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Rotational Motion &
Angular Momentum
Rotational Motion
Every quantity that we have studied with
translational motion has a rotational
counterpart
TRANSLATIONAL ROTATIONAL
Displacement x Angular Position
Velocity v Angular velocity
acceleration a Angular acceleration
Mass m Inertia I
Momentum p Angular Momentum L
Force F Torque
Angular Position
r
s
Arclength
Radius (from axis of
rotation)
Measured in radians – all angular quantities will be measured in
radians (NOT degrees)
• In translational motion,
position is represented by a
point, such as x.
• In rotational motion,
position is represented
by an angle, such as ,
and a radius, r.
x
linear
0 5
x = 3
r
0
p/2
p
3p/2angular
Angular Position
Angular
Displacement
• Linear displacement
is represented by the
vector Dx.
• Angular displacement is represented by D, which is not a vector, but behaves like one for small values.
x
linear
0 5
Dx = 4
D 60
0
p/2
p
3p/2angular
Angular Displacement
0DWhich direction is positive (by
convention)?
Positive – Counterclockwise
Negative - Clockwise
Compare to
D𝒙 = 𝒙 − 𝒙𝒐
Tangential vs. angular displacement
• A particle that rotates through an angle Dalso translates through a distance s, which is the arc length defining its path.
• This distance s is related to the angular displacement D by the equation s = rD, as we saw in a previous slide.
rD
ss
D
EXAMPLE: (a) What is the angular position, , if we go
around a circle two times?
Ans. = s
r=
2(2p𝑟)r
= 4π
(b) Say you go a quarter turn more, what is D?
Ans. D = o = 9𝜋
2− 4π =
𝜋
2(c) What is the arclength covered in total between
Probls. (a) and (b) if the radius of the circle is 3 m?
Ans. s = r = (3 m)(4p + 𝜋
2) =3m(
9𝜋
2) =
27𝜋
2m
P.O.D. 1: Two
synchronous
communications
satellites are put into an
orbit whose radius is
r = 4.23 x 107 m. The
two adjacent satellites
have an angular
separation of 2. Find
the arc length, s, that
separates the satellites.
Speed and velocity
• The instantaneous velocity of a particle in a circular path has
magnitude vT = 𝐝𝐬
𝐝𝐭and is tangent to
the circle.
• The same particle rotates with an avg. angular velocity
= D𝛉
D𝐭=
𝛉𝒇−𝛉𝒊
D𝐭• or instantaneous angular velocity
= 𝐝𝛉
𝐝𝐭• Tangential and angular speeds are
related by the equation
vT = r .
rD
s
vT
vT
rr
Angular Velocity
tavg
D
D
dt
dinst
Units – rad/s, rev/s
Direction – Same as displacement
(positive is counterclockwise, negative
is clockwise)
Magnitude – Angular Speed
Compare to
𝒗𝒂𝒗𝒈 =∆𝒙
∆𝒕and
𝒗𝒊𝒏𝒔𝒕 = 𝒅𝒙
𝒅𝒕
Derivation of tangential velocity formula
rv
rdt
d
dt
ds
rs
Take the derivative of each
side of s = r with respect
to t.
Substituting = 𝐝𝛉
𝐝𝐭and vT =
ds
dt
Derivation of tangential velocity formula
rvT
•OBSERVATIONS:
•All points on object have the same angular speed,
•Those points farther from the center have a greater
linear (tangential) speed, vT
•These points have to travel farther in the same time
tavg
D
D
P.O.D. 2: A gymnast on a high bar swings
through two revolutions in a time of 1.90 s.
(a) Find the average
angular velocity (in
rads/s) of the gymnast.
(b) What is her angular
displacement (in rad)
after t = 1 s.
(c) Find her tangential
velocity if the gymnast
is 1.8 m tall.
Acceleration• Tangential acceleration is given by
aT = 𝐝vT𝒅𝒕
• Instantaneous angular
acceleration of this particle
is given by
= 𝐝𝒅𝒕
• Average Angular acceleration of this
particle is given by
= DD𝒕
• Tangential and angular accelerations
are related by the equation
aT = r
rD
s
aT
aT
r
Angular Acceleration
• Angular Acceleration is how
the angular velocity changes
with time
• This is not the centripetal
acceleration – that tells about a
translational acceleration
tavg
D
D
dt
dinst
Units – rad/s2
Compare to
𝒂𝒂𝒗𝒈 =∆𝒗
∆𝒕and
𝒂𝒊𝒏𝒔𝒕 = 𝒅𝒗
𝒅𝒕
Sample Problem
A DVD spins at an angular velocity of 3 rad/s and 5
seconds later is spinning at a velocity of 8 rad/s. Its
radius is 0.02 m. Find
(a) Angular acceleration,
=∆𝜔
∆𝑡=𝜔𝑓−𝜔𝑖
∆𝑡=
(a) Tangential acceleration, aT
(a) Centripetal acceleration, ac, at your final angular
velocity
P.O.D. 3: A jet awaiting clearance for takeoff is
momentarily stopped on the runway. The fan blades are
rotating with an angular velocity of 110 rad/s, where the
negative sign indicates a clockwise rotation. As the
plane takes off, the angular velocity of the blades
reaches 330 rad/s in a time of 14 s.
(a) Find the angular acceleration, assuming it to be
constant.
(b) Find the tangential acceleration, if the fan blade has
a radius of 1.5 m.
Centripetal Acceleration
rr
vac
22
A pendulum is swinging back and
forth.
At the bottom of the swing the
force of gravity is pulling it
downwards but it doesn’t fall down.
This means there must be a force
pulling upwards to balance it out.
This is the centripetal force.
Since F = ma, the center-seeking
acceleration is called centripetal
acceleration and is given by:
Fc
Fg
Derivation of
Acceleration
ra
rdt
dv
rdt
d
dt
dv
rv
T
T
T
T
Tangential Acceleration
Take the derivative of each side of
v = r with respect to t.
Substituting = 𝐝𝐝𝐭
Substituting a = dv
dt
Tangential Acceleration vs. Angular Acceleration
vs. Centripetal Acceleration
• The net acceleration is the sum
of the tangential and
centripetal accelerations.
2 2
r ta a a
rr
vaa 2
2
cr
aT = r
Sample Problem
A compact disk rotates about an axis
through its center according to
t6t3
1)t( 3
(a) Determine its angular velocity and angular
acceleration at time t = 5 seconds.
Ans. To find the angular velocity, take the first derivative.
To find the angular acceleration, take the second derivative
srads196)5(6t
dt
d)t(' 22
2srads10)5(2t2
dt
d)t(''
Sample Problem (Cont.)
A compact disk rotates about an axis
through its center according to
t6t3
1)t( 3
(a) What is the linear speed of a point 20 cm from the
center at t = 5 s?
(b) What is the linear acceleration at 0.5 m at t = 5 s?Ans.
To find the linear speed, use the relationship, where (5) we
obtained from the previous slide:
v = r = (19 rad/s)(0.20 m) = 3.8 m/s.
To find the linear acceleration, use the relationship
ac = 2r = (19 rad/s)2(0.20 m) = 72.2 m/s2
P.O.D. 4: The gyroscope of a plane is spinning
according to 𝜽 𝒕 = 𝟑𝒕𝟑 + 𝟔𝒕𝟐 ½t.
(a) Find the average angular velocity of the gyroscope between t
= 2 s and t = 5 s.
(b) Find the instantaneous angular acceleration at t = 2 s.
(c) Find the linear speed, vT, at t =
2 s of a point on the gyroscope
if it has a radius of 0.05 m.
(d) Find the linear (tangential)
acceleration, aT, at t = 2 s of
the gyroscope.
(e) Find the centripetal
acceleration, ac, of a point on
the edge of the gyroscope.
Constant Angular Acceleration
• Our kinematics equations have angular
equivalents
• Just as with their linear counterparts, these only
work for constant acceleration
First Kinematic Equation
• v = vo + at (linear form)
– Substitute angular velocity for linear velocity.
– Substitute angular acceleration for linear
acceleration.
• = o + t (angular form)
Second Kinematic Equation
• x = xo + vot + ½ at2 (linear form)
– Substitute angle for position.
– Substitute angular velocity for linear velocity.
– Substitute angular acceleration for linear
acceleration.
• = o + ot + ½ t2 (angular form)
Third Kinematic Equation
• v2 = vo2 + 2a(x xo)
– Substitute angle for position.
– Substitute angular velocity for linear velocity.
– Substitute angular acceleration for linear
acceleration.
• 2 = o2 + 2( o)
Sample Problem
An automobile starts from rest and for
20.0 s has a constant linear
acceleration of 0.8 m/s2 to the right.
During this period, the tires do not slip.
The radius of the tires is 0.330 m. At
the end of the 20.0 s interval what is
the angle through which each wheel
has rotated?
Ans. The angular acceleration can be found from the formula
a = r 𝜶 =𝒂
𝒓 𝜶 =
𝟎.𝟖𝒎
𝒔𝟐
𝟎.𝟑𝟑𝟎 𝒎= 2.42 rad/s2
The angular acceleration should be negative because the tire
spins clockwise.
To find the angular displacement:
(t) = ot + ½t2 = 0(20 s) + ½(2.42 rad/s2)(0.20 s)2 = 484 rad
P.O.D. 5: The blades of an electric
blender are whirling with an angular
velocity of +375 rad/s while the puree
button is pushed in. When the blend
button is pressed, the blades
accelerate and reach a greater angular
velocity after the blades have rotated
through an angular displacement of
+44.0 rad (seven revolutions). The
angular acceleration has a constant
value of +1740 rad/s2.
(a) Find the final angular velocity
of the blades.
(b) Find the angular displacement
of the blades after 10
seconds.
(c) Find the change in tangential
velocity of the blades from the
puree to the blend position if
they have a radius of 0.02 m.
ANGULAR
MOMENTUM!, INERTIA,
ETC.
Angular Momentum
v
m
Angular momentum depends on linear momentum and the distance
from a particular point. It is a vector quantity with symbol L. If rand v are then the magnitude of angular momentum w/ resp. to
point Q is given by L = rp = mvr. In this case L points out of the
page. If the mass were moving in the opposite direction, L would
point into the page. The SI unit for angular momentum is the
kg m2 / s. (It has no special name.)
Angular momentum is a conserved
quantity. A torque is needed to change L,
just a force is needed to change p.
Anything spinning has angular has angular
momentum. The more it has, the harder it
is to stop it from spinning.
Q
r
Angular Momentum: General Definition
If r and v are not then the angle between
these two vectors must be taken into account. The
general definition of angular momentum is given
by a vector cross product:
L = r pThis formula works regardless of the angle. From cross products, the magnitude of the
angular momentum of m relative to point Q is: L = r p sin = m v r. In this case, by the
right-hand rule, L points out of the page. If the mass were moving in the opposite
direction, L would point into the page.
r
v
m
Q
Moment of Inertia vs. Angular Momentum
Any moving body has inertia. (It wants to keep moving at constant
v.) The more inertia a body has, the harder it is to change its linear
motion. Rotating bodies possess a rotational inertia called the
moment of inertia, I. The more rotational inertia a body has, the
harder it is change its rotation. For a single point-like mass w/ respect
to a given point Q, I = mr 2.
I = mr 2
m
r
For a system, I = the sum of each mass
times its respective distance from the
point of interest.
r2
m2
r1
m1
I = mi ri2 = m1 r1
2 + m2r22
Q
Q
Moment of Inertia of various shapes
Moment of Inertia ExampleTwo merry-go-rounds have the same mass and are spinning with the
same angular velocity. One is solid wood (a disc), and the other is a
hollow metal ring. Which has a bigger moment of inertia relative to
its center of mass?
mm
r
r
Ans. I is independent of the angular speed. Since their
masses and radii are the same, the ring has a greater
moment of inertia. This is because more of its mass is
farther from the axis of rotation. Since I is bigger for the
ring, it would more difficult to increase or decrease its
angular speed.
Torque & Angular Acceleration
Newton’s 2nd Law, as you know, is Fnet = ma
The 2nd Law has a rotational analog: net = I
A force is required for a body to undergo acceleration. A “turning force”
(a torque) is required for a body to undergo angular acceleration.
The bigger a body’s mass, the more force is
required to accelerate it. Similarly, the
bigger a body’s rotational inertia, the more
torque is required to accelerate it angularly.
Both m and I are measures of a body’s
inertia
(resistance to change in motion).
Example: The torque of an Electric Saw Motor
The motor in an electric saw brings the circular
blade up to the rated angular speed of 80.0 rev/s
in 240.0 rev. One type of blade has a moment of
inertia of 1.41 x 10-3 kgm2. What net torque
(assumed constant) must the motor apply to the
blade?
Ans. First we need to convert our values into rad/s for calculation
purposes.
o t240 rev x 2p = 1508 rads ? 80 revs/s x 2p = 503 rad/s 0 rad/s ?
We can find the angular acceleration from 2 = o2 + 2
Solving for : = 2−o
2
2= (503 rad/s)2−02
2(1508 rad/s)= 83.89 rad/s2
= I = 1.41 x 10-3 kgm2 83.89 rad/s2 = 0.118 Nm
P.O.D. 6: A Chinese star of mass 0.025 kg and radius 0.03 m is
thrown by Dwight from The Office at his adversary, Jim Halpert.
The Chinese star is thrown from rest. If its final angular velocity is
15 revs/s after 3 sec,
(a) find the angular acceleration of the Chinese star (in rad/s2).
(b) Find the torque of the Chinese star (Assume the Chinese star is
hoop-shaped).
Linear Momentum vs. Angular Momentum
If a net force acts on an object, it must accelerate, which means its
momentum must change. Similarly, if a net torque acts on a body, it
undergoes angular acceleration, which means its angular momentum
changes. Recall, angular momentum’s magnitude is given by
L = mvr
r
v
m
So, if a net torque is applied, angular velocity must
change, which changes angular momentum.
Proof: net = r Fnet = r m a
= r mDvt
= DLt
So net torque is the rate of change of angular momentum, just as net
force is the rate of change of linear momentum.continued on next slide
(if v and r are perpendicular)
Linear & Angular Momentum (cont.)
Here is yet another pair of similar equations, one linear,
one rotational. From the formula v = r , we get
L = mv r = m ( r) r = m r2 = I
This is very much like p = mv, and this is one reason I is
defined the way it is.
In terms of magnitudes, linear momentum
is inertia times speed, and angular
momentum is rotational inertia times
angular speed.
L = I
p = m v
Comparison: Linear & Angular Momentum
Linear Momentum, p
• Tendency for a mass to continue
moving in a straight line.
• Parallel to v.
• A conserved, vector quantity.
• Magnitude is inertia (mass)
times speed.
• Net force required to change it.
• The greater the mass, the greater
the force needed to change
momentum.
Angular Momentum, L
• Tendency for a mass to continue
rotating.
• Perpendicular to both v and r.
• A conserved, vector quantity.
• Magnitude is rotational inertia
times angular speed.
• Net torque required to change it.
• The greater the moment of
inertia, the greater the torque
needed to change angular
momentum.
Example: Spinning Ice SkaterSuppose Mr. Stickman is sitting on a stool that swivels holding a pair of
dumbbells. His axis of rotation is vertical. With the weights far from that axis,
his moment of inertia is 600 kgm2 and he is spinning at an angular velocity of
20 rad/s. When he pulls his arms in as he’s spinning, the weights are closer to the
axis, so his moment of inertia gets to
400 kgm2. What will be his angular
velocity at this moment?
Ans.
I = L = I 600 kgm2(20 rad/s) = 400 kgm2
12,000 = 400
30 rad/s =
P.O.D. 7: An artificial satellite
(m = 1500 kg) is placed into an
elliptical orbit about the earth.
Telemetry data indicate that its
point of closes approach (called
the perigee) is rp = 8.37 x 106 m
from the center of the earth,
while its point of greatest
distance (called the apogee) is
rA = 25.1 x 106 m from the
center of the earth. The speed
of the satellite at the perigee is
vp = 8450 m/s.
(a) Find its speed vA at the
apogee.
(b)Find its angular momentum
at any point in its orbit.
Rotational Kinetic Energy
• A particle in a rotating object has rotational kinetic energy:
Ki = ½ mivi2, where vi = i r (tangential velocity)
2 2
2 2 2
1
2
1 1
2 2
R i i i
i i
R i i
i
K K m r
K m r I
The whole rotating object has a rotational kinetic energy given by:
Rotational Kinetic
Energy example:
A thin walled hollow cylinder (mass
= mh, radius = rh) and a solid
cylinder ( mass = ms, radius = rs)
start from rest at the top of an
incline. Both cylinders start at the
same vertical height ho. All heights
are measured relative to an
arbitrarily chosen zero level thatpasses through the center of mass of a cylinder when it is at the
bottom of the incline. Ignoring energy losses due to retarding
forces, determine which cylinder has the greatest translational
speed upon reaching the bottom.
Rotational Kinetic Energy example (cont.):
Ans. At the top of the incline the cylinder have only gravitational potential
energy. At the bottom of the incline this energy has converted into
translational kinetic and rotational kinetic energy.
Ein = Eout
GPEin = TKEout + RKEout
mgh = ½mvf2 + ½If
2
The angular velocity can be related to the linear velocity vf by f = 𝐯𝐟
𝐫
Substituting the given values and for the angular velocity:
mhgho = ½mhvf2 + ½I(
𝐯𝐟
𝐫𝒉)2
For the hollow cylinder, the moment of inertia is given by: I = mr2
Substituting: mhgho = ½mhvf2 + ½(mhrh
2)(vf
rℎ)2
Simplifying: gho = ½vf2 + ½rh
2(𝐯𝐟
𝐫𝒉)2
gho = ½vf2 + ½rh
2𝐯𝒇𝟐
𝒓𝒉𝟐
gho = ½vf2 + ½𝐯𝒇
𝟐
gho = vf2
𝐠𝐡𝐨= vf
Rotational Kinetic Energy example (cont.):
Ans. At the top of the incline the cylinder have only gravitational potential
energy. At the bottom of the incline this energy has converted into
translational kinetic and rotational kinetic energy.
Ein = Eout
GPEin = TKEout + RKEout
mgh = ½mvf2 + ½If
2
The angular velocity can be related to the linear velocity vf by f = 𝐯𝐟
𝐫
Substituting the given values and for the angular velocity:
msgho = ½msvf2 + ½I(
𝐯𝐟
𝐫𝒔)2
For the solid cylinder, the moment of inertia is given by: I = ½mr2
Substituting: msgho = ½msvf2 + ½(½msrs
2)(vf
r𝑠)2
Simplifying: gho = ½vf2 + ½ ½rs
2(𝐯𝐟
𝐫𝒔)2
gho = ½vf2 + ¼rs
2𝐯𝒇𝟐
𝒓𝒔𝟐
gho = ½vf2 + ¼𝐯𝒇
𝟐
gho = ¾vf2
𝟒
𝟑𝐠𝐡𝐨= vf
PROBLEM 8: