Download Report

Rotational Motion
Angular Quantities
In purely rotational motion, all
points on the object move in
circles around the axis of
rotation (“O”). The radius of
the circle is r. All points on a
straight line drawn through the
axis move through the same
angle in the same time. The
angle θ in radians is defined:
(8-1a)
where l is the arc length.
Example: A particular bird’s eye can
just distinguish objects that subtend
an angle no smaller than about
3x10-4 rad.
a)  How many degrees is this?
b)  How small an object can the bird
just distinguish when flying at a
height of 100 m?
Solution:
a)
3600=2π rad, therefore it is:
3x10-4 (360/2π)= 0.0170
b) l=rϴ
l= (100 m)(3x10-4 rad)=3x10-2 m= 3 cm
Angular displacement:
The average angular velocity is
defined as the total angular
displacement divided by time:
The instantaneous angular
velocity: units:( radians/s, or s-1)
*Radians are dimensionless, as they are defined
by the ratio of two lengths*
The angular acceleration is the rate at which the
angular velocity changes with time:
The instantaneous acceleration:units: (radians/
s2, or s-2)
Every point on a rotating body has an angular
velocity ω and a linear velocity v.
They are related:
Therefore, objects
farther from the axis
of rotation will move
faster.
Example: A rotating carousel has one child
sitting on a horse near the outer edge and
another child seated on a lion halfway out
from the center.
How do their linear and angular speeds
compare?
Answer:
The rotate with the same rate ( i.e. their
times for a complete revolution are the
same, ω= constant). The child sitting on the
horse rotates at a greater radius, therefore
his speed would be greater (v=rω)
Example: In the preceding question, how
would the centripetal acceleration of
each child (at different radii) compare?
Answer: As shown in the following
slide, for a constant angular speed the
centripetal acceleration increases with
radius. In fact it depends on the r2.
If the angular velocity of a
rotating object changes, it
has a tangential
acceleration:
Even if the angular
velocity is constant, each
point on the object has a
centripetal acceleration:
Here is the correspondence between linear
and rotational quantities:
The frequency is the number of complete
revolutions per second:
Frequencies are measured in hertz.
The period is the time one revolution takes:
Example:
a)  What is the linear speed of a child seated 1.2
m from the center of steadily rotating merrygo-round that makes one complete
revolution in 4.0s?
b)  What is her acceleration?
Solution:
a)  f=1/T= 1 rev/ 4/0s = 0.25 rev/s= 0.25 Hz
ω= 2πf=2(π)(0.25)= 1.6 rad/s
v=rω= (1.2m)(1.6 rad/s)= 1.9 m/s
b) Since ω= constant, atan= 0
aR= ω2r=(1.6)2(1.2)= 3.0 m/s2
Kinematics for Uniformly Accelerated
Rotational Motion
The equations of motion for constant angular
acceleration are the same as those for linear
motion, with the substitution of the angular
quantities for the linear ones.
Example:
a) A centrifuge rotor is accelerated from rest to
20,000 rpm in 5.0 min. What is its average
angular acceleration over this time period?
Solution:
ω=2πf=(2π)[(20000)/(60)]= 2100 rad/s
Δω ω − ω 0 2100 rad/s-0
= 7.0 rad/s2
α=
=
=
Δt
Δt
300s
b) Through how many revolutions has the centrifuge
rotor on part a) turned during its acceleration period
(5 minutes or 300 s)?
Solution:
It rotated from rest ω0=0. The final angular speed was
ω= 2100 rad/s.
The angular acceleration was α= 7.0 rad/s2.
1 2
θ = ω 0 t + α t = 0+ 0.5(7)(300)2= 3.15x105 rad
2
The number of revolutions = 3.15x105/(2π)= 5.0x104 rev
Rolling Motion (Without Slipping)
In (a), a wheel is rolling without
slipping. The point P, touching
the ground, is instantaneously
at rest, and the center moves
with velocity v.
In (b) the same wheel is seen
from a reference frame where C
is at rest. Now point P is
moving with velocity –v.
The linear speed of the wheel is
related to its angular speed:
Example: A bicycle slows down from 8.40 m/s to rest
over a distance of 115 m. Both wheels of the bike have
diameter of 68.0 cm. Determine:
a)  The initial angular velocity of the wheels.
b)  The total number of revolutions each wheel rotates
in coming to rest.
c)  The angular acceleration of the wheels.
d)  The time it took to come to a stop.
Solution:
υ0
a) ω 0 =
= 8.40/0.340= 24.7 rad/s
r
b) A complete revolution corresponds to 2π radians. A
complete revolution would correspond to linear distance
equivalent to the circumference of the circle ( 2πr)
The number of revolutions is = 115/[2π(0.34)]= 53.8 rev
2
2
ω
−
ω
0
c) α =
= (0- 24.72)/[2(2π)(53.8)] = -0.902 rad/s2
2θ
d) You could find the linear acceleration (a) from
a=rα , then use a kinematic equation to solve for t: υ = υ 0 + at
Or one could just use the known angular quantities.
ω − ω0
=[0 – 24.7]/[-0.902]= 27.4 s
t=
α
Rotational Dynamics Torque & Rotational
Inertia
Knowing that
we see that
, and
τ = rF
for a single particle.
What about an extended object?
As the angular acceleration is the
same every particle that makes
up the whole object, we can
write:
The quantity
is called the
rotational inertia (or moment of inertia)
of an object.
Στ = Iα
One notices that the comparison between Force and
Torque ( and mass and inertia). The rotational inertia in
the torque equation is similar to mass in newton’s 2nd
law:
F = ma
τ = Iα
Net
Net
The distribution of mass matters (for rotational inertia)–
the two objects have the same mass, but the one on the
left has a greater rotational inertia, as so much of its
mass is far from the axis of rotation.
Can you guess what type of shape might have the
greatest rotational inertia? Think of a mass distribution
that could maximize the rotational inertia.
The rotational
inertia of an
object depends
not only on its
mass
distribution but
also the location
of the axis of
rotation –
compare (f) and
(g), for example.
Example: Two weights of mass 5.0 kg and 7.0 kg are
mounted 4.0 m apart on a light rod (whose mass can be
ignored). Calculate the moment of inertia of the system:
a)  When rotated about an axis halfway between the
weights.
b)  When the system rotates about an axis 0.50 m to the left
of the 5.0 kg mass.
0.5m
4.0 m
b)
a)
7.0 kg
5.0 kg
axis of rotation
4.0 m
5.0 kg
axis of rotation
Recall:
, where r is the perpindicular
distance from the axis of rotation.
7.0 kg
Solution:
a)
I= (5.0 kg)(2.0m)2 + (7.0 kg)(2.0 m)2 = 48 kg・m2
b)
I= (5.0 kg)(0.50m)2 + (7.0 kg)(4.5 m)2 = 143 kg・m2
Again it is worth pointing out that the rotational inertia of
an object (often called a rigid body) depends on the its axis
of rotation.
Solving Problems in Rotational Dynamics
1.  Draw a diagram.
2.  Decide what the system comprises.
3.  Draw a free-body diagram for each object
under consideration, including all the forces
acting on it and where they act.
4.  Find the axis of rotation; calculate the torques
around it.
5. Apply Newton’s second law for rotation. If
the rotational inertia is not provided, you
need to find it before proceeding with this
step.
6. Apply Newton’s second law for translation
and other laws and principles as needed.
7. Solve.
8. Check your answer for units and correct
order of magnitude.
Rotational Kinetic Energy
The kinetic energy of a rotating object is given
by
By substituting the rotational quantities, we find
that the rotational kinetic energy can be written:
A object that has both translational and
rotational motion also has both translational and
rotational kinetic energy:
Example: What will be the speed of a solid sphere of
mass M and radius R when it reaches the bottom of an
incline if it starts from rest at a vertical height H an
rolls without slipping?
R
H
ϴ
Solution: Using conservation of energy once
can state the following
1
1 2 1
1 2
2
2
M υ1 + mgh1 + Iω1 = M υ 2 + mgh2 + Iω 2
2
2
2
2
One is comparing the total energy at the top of
the incline to that at the bottom (h2=0). What is
new, is the accounting of the rotational energy.
2
υ
Given: I Sphere = MR 2 , υ = rω, ω =
5
r
In terms of the variables given one can
state:
2 $
!
!
$
1
1
2
υ
MgH = M υ 22 + # MR 2 &# 22 &
%" R %
2
2"5
Solving for υ2:
10
υ2 =
gH
7
When using conservation of energy, both
rotational and translational kinetic energy must
be taken into account.
All these objects have the same potential energy
at the top, but the time it takes them to get down
the incline depends on how much rotational
inertia they have.
The torque does work as it moves the wheel
through an angle θ:
Angular Momentum and Its Conservation
In analogy with linear momentum, we can define
angular momentum L:
For a point mass L=rp= rmv. The angular
momentum of a point mass is actually the
vector cross product of r and the linear
  
momentum(enrichment).
L=r×p
We can then write the total torque as being the
rate of change of angular momentum.
If the net torque on an object is zero, the total
angular momentum is constant.
Therefore, systems that can change their
rotational inertia through internal forces will also
change their rate of rotation:
Example: A mass m attached to the end of a string revolves in a
circle on a frictionless tabletop. The other end of the string
passes through a hole in the table. Initially, the mass revolves
with a speed v1=2.4 m/s in a circle of radius r1=0.80 m. The
string is then pulled very slowly through the hole so the radius
is reduced to r2=0.48 m. What is the speed , v2, of the mass now?
Solution:
Angular momentum is conserved in this case even with
the addition of the force (torque) with the string. The
additional torque created by the increase in tension of
the string is zero about the axis of rotation since ϴ=0
(lever arm length=0).
Conservation of angular momentum states: L = I1ω1 = I 2ω 2
We will consider this to be a point mass then:
2
2
mr
ω
=
mr
∴
1
1
2 ω2
r1
υ 2 = υ1
r2
also
I = mr 2
υ = rω
= 4.0 m/s
A more direct way could have been used. For a point mass
L=rp=rmv. One obtains r1v1=r2v2 immediately.
Vector Nature of Angular Quantities
The angular velocity vector points along the axis
of rotation; its direction is found using a right
hand rule:
Angular acceleration and
angular momentum vectors
also point along the axis of
rotation.