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Rotational Motion
Every quantity that we have studied with
translational motion has a rotational
counterpart
TRANSLATIONAL ROTATIONAL
Displacement x Angular Displacement
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
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
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: A Boy
on a horse is
hunting a goose
around a strange
world of radius 25
m. The angular
separation
between the
hunter/hunted is a
constant 5. What
is the angular
distance (in m)
between the
boy/horse and the
goose?
Speed and velocity
• The same particle rotates with an avg. angular velocity given by
= D𝛉
D𝐭• Linear (Tangential) and angular
speeds are related by the equation
v = r
rD
s
vT
vT
Angular Velocity
tavg
D
D
Units – rad/s, rev/s
Direction – Same as
displacement
(positive is
counterclockwise,
negative is
clockwise)
Compare to
𝒗𝒂𝒗𝒈 =∆𝒙
∆𝒕
EXAMPLE: A space station ring of radius 500 m
spins twice in 12 minutes.
(a) Find its angular velocity
(in rads/s).
(b) Find the linear
(tangential) velocity (in
m/s) of a point on the outer
edge of the ring.
Ans. (a) Twice means two revolutions = 2(2p) = 4 p radians
= 𝜃
𝑡=
4𝜋
12 min × 60 Τ𝑠𝑒𝑐𝑚𝑖𝑛
= 0.017 rad/sec
(b) tangential velocity, vT = r = 0.017 (500 m) = 8.73 m/s
P.O.D. 2: A figure skater spins through five
revolutions in a time of 2.4 s.
(a) Find her angular velocity (in rads/s).
(b) Find the linear (tangential) velocity (in m/s) of her
foot if it is 0.3 meters away from her body.
Acceleration
• The same particle rotates with an avg. angular acceleration given by
= DD𝐭
• Linear (Tangential) and angular accelerations are related by the equation
a = r
rD
s
vT
vT
Angular Acceleration
• Angular
Acceleration is
how the angular
velocity changes
with time
tavg
D
D
Units – rad/s2
Compare to
𝒂𝒂𝒗𝒈 =∆𝒗
∆𝒕
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
Sample Problem
A compact disk rotates about an
axis according to the formula:
(t) = t2 – 6
(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.
(a) To find the linear speed, use the relationship, where (5)
we obtained from the function (t) = t2 – 6 (5)= 52 –6=19v = r = (19 rad/s)(0.20 m) = 3.8 m/s.
(b) To find the linear acceleration, use the relationship
ac = 2r = (19 rad/s)2(0.20 m) = 72.2 m/s2
P.O.D. 3: Diana Prince starts rotating with an angular
velocity of 5 rad/s, where the negative sign indicates a
clockwise rotation, to transform into Wonder Woman.
After 5 seconds she has completed the transformation
and her angular velocity is 2 rad/s.
(a) Find Wonder
Woman’s angular
acceleration (in
rad/s2).
(b) Find the centripetal
acceleration (in m/s2)
of her indestructible
bracelets (at = 2
rad/s) if they are 0.8
m away from her
body when she is
spinning.
Constant Angular Acceleration
• Our kinematics equations have angular
equivalents
• Just as with their linear counterparts, these only
work for constant acceleration
First Kinematic Equation
• x = vot + ½ at2 (linear form)
– Substitute angle for position.
– Substitute angular velocity for linear velocity.
– Substitute angular acceleration for linear
acceleration.
= ot + ½ t2 (angular form)
Second Kinematic Equation
• v = vo + at (linear form)
– Substitute angular velocity for linear velocity.
– Substitute angular acceleration for linear
acceleration.
= o + t (angular form)
Sample Problem
A bicycle starts from rest and for 10.0
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 0.50 m. At the
end of the 10.0 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
𝟎.𝟓𝟎 𝒎= 1.6 rad/s2
The angular acceleration should be negative because the tire
spins clockwise.
To find the angular displacement:
(t) = ot + ½t2 = 0(10 s) + ½(1.6 rad/s2)(10 s)2 = 80 rad
P. O. D. 4: an extreme diver rotates at an angular velocity of +3.75 rad/s
while doing his first set of flips. When he does his second set of flips, he
accelerates and reaches a greater angular velocity of +4.4 rads/s. If his
angular acceleration is 1.7 rad/s2...
(a) How long does this flipping part of the dive take (in s)?
(b) Find his angular displacement after the time found in (a).
Angular Momentum
r
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
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 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
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
PROBLEM 6: A Chinese star of mass 0.025 kg and radius 0.03 m is
thrown by Bruce Lee at his adversary. 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.
(b) Find the torque of the Chinese star (Assume the Chinese star is
hoop-shaped).
Linear Momentum vs. Angular MomentumIf 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 =
PROBLEM 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