It is the standard
problem-solving flow, so be prepared to do so.
Equilibrium Example 5
S is normal force here
Example 5 - Bodybuilding
The arm is horizontal and weighs 31.0 N. The deltoid muscle can
supply 1840 N of force. What is the weight of the heaviest
dumbbell he can hold?
Example 5 Bodybuilding
Στ = -Wala - Wdld + MlM = 0
•
•
Method 1: lM = 0.150 sin 13.0, and just use M
Method 2: Mperp = M sin 13.0, and just use lM =0.150
Example 5 Bodybuilding
Στ = -Wala - Wdld + MlM = 0
 Wa  a  M M
Wd 
d
 31.0 N 0.280 m   1840 N 0.150 m sin 13.0

 86.1 N
0.620 m
Equilibrium example 6
• Consider a boom crane
whose components must be
strong enough to withstand
any force a client might apply.
• We need to know the required
tensions in the cables.
• We need to know the strength
of the pin.
pin
Equilibrium example 6
Translational and rotational equilibrium
• Here are the variables:
• And an extended FBD is
the way to go:
• Let x be the distance FH
mg is from the red pin (wall).
• Note that the weight of
FV
the boom itself acts as if
all of its mass is located
at its center, which is a
distance of L / 2 from the FH
pin.
• In general M, m, L, and 
FV
will be known.
M
T
θ
mg
m
Mg
T
xθ
L/2
mg
Mg
Equilibrium example 6
• Suppose M = 400. kg, m = 200. kg,
T
30
L = 20.0 m, x = 15.0 m, and  = 30.
15
10
• Then our diagram reduces to:
60
FH
60
2000
• Note that the angles between
the black forces and the boom are 60.
 Why?
4000
• From F = 0 we see that FV = 6000 N.
• We also see that FH = T. Why?
FV
• For the torques, let’s choose the location
of the red pin forces as our pivot, cw = (+) as shown.
• From  = 0 we see that
400010 sin 60 + 2000 15 sin 60 – T sin 30 = 0
T = 121000 N = FH.
Stability of equilibrium
Translational and rotational equilibrium – stability
• Consider the following three scenarios:
• Two bowls and one flat surface.
• A marble is carefully placed on each surface so that it
remains at rest: All marbles are in static equilibrium.
• Each ball is displaced a small amount.
• The three different types of equilibrium are illustrated.
NEUTRAL
UNSTABLE
STABLE
EQUILIBRIUM
EQUILIBRIUM EQUILIBRIUM
FYI
• Note that the stable equilibrium has a restoring force.
Translational vs. rotational motion
Extended bodies
• Up to this point we have talked about
moving particles, and moving bodies
comprised of many particles (atoms)
moving as a group without rotation. Albert the physics cat
• In this topic we will discuss the characteristics of a
set of particles, moving as a group with rotation.
• In order to make our analysis easier, we will review
the idea of the center of mass (cm) - the “balance
point” of an extended body, or set of particles.
• To illustrate cm, consider Albert the physics cat who
has been thrown as shown:
Translational vs. rotational motion
Extended bodies
• Suppose we place a blue dot on Albert’s cm (his
balance point) and a red dot Albert’s tail and we give
him another toss:
• Note that Albert’s center of mass follows a perfect
parabolic trajectory, whereas his tail does not.
• Furthermore, every point on Albert will have a
different equation of motion.
• Add to this yet another level of complexity: Albert can
change his shape!
• We are entering a whole new world of complexity…
(but don’t panic, we only do rigid bodies in this class,
no shape shifters!)
Translational vs. rotational motion
Extended bodies
• We call Albert a non-rigid extended body
because he can change his shape.
• A wrench, on the other hand, is a rigid
extended body, because its shape does
not change.
• A wrench can be translated (moved without
rotation)…
• Note that every point in the wrench has the same
velocity (this includes speed and direction).
• This is why in the past we could treat an extended
mass in translation as a single particle.
Translational vs. rotational motion
Extended bodies
• A wrench can be rotated without translation.
• Note that every point in the wrench has a
different velocity (speed and direction).
• We have already studied this sort of
circular motion in Topic 6.
• And if we rotate and translate a body, we get this:
• Just as we studied pure translational dynamics in the
core, we will now study pure rotational dynamics.
Translational vs. rotational kinetic energy
Officially we study kinetic energy in January (preview now)
• Consider a bowling ball on a table top:
Stationary
EK = 0.
Spinning in place
(perhaps on ice)
EK ≠ 0.
TOP VIEW
• Neither ball is rolling, so both have a translational
kinetic energy equal to zero.
• The second ball has only rotational kinetic energy.
Rotational Motion and Moment of Inertia
Rotational inertia I (moment of inertia)
• Even though the center of
mass of the spinning bowling
ball is not moving, each
particle in the ball not in the
center has a tangential
velocity and thus has kinetic
energy.
• In translation every mass
particle has the same velocity.
• Not so in rotation. Each mass has a velocity that is
proportional to its radius from the axis of rotation,
recall v=rω
Rotational Dynamics Preview from our Energy unit in January
– Moment of Inertia Kinetic energy = KE = EK = ½ mv2
Rotational inertia I (moment of inertia)
• In fact, if you recall that for
circular motion v = r , we see
that for each particle in a
rotating extended mass
EK = (1/2)mv 2
= (1/2)m(r )2
= (1/2) (mr 2) 2.
• Given that the  is the same
for all particles in a rigid extended
body, clearly the total kinetic energy is given by
𝑬𝑲 = 𝑰𝝎𝟐 with I = mr 2.
where I is the rotational moment of inertia
moment of
inertia I
NOTE THAT HONORS USES KE AND SL USES Ek FOR KINETIC ENERGY
Rotational (I) vs. translational inertia (m)
Rotational inertia I (moment of inertia)
• It turns out that the rotational inertia I has the same
function in rotation as the translational inertia m has
in translational motion. We will soon see that all the
translational kinematic and dynamic equations can be
directly translated into their rotational counterparts by
simple substitutions – one of which will be I  m!
PRACTICE: Find the moment of inertia of the
dumbbell about its center. Each end has a
mass of 15.0 kg. Assume the 30.0-cm
handle is massless.
SOLUTION:
• Each mass is 0.15 m from the center of rotation. Thus
I = mr 2 = 15(0.15)2 + 15(0.15)2 = 0.675 kg m2.
Rotational kinetic energy and
moment of inertia
Kinetic energy will not
appear on this test, just a
preview, but you do need
to know moment of inertia
PRACTICE: If the mass of the previous example rotates
once in 2.0 seconds, what is its rotational kinetic
energy?
SOLUTION:
• Use the I we just calculated and EK = (1/2) I 2.
 =  / t = 2 rad / 2 s = 3.14 rad s-1.
I = 0.675 kg m2.
• Thus
EK = (1/2) I 2
= (1/2)0.6753.142
= 3.3 J.
FYI
• You can verify that the unit is indeed J.
Rotational inertia I (moment of inertia)
PRACTICE: Find the moment
of inertia I of the dumbbell about
one of its ends. Each end has
a mass of 15.0 kg. Assume the
30.0-cm handle is massless.
SOLUTION:
• One mass is 0.00 m from the center of rotation. The
other mass is 0.30 m from the center of rotation. Thus
I = mr 2 = 15(0.00)2 + 15(0.30)2 = 1.35 kg m2.
FYI
• Note that the moment of inertia depends not only on
the mass distribution (hence the geometry) but also
on the axis of rotation. Be wary!
PUT ON
YOUR
STUDY
BUDDY
Note
that to
derive
moment
of
inertia
from
scratch,
use this
equation
I = mr 2
Rotational inertia I (moment of inertia)
– formulas for common shapes
Rotational inertia I – example 1
PRACTICE: Find the moment
of inertia of a 7.27-kg bowling ball
about its center of mass. A
regulation bowling ball has a
diameter of 22 cm. If it revolves
twice each second, what is its
rotational kinetic energy?
SOLUTION:
• Use the rotational inertia formula for a solid sphere.
I = (2/5)MR2 = (2/5)7.270.112
= 0.035 kg m2.
EK = (1/2)I2 = (1/2)0.035(2 / 0.5)2
= 2.8 J
Rotational inertia I (moment of inertia)
– formulas for common shapes
Rotational inertia I – example 2
PRACTICE: Find the moment
of inertia of a 125-gram meter
stick about its end.
SOLUTION:
• Use the rotational inertia formula for a thin rod about its
end.
I = (1/3)ML2 = (1/3)0.1251.002
= 0.042 kg m2.
FYI
• Note that the moment of inertia about the end of the
ruler is more than that about its center. Why?
• Because the mass making up the ruler is, on average,
farther from the pivot point in the former case.
Finding Moments of Inertia – Example 3
Example 3
The Moment of Inertia Depends on Where the Axis Is.
• Two particles each have mass m and are fixed at the
ends of a thin rigid rod. The length of the rod is L. Find
the moment of inertia when this object rotates relative to
an axis that is perpendicular to the rod at (a) one end and
(b) the center.
For this
question, we are
only including
the two particles,
and we are
assuming the
rod is
“massless”
Finding Moments of Inertia – example 3
For figure (a):
I   mr   m r  m r  m0  mL 
2
2
1 1
2
2 2
m1  m2  m
I  mL2
For a massless rod with two
particles spun from one end
2
2
r1  0 r2  L
Finding Moments of Inertia – example 3
For Figure (b):
 
I   mr 2  m1r12  m2 r22  mL 2  mL 2
2
m1  m2  m
I  mL
1
2
2
For a massless
rod with two
particles spun
from its center
2
r1  L 2 r2  L 2
Rotational inertia I – the parallel axis theorem
• Suppose instead of rotating the ruler about its end (for
which we have a formula) or its center (for which we also
have a formula), we wish to rotate it about a point onequarter of a meter from the end (for which we don’t have a
formula.
• Instead of having an infinite number of formulae for each
extended mass shape, we have the parallel axis
theorem, presented without proof here:
IP = ICM + Md 2
parallel axis theorem
FYI
• To use the Parallel Axis Theorem you need two things:
(1) ICM, the moment of inertia relative to the Center of
mass
(2) the distance d that the new parallel axis is from
the center of mass axis.
Rotational inertia I – the parallel axis
theorem (also known as PAT)
IP = ICM + Md 2
parallel axis theorem
PRACTICE: Find the moment of
inertia of a rod about its end if it
has a solid sphere on the other end.
pivot point
SOLUTION:
• Start with the formula for a thin rod about its end:
IROD = (1/3)ML2 = (1/3)128.02 = 256 kg m2.
• For the solid sphere (R=1.0m in diagram)
ICM = (2/5)MR2 = (2/5)151.02 = 6.0 kg m2.
• Using the PAT for the sphere, where d = 8.0+1.0 = 9.0 m
IP = ICM + Md 2 = 6.0 + 159.02 = 1221 kg m2.
• Finally, ITOT = IROD + IP = 256 + 1221 = 1500 kg m2.
Linear and angular displacement and velocity
(review from Rotational Kinematics Unit 2)
• Recall that arc length is given by the following simple
relationship:
s = r
where  is in radians linear and angular
displacement
• Recall that v = s / t and that  =  / t. Then the
following is true:
v = s / t
definition of linear velocity
= (r ) / t substitution
= r  / t
r constant during rigid body rotation
= r .
definition of angular velocity
v = r
where  is in radians
per second
linear and angular
velocity
Linear and angular displacement and
velocity (review of rotational kinematics)
• Angular velocity implies a direction because it’s a
vector. It is given by yet another right hand rule:
• Grasp the axis of rotation with the right hand, with
your fingers curled in the direction of rotation. Your
extended thumb points in the direction of .
Linear and angular displacement and
velocity – typical test question example
PRACTICE: Find the angular
velocity of Earth.
SOLUTION:
•  = 2 rad.
• t = 24 h (3600 s h-1) = 86400 s.
• From  =  / t we see that
 = 2 rad / 86400 s
= 7.2710-5 rad s-1.
• This small angular speed is why
we can’t feel the earth spinning.
• From the right hand rule for spin
we see that the angular velocity points north.

Linear and angular acceleration
(review)
• Recall that acceleration was defined as a = v / t.
In a similar manner we define angular acceleration
 as
angular
 =  / t
where  is in radians
per second squared acceleration
• But since v = r  we can then write
a = v / t
definition of linear acceleration
= (r ) / t substitution
= r  / t
r constant during rigid body rotation
= r .
definition of angular acceleration
at = r 
where  is in radians
per second squared
linear and angular
acceleration
Centripetal and tangential acceleration
• Recall that centripetal acceleration ac was a centerpointing acceleration given by
ac = v 2/ r = r2
centripetal acceleration
• The formula at = r  represents
at
the tangential acceleration.
• The tangential and
ac
a
centripetal accelerations
are mutually perpendicular.
• The net acceleration is
the vector sum of ac and at.
• Note that a2 = ac2 + at2.
• Once the wheel reaches
operational speed, at = 0 (vt ≠0) and only ac remains.
Rotational Kinematics (review from Unit 2)
Rotational kinematics
• Recall the kinematic equations:
s = ut + (1/2)at 2
kinematic
equations
v = u + at
(translational)
v 2 = u 2 + 2as
• And the following conversions :
s = r
translational /
v = r
rotational
conversions
a = r
• Those two sets of equations are used to derive:
f = it + (1/2)t 2
kinematic
f = i + t
equations
(rotational)
f2 = i2 + 2
Rotational Kinematics review –
example 1
f = it + (1/2)t 2
f = i + t
f2 = i2 + 2
kinematic
equations
(rotational)
PRACTICE: Find the angular
acceleration of a bench grinder’s
cutting wheel if it reaches
2500 rpm in 3.5 s.
SOLUTION: First convert to rad/s
i = 0.
f = (2500 rev min-1)(2 rad rev-1)(1 min / 60 s)
= 261.8 rad s-1.
 =  / t = (262 – 0) / 3.5 = 75 rad s-2
Or use f = i + t and solve for , same thing
Rotational Kinematics review –
example 1 (continued)
f = it + (1/2)t 2
f = i + t
f2 = i2 + 2
PRACTICE: Find the angle
through which the cutting wheel
rotates during its acceleration.
SOLUTION: You can use the
first or the last formula.
f = it + (1/2)t 2
f = 03.5 + (1/2)75 3.52
f = 460 rad.
kinematic
equations
(rotational)
Rotational Kinematics review –
example 1 (continued)
f = it + (1/2)t 2
f = i + t
f2 = i2 + 2
PRACTICE: Find the tangential
acceleration of the edge of the
5.0-cm radius cutting wheel
during and after acceleration.
SOLUTION: at = r.
• During acceleration  = 75:
at = r  = 0.05075 = 3.8 m s-2.
• After acceleration  = 0:
at = r  = 0.0500 = 0.0 m s-2.
kinematic
equations
(rotational)
Rotational kinematics – example 1
(continued)
f = it + (1/2)t 2
f = i + t
f2 = i2 + 2
ac
kinematic
equations
(rotational)
PRACTICE: Find the net
acceleration of the edge of the
5.0-cm radius cutting wheel
at t = 0.08 s.
SOLUTION: at = r.
• During acceleration at = 3.8 m s-2 (from last slide).
• At t = 0.08 s, f = i + t = 0 + 76.90.08 = 6.2 rad s-1.
aC = r 2 = 0.0506.22 = 1.9 m s-2.
aNET2 = aC2 + at2 = 1.92 + 3.82  aNET = 4.2 m s-2.
at
a
Newton’s Second Law for Rotational
Motion About a Fixed Axis τ = Iα
FT  maT
aT  r
τ = FT r = (maT) r = m(rα)r
τ = (mr2) α
τ = (mr2) α = Iα
where
I = moment of inertia
τ = Iα
Newton’s Second Law for Rotational
Motion About a Fixed Axis
THIS IS THE ROTATIONAL VERSION OF NEWTON’S SECOND LAW FOR
A RIGID BODY ROTATING ABOUT A FIXED AXIS
 Moment of   Angular

  

Net external torque  
 inertia
  accelerati on 
PUT ON STUDY BUDDY
  I 
Requirement: Angular
acceleration
must be expressed in radians/s2.
 
I   mr
2
Newton’s Second Law for Rotational Motion About a
Fixed Axis – Example 1
Example 1 Hoisting a Crate – challenging but typical test problem
The combined moment of inertia of the dual pulley is 50.0 kg·m2. The
crate weighs 4420 N. A tension of 2150 N is maintained in the cable
attached to the motor. Find the angular acceleration of the dual
pulley.
Crate cable
winds around
smaller inner
pulley with lever
arm 0.200 m.
Motor cable
winds around
larger outer
pulley with lever
arm 0.600 m.
Newton’s Second Law for Rotational Motion – example 1
T2 and T2 equal in magnitude
On the crate:

F

T
 y 2  mg  may
a y   2
^^ because
crate’s rope
goes around
smaller inner
pulley with
radius l2
On the pulley:
  T 
T2  mg  may
1 1
 T2 2  I
Newton’s Second Law for Rotational Motion
– example 1
substituting in for T2 …
T1 1  mg  may  2  I
a y   2
T11  mg  m 2  2  I
T1 l1 – mgl2 = α (I + ml22 )
T1 1  mg 2

I  m 22

2150 N 0.600 m   451 kg 9.80 m s 2 0.200 m 

 6.3 rad
2
2
46.0 kg  m  451 kg 0.200 m 
s2
Newton’s 2nd law for rotational
dynamics – example 2
EXAMPLE: Consider a disk-like pulley of
mass m and radius R. A string is connected
to a block of mass M, and wrapped around
the pulley. What is the acceleration of the
block as it falls?
SOLUTION: We can insert the forces into our
diagrams, important dimensions, and accelerations.
• Clearly the acceleration of the pulley is angular: 
• While the acceleration of the block is linear: a
• Recall the relationships between then angular
and the linear variables: a = R or  = a / R.
• For the disk, I = (1/2)mR2 so that
α
m
R
M
T
T
a
 = I = Ia / R = (1/2)mR2a / R = (1/2)mRa.
Mg
Rotational dynamics – example 2
(continued)
α
EXAMPLE: Consider a disk-like pulley of
m
mass m and radius R. A string is connected
R
to a block of mass M, and wrapped around
T
the pulley. What is the acceleration of the
block as it falls?
SOLUTION: (recall torque = force x distance)
• But  = RT so that
RT = (1/2)mRa  T = (1/2)ma.
T
• For the falling mass use ΣF = ma
T – Mg = -Ma negative because >>>a
M
• Finally substitute in for T,
Mg
(1/2)ma – Mg = -Ma
a = Mg / [M + (m / 2)].
Rotational kinematics and dynamics
• How am I going to remember all of this?!?
v 2 = u 2 + 2as
F = ma
It’s EASY!!
s
v
a
mI
F
Rolling motion (reminder from rotational
kinematics unit – this slide presented then)
• A special subset of dynamics is called rolling motion,
where by “rolling” we mean “rolling without slipping.”
• A rolling wheel is shown here:
s vt
vCM
s
• If the wheel’s cm has traveled a distance s, so has a point
on its circumference (it is not slipping).
• The speed of the wheel is given by vCM = s / t.
• The speed of a point on the circumference, the tangential
speed, is also given by vt = s / t.
Rolling motion (review from kinematics)
• A special subset of dynamics is called rolling
motion, where by “rolling” we mean “rolling without
slipping.”
• A rolling wheel is shown here: vt
s
vCM
s
• Since vt = R = vCM, we can write:
vCM = R
condition for rolling motion
FYI
• This only holds for rolling without slipping.
Rolling motion (review from kinematics)
v = vCM
v = R
v = 2R
v = vCM
v=0
v = R
v = vCM
v = R
TRANSLATIONAL
MOTION
ROTATIONAL
MOTION
v=0
ROLLING
MOTION
• Consider the following three scenarios:
• In pure translation, all points move at vCM.
• In pure rotation, all points move at v = r. Note the velocities
of top and bottom (r = R), in particular.
• In rolling motion, the two pure aspects are summed.
Rolling motion – example 1
EXAMPLE: A 13-cm radius hoop is rolling without
slipping at a speed of 25 ms-1. What are the linear
speeds of the top of the wheel, the middle of the wheel,
and the bottom of the wheel? If the hoop fails because
of centripetal accelerations, where will it be most likely
to fail first?
SOLUTION:
• The top is traveling at 2vCM = 50 ms-1.
• The middle is traveling at vCM = 25 ms-1.
• The bottom is traveling at 0 ms-1.
Because aC = v2 / r, clearly it will fail at
the top first. In fact, at the top,
aC = v2 / r = 502 / 0.26 = 9600 ms-2 = 960 G!
Center of
Gravity
DEFINITION OF CENTER OF GRAVITY
• The center of gravity of a rigid body is the point at which
its weight can be considered to act when the torque due
to the weight is being calculated.
• When an object has a symmetrical shape and its weight
is distributed uniformly, the center of gravity lies at its
geometrical center.
Center of Gravity
W1 x1  W2 x2  
xcg 
W1  W2  
• Did you notice that equation is
basically the center of gravity is
equal to the sum of the torques
over the sum of the forces? It’s
basically the average lever arm.
• If you use the center of gravity
as a balance point as shown,
the object should balance and
not tilt either way.
Center of Gravity
Example 1: The Center
of Gravity of an Arm
The horizontal arm is
composed of three parts:
the upper arm (17 N),
the lower arm (11 N), and
the hand (4.2 N).
Find the center of gravity
of the arm relative to the
shoulder joint.
Center of Gravity
example 1
upper arm (17 N)
lower arm (11 N)
hand (4.2 N)
W1 x1  W2 x2  
xcg 
W1  W2  
xcg

17 N 0.13 m   11 N 0.38 m   4.2 N 0.61 m 

 0.28 m
17 N  11 N  4.2 N
• The center of gravity is 0.28 m from the shoulder.
Center of Gravity – example 2
Conceptual Example 2: Overloading a Cargo Plane
This accident occurred because the plane was overloaded toward
the rear. How did a shift in the center of gravity of the plane cause
the accident?
Center of Gravity
Finding the center of gravity of an irregular shape can be done by first
suspending the object so there is no rotation (no torque) like in (b), then
choosing a 2nd point P2 and suspending it with no rotation like in (c). The
intersection of those two lines of force from the weight is the center of mass..