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Lecture 23: Rotation
Motivation for Rotation
• Want to describe motion of extended
objects
– How does a hammer fly?
• Split motion into:
– Motion of COM
– Motion around COM
That is rotation in 3D
Rotations
• Angular variables: position angle
(Simulation), angular displacement, angular
velocity (angular frequency), angular
acceleration
• Measure in radians!
• Basically a re-writing of linear kinematics
and dynamics can “transcribe” old
equations
What is angular velocity?
• LI: Angular velocity is the ratio of angular
displacement of an object to the time
interval and can be expressed by the
equation ω2-ω1/ t2-t1 where ω is the angle
of displacement and t is time.
Example: Radians!
• How many complete rotations did a rigid
body make when its angular position is
θ = 45.0? (See also the simulation Rigid Body Rotating the Z
Axis)
• Answer: θ = 45 is in radians, and one
rotation corresponds to an angle of 2π, so n
= int(45/(2π))=7.
A ladybug sits at the outer edge of a
merry-go-round, and a gentleman bug
sits halfway between her and the axis of
rotation. The merry-go-round makes a
complete revolution once each second.
The gentleman bug’s angular speed is …
• … half the ladybug’s.
• … the same as the ladybug’s.
• … twice the ladybug’s.
• … impossible to determine.
Vector Nature of rotational
quantities
• Angular velocity: Use right-hand-rule
– Fingers trace direction of rotation, thumb gives
direction of angular velocity
– Magnitude is change of angular position over
elapsed time
Example: A data bit on a CD that is turning
counter-clockwise
• Initially in z direction,
turns into x direction
after 1/10 sec
• θ(0s) = - π/2
• θ(0.1s) = 0
• ωavg = (0 – (–π/2))/0.1s
= +5π Hz
• Direction: RHR: j (+y)
y
z
x
A ladybug sits at the outer edge of a merry-go-
round, that is turning and slowing down. The
vector representing her angular velocity is in
the …
• -z direction
• +z direction
• +y direction
• zero
z
x
y
Angular acceleration
• Two acceleration components:
– Radial: the centripetal acceleration of uniform
circular motion
– Tangential: changes angular velocity in non-
uniform circular motion
• The latter is the angular acceleration
α = dω/dt = d2θ/dt2
• Direction: parallel to ω if speeding up
rotation, anti-parallel if slowing down
A ladybug sits at the outer edge of a merry-go-
round, that is turning and slowing down. At
the instant shown, the bug’s angular
acceleration is in the …
• -z direction
• -y direction
• +y direction
• +z direction
z
x
y
A ladybug sits at the outer edge of a merry-
go-round, that is turning and slowing down.
At the instant shown, the radial component of
the bug’s radial acceleration is in the …
• -x direction
• +y direction
• +z direction
• Zero
z
x
y
Transliteration of earlier
kinematics equations
• Position
• Velocity
• Velocity squared
• Cf. the Comparison between linear and
rotational variables
Example: Speeding up the bike
wheel
• If the bike wheel is brought up to 2 rev/s, in
4 revolutions, what is the (constant) angular
acceleration?
Relation between linear and
angular variables
• Displacement along circular arc s = r θ
• ds/dt yields v = r ω
• dv/dt at constant r yields a = r α (tangential)
• Also: ar = v2/r = r ω2
• See Simulation
Lecture 24: Rotation & Rigid
Bodies
Rotational Motion Example
• Car motor speeds up from rest to 10,000 rpm in
5s, has diameter of 4cm.
• Final frequency: f = 10,000 1/min *(1/60 min/s)
= 167 Hz
• Angular speed: ω = 2πf = 1050 Hz
• Linear velocity: v = R ω = (0.02m)(1050 Hz)
=21 m/s
Rotational Motion Example
• Car motor speeds up from rest to 10,000 rpm in
5s, has diameter of 4cm.
• Linear acceleration: a = 21 m/s / 5s = 4.2 m/s2
• Angular acceleration: α=Δω/Δt = 1050Hz/5s =
210 s-2
• Check: a= R α = (0.02m)(210s-2) = 4.2 m/s2
Rotational Kinetic Energy and
moment of inertia I
• We can rewrite (translational) kinetic
energy K=1/2 m v 2 in our new, rotational
language
• For an extended object (rigid body) we have
• K=∑1/2 mi vi2
• Use vi = ri ω to obtain (ωi = ωj =ω )
K=∑1/2 mi ri2 ω2 = 1/2 (∑ mi ri
2 ) ω2
= ½ I ω2
Moment of Inertia I
• Is a property of a rotating object
• Need to know axis of rotation to be able to
compute it
• Measure for how much mass sits how far
away from the rotation axis
• Demo: Difference of rotation of solid disk
vs ring
Calculating Rotational Inertia
• I = ∫ r2 dm
• Parallel-Axis Theorem
I = ICOM + Mh2
Example
• Two balls of masses 2 kg & 4kg sitting 1m off
to the left and the right of the y axis (held by a
massless rod).
– What is the MOI when rotating around the COM ?
• COM is at x = (–2+4)/6 m = +1/3 m
• I = (4/3m) 2 2kg + (2/3m)2 4kg = (32 +16)/9 kg m2
= 5.33 kg m2
– What is the MOI when rotating around the y-axis?
• I’ = (1m) 2 2kg + (1m)2 4kg = 6 kg m2
• Note: I = ICOM < I’
Example
• Two balls of masses 2 kg & 4kg sitting 1m
off to the left and the right of the y axis (held
by a massless rod).
– What is the RI when rotating around the y-axis?
• I’ = (1m) 2 2kg + (1m)2 4kg = 6 kg m2
• Alternative method: parallel axis theorem:
– I’ = I + Mh2 = 5.33 kg m2+ 6kg(–1/3m)2
= 6 kg m2
Determining Rotational Inertia
• Integral
• Table
• Parallel-axis
theorem
Which configuration has a
greater MOI, assuming same
length and mass of rods?
• Left
• Right
• Same
• Cannot tell
Which object will roll faster
down the incline?
• Hollow Ring
• Disk
• Both the same
• Cannot tell
Example: Thin Rod
• Use dm = M/L dx (linear mass density M/L)
• I = ∫ x2 dm = ∫ x2 (M/L) dx
• Integrate from -L/2 to + L/2 for rotation
about COM
• I = M/(3L)[(L/2)3- (-L/2)3] = ML2/12
Lecture 24: Torque
Kinetic Energy of Rolling
• The energy is split up into linear kinetic
energy and rotational kinetic energy, see
simulation
• Two forms of inertia: mass and rotational
inertia
Objects of different I (but same R,M)
experience a different acceleration!
Example: Rotational Kinetic
Energy
• Two masses of 2 kg are connected by a
massless 1m rod and rotated around their center
of mass with a period of 2s. Calculate the
rotational kinetic energy of this configuration.
• Use Eqs. (10-33) & (10-34): I = 2kg(-0.5m)2+
2kg(0.5m)2 = 1 kg m2
ω = 2π/T = π Hz, so
• K = ½ I ω2 = ½ π2 J = 4.93 J
Torque
• Torque is force times lever arm
• Lever arm is distance to rotation axis along
a direction perpendicular to the force
• Later: τ = r x F
• | τ | = |r| |F| sin φ
Torque Example
• In the simulation Torque how large does the red
force have to be (if the red position is negative 1m
and all other quantities at their initial values) such
that the sum of the torques produced by the blue
and the red forces is zero, i.e. that there is no net
force, and hence no net angular acceleration, and
hence no rotational motion of the bar?
• Fred = 10N• Torque = force times lever arm. If the lever arm is half as
long, we need twice as much force: f = 10 N
Torque example
• A force of 100N acts at a distance 40 cm away
from the pivot point at an angle 120 degrees:
torque is 40Nm sin 240°= -20Nm (clockwise)
• Note that the angle is the angle of F wrt r, not
the other way around.
• Positive torque means counterclockwise
torque
Newton II for Rotation
• Transliterate:
• F = ma becomes τ = I α
Rolling = Rotation plus
Translation
• Rolling is a combination of motion of the
COM and rotation about the COM
• Point of contact remains stationary, while
point on top of wheel moves with twice the
velocity of the COM.
Rolling as pure Rotation
• Rolling can also be viewed as a pure
rotation around the point of contact with
floor, see simulation
• Need parallel axis theorem to calculate
correct MOI
• Two parts represent contributions from
rotation and translation to KE
Forces of rolling• Friction is static, since point of contact is
stationary.
Example: Acceleration of solid cylinder
rolling down an incline without slipping
• Properties of cylinder: R=0.02m, m =1kg
• Incline: 30 degrees
• Demo: inclined plane with 1kg weight
• Q: Difference
between rolling
and sliding the
weight? (Static fr.
only!)
Lecture 25: Torque & Angular
Momentum
• New homework scheme: easy, medium,
hard exam preparation
Example: Acceleration of solid cylinder
rolling down an incline without slipping
• Properties of cylinder: R=0.02m, M =1kg
• Incline: 30 degrees
• Q: What is the disk’s acceleration and the
magnitude of friction?
Torque as a Vector
Vector Product
• 𝑎 𝑥 𝑏 = (𝑎𝑦 𝑏𝑧 −𝑎𝑧 𝑏𝑦 ) 𝑖 +
𝑎𝑧 𝑏𝑥 −𝑎𝑥 𝑏𝑧 𝑗 +(𝑎𝑥 𝑏𝑦 −𝑎𝑦 𝑏𝑥 ) 𝑘
• Proof by writing 𝑎 & 𝑏 in unit-vector notation
• Column vector method: hide row in
consideration
• Determinant method
Example:
• A = i -2j + 3k
• B = 2i – k
𝐴 𝑥 𝐵 =1−23
𝑥20−1
=
−2 −1 − 3(0)
3 2 − 1(−1)
1 0 − −2 2
=274
• Which has length √69 = 8.3066
Check
• Do with: |A||B|sin β same result
• A = i -2j + 3k |A| = √14
• B = 2i – k |B| = √5
• A•B = |A||B| cos β = 2–3= –1 β = 96.86̊
• |A x B| = √5 √14 sin 96.86 ̊ = 8.3066
Direction
• AxB is perpendicular to A and
perpendicular to B!
Check:
A•(AxB) = 0
B•(AxB) = 0
A ladybug sits at the outer edge of a merry-go-
round, that is turning and slowing down due to
a force exerted on its edge. At the instant
shown, the torque on the disc is pointing in …
• -z direction
• -y direction
• +y direction
• +z direction
z
x
y
F
What is the direction of the torque
produced by a force pointing in the SW
direction and at a point 2m directly
below the origin?
• a. SW b. Down
• c. NE d. NW
What is the direction of the torque
produced by a force pointing in the SW
direction and at a point 2m directly
below the origin?
• a. SW b. Down
• c. NE d. NW
• Answer: torque = r x F, where r is in –k,
and F in –i – j direction, so –i+j or NW or d)
Lecture 26 – Angular Momentum
What is A x A ?
• Zero
• A
• -A
• A2
Angular Momentum
Example: 2kg Ball located at
(1m,1m) falling from 1m. What
is its initial and final angular
momentum around the origin?
• Initially v=0, so p=0, so l=0
• Final: v2 =2g(1m) p=mv l = r x p
A ladybug sits at the outer edge of a merry-go-
round, that is turning and slowing down due to
a force exerted on its edge. The angular
momentum of the bug is pointing in …
• -z direction
• -y direction
• +y direction
• +z direction
z
x
y
F
A ladybug sits at the outer edge of a merry-go-
round, that is turning and slowing down due to
a force exerted on its edge. The angular
momentum of the bug is pointing in …
+z direction
• Angular
momentum goes
with velocity, not
with force: r = r i,
v=v j L=L k,
z
x
y
F
What is the direction of the angular
momentum (around the origin) of a particle
located 2m directly above the origin (i.e. 2m
along the z axis) with a momentum vector in
the SW (i.e. –i–j) direction?
• a. NE b. NW
• c. 45 degrees upward from West
• d. None of the above.
What is the direction of the angular
momentum (around the origin) of a particle
located 2m directly above the origin (i.e. 2m
along the z axis) with a momentum vector in
the SW (i.e. –i–j) direction?
• a. NE b. NW
• c. 45 degrees upward from West
• d. None of the above.
• Answer: l = r x p, where r is in k, and p in –i – j
direction, so i-j or SE so d)
Angular Momentum of Rigid
Body about fixed axis
• L = sum over angular momenta of parts
• Use l = r p = m r v = m r (r ω)
L = I ω
This also works as a vector equation, so L and
ω point in the same direction
Example: DVD spinning in
Player
• R = 6cm, M = 16g , f = 1200 rpm
• MOI is that of a disk: I = ½ MR2
= 5.76 x 10-5 kg m2
• ω=2πf = 2π 20Hz = 40π Hz
• Angular Momentum
L = I ω = 7.24 x 10-3 kg m2/s
Dynamics Transliteration
• Mass Moment of inertia
• Force Torque
• Momentum Angular Momentum
• Newton II
• F =dP/dt τ = dL/dt
• Momentum Conservation Angular Momentum
Conservation
• Work-KE theorem Work-rotational KE theorem
Angular Momentum
Conservation
• Angular momentum is conserved if no
external net torque is present
• Demo: wheel spinning on rope
• Demo: turntable with weights & bike wheel
In a Demo someone is standing
on a turntable with outstretched
arms, spinning around. If the
person pulls in the arms …
• … the rotation becomes faster
• … the rotation becomes slower
• … the rotation stays the same
• Impossible to tell
In a demo someone is standing on a
turntable with a bike wheel spinning
around in the same direction that the
person is spinning. If the person
turns over the bike wheel …
• … the rotation becomes faster
• … the rotation becomes slower
• … the rotation stays the same
• Impossible to tell
Example for Calculation
• An open door of mass 40 kg and dimension
1m x 2m is hit centrally by a 0.5kg sticky
clay ball. The ball impacts the door at a
right angle with speed 12 m/s. How long
does it take to close the door?
• Is angular momentum conserved here?
– What is the system?
Lecture 27 – Angular Momentum
& Exam Prep
Rate change of angular
momentum equals torque
• A counter-clockwise spinning bike wheel has
its axis pointing in the x direction. If we try to
change the axis so it point s up (+z direction),
how does the wheel react?
• … the rotation rate changes
• … the rotation & axis stays the same, L is conserved
• … the axis is moving up/in the +z direction
• … the axis is moving in the -y direction
Experience and Analyze
• We’ll let everyone try it out
• Meanwhile we analyze!
Analyze!
• A counter-clockwise spinning bike wheel
has its axis pointing in the x direction. In
which direction does the angular
momentum vector point?
• … x
• … -x
• … z
• … -z
Analyze!
• A counter-clockwise spinning bike wheel has
its axis pointing in the x direction. In which
direction does the angular momentum vector
point?
• … x
• Spins around the x axis, so either +x or –x
• Spins mathematically positive (CCW)
Analyze!
• A counter-clockwise spinning bike wheel
has its axis pointing in the x direction. If we
try to change the axis so it points up (+z
direction), what direction has the force
applied?
• … x
• … y
• … z
• … -z
Analyze!
• A counter-clockwise spinning bike wheel
has its axis pointing in the x direction. If we
try to change the axis so it points up (+z
direction), what direction has the force
applied?
• … z
• Straightforward: if +z is up and the force is up, the
force is in the +z direction
• Note that reference frame has to be specified!
Analyze!
• A counter-clockwise spinning bike wheel
has its axis pointing in the x direction. If we
try to change the axis so it point s up (+z
direction), what direction has the torque
applied?
• … x
• … y
• … -y
• … -z
Analyze!
• A counter-clockwise spinning bike wheel
has its axis pointing in the x direction. If we
try to change the axis so it point s up (+z
direction), what direction has the torque
applied?
• … -y
• Torque is perp. to r (+x) and to F(+z), so +y or –y
• RHR: -y
Analyze!
• A counter-clockwise spinning bike wheel
has its axis pointing in the x direction. If we
try to change the axis so it point s up (+z
direction), what direction has the resulting
change of angular momentum?
• … x
• … y
• … -y
• … -z
Analyze!
• A counter-clockwise spinning bike wheel
has its axis pointing in the x direction. If we
try to change the axis so it point s up (+z
direction), what direction has the resulting
change of angular momentum?
• … -y
• Change of angular momentum ΔL is in direction
of torque (-y)
Rate change of angular
momentum equals torque
• A counter-clockwise spinning bike wheel has
its axis pointing in the x direction. If we try to
change the axis so it point s up (+z direction),
how does the wheel react?
• … the rotation rate changes
• … the rotation & axis stays the same, L is conserved
• … the axis is moving up/in the +z direction
• … the axis is moving in the -y direction
Rate change of angular
momentum equals torque
• A counter-clockwise spinning bike wheel has
its axis pointing in the x direction. If we try to
change the axis so it point s up (+z direction),
how does the wheel react?
• … the axis is moving in the -y direction
• Change of L is change of axis in direction of torque
