Download Rotational Motion I

There are 2 types of pure unmixed motion:
 Translational - linear motion
 Rotational - motion involving a rotation or
revolution around a fixed chosen axis( an axis
which does not move).
We need a system that defines BOTH types of
motion working together on a system.
Rotational quantities are usually defined with
units involving a radian measure.
Example: Unscrewing a screw or bolt
= 5 rad/sec k
Counter Clockwise rotations are defined as
out of the board ( POSITIVE K motion on the
"z" axis)
Example: Tightening a screw or bolt
= -5 rad/sec k
Clockwise rotations are defined as into the
board ( NEGATIVE K motion on the "z" axis)


1.
2.
3.
Circular motion about
AXIS
Three different measures of
angles:
Degrees
Revolutions (1 rev. = 360
deg.)
Radians (2p rads = 360
deg.)
An automobile wheel has a radius of 42 cm. If a car drives 10 km, through
what angle has the wheel rotated?
a) In revolutions
b) In radians
c) In degrees

Can be given in



Revolutions/s
Radians/s --> Called w
Degrees/s
 f i
w
in radians
t

Linear Speed at r
v  wr
A race car engine can turn at a maximum rate of 12 000 rpm. (revolutions per
minute).
a) What is the angular velocity in radians per second.
b) If helipcopter blades were attached to the crankshaft while it turns with this
angular velocity, what is the maximum radius of a blade such that the speed of the
blade tips stays below the speed of sound.
DATA: The speed of sound is 343 m/s
x
v
 translational velocity
t

w
 rotational velocity
t
Since velocity is defined as the rate of
change of displacement. ANGULAR
VELOCITY is defined as the rate of
change of ANGULAR DISPLACEMENT.
NOTE:
Translational motion tells you TWO THINGS
• magnitude of the motion and the units
• • direction on the given axis
Example: v =3m/s
This tells us that the magnitude is 3 m/s, the positive axis direction




Denoted by a
w f  wi
a
t
w must be in radians per sec.
Units of angular acceleration are rad/s²
Every portion of the object has same angular
speed and same angular acceleration
v
 translatio nal acceleration
t
w
a 
 rotational acceleration
t
a
Once again, following the same lines of
logic. Since acceleration is defined as the
rate of change of velocity. We can say the
ANGULAR ACCELERATION is defined
as the rate of change of the angular
velocity.
Also, we can say that the ANGULAR
ACCELERATION is the TIME
DERIVATIVE OF THE ANGULAR
VELOCITY.
All the rules for integration apply as
well.

Distance
s  r

Speed
v  wr


Different points on the
same object have
different linear
motions!
Acceleration
a  ar
Only works when , w and a are in radians!
Rotational Motion

w
 
i
w f 
2
Linear Motion
t

v v 
x 
t
i
f
2
1 2
  w i t  at
2
1 2
x  v i t  at
2
w  wi  a t
v  v i  at
w  w  2a  
v  v  2a  x
2
2
i
2
2
i
A pottery wheel is accelerated uniformly from rest to a rate of 10 rpm in 30
seconds.
a.) What was the angular acceleration? (in rad/s2)
b.)How many revolutions did the wheel undergo during that time?
A coin of radius 1.5 cm is initially rolling with a rotational speed of 3.0 radians
per second, and comes to a rest after experiencing a slowing down of a = 0.05
rad/s2.
a.) Over what angle (in radians) did the coin rotate?
b.) What linear distance did the coin move?
A turntable capable of
angularly accelerating at
12 rad/s2 needs to be
given an initial angular
velocity if it is to rotate
through a net 400
radians in 6 seconds.
What must its initial
angular velocity be?
a  12 rad / s
  400 rad
2
t  6s
wo  ?
  wo 2t  1 at 2
2
2
400  wo (6)  (0.5)(12)(6)
wo 
30.7 rad/s

Note that weight doesn’t act at a single point – it is
distributed over the entire body

However, we can always calculate the torque due to a
body’s weight by assuming that the entire force of gravity
(weight) is concentrated at a single point called the center
of gravity

If the acceleration of gravity has the same value at all
points on a body, its center of gravity is identical to its
center of mass (where all mass of object is considered
concentrated at a single point)


m2
m1
For a system of point particles:
The center of mass of the system is the point in space
having coordinates (xCM, yCM) defined by:
m3
mx

m x  m x  m x 


m  m  m 
m
i i
xCM
1 1
2 2
1
2
3 3
i
3
i
i
my

m y  m y  m y 


m  m  m 
m
i
yCM
1 1
2
1
2
2
3 3
i
i
3
i
i

Average position of set of particles, weighted by their
masses

Coordinates where all the mass of entire system can be considered to
be concentrated


Sometimes can use symmetry to deduce position of center of
mass
Where would center of mass be for the system of 3 equallyspaced point masses if m1 = m2 = m3 ?
m1
m2
m3
center of mass


For solid bodies (which are a continuous distribution of
matter), the sums given earlier would be impractical
Center of mass can usually be deduced from symmetry
Homogeneous
solid sphere:
CM at center of sphere
Donut:
CM at center of donut hole
A)
B)
C)
D)
3
4
5
6
Just like massive bodies
tend to resist changes in
K  1 mv2 , vt  rw
2
their motion ( AKA "Inertia") . Rotating
K  1 m ( rw ) 2
2
bodies also tend to resist
changes in their motion.
K  1 mr 2w 2 , I   mr 2
2
We call this
ROTATIONAL
K rot  1 Iw 2
2
INERTIA. We can
determine its expression
by looking at Kinetic
Energy.
We now have an expression for the rotation of a
mass in terms of the radius of rotation.
We call this quantity the MOMENT OF INERTIA (I)
with units kgm2
I   mr
Consider 2 masses, m1 & m2,
rigidly connected to a bar of
negligible mass. The system
rotates around its CM.
2
This is what we would see if m1 = m2.
Suppose m1>m2.
m1
m2
r1
vt  rw
m1
m2
r2
Since it is a rigid body, the have the SAME
angular velocity, w. The velocity of the center, vcm
of mass is zero since it is rotating around it. We
soon see that the TANGENTIAL SPEEDS are
NOT EQUAL due to different radii.
I   mr
2
Let's use this equation to analyze
the motion of a 4-m long bar with
negligible mass and two equal
masses(3-kg) on the end rotating
around a specified axis.
m1
m2
EXAMPLE #1 -The moment of Inertia when they are rotating around the
center of their rod.
I   mr 2  mr 2  mr 2
I  (3)( 2) 2  (3)( 2) 2 
24 kgm2
EXAMPLE #2-The moment if Inertia rotating at
one end of the rod
2
2
2
I   mr  mr  mr
I  (3)(0) 2  (3)( 4) 2 
48 kgm2
m1
m2
Now let’s calculate the moment of Inertia rotating at
a point 2 meters from one end of the rod.
m1
m2
2m
I   mr 2  mr 2  mr 2
I  (3)( 2) 2  (3)(6) 2 
120 kgm2
As you can see, the FARTHER the axis of rotation is from the center of mass,
the moment of inertia increases. We need an expression that will help us
determine the moment of inertia when this situation arises.
This theorem will allow us to calculate the moment of
inertia of any rotating body around any axis,
provided we know the moment of inertia about the
center of mass.
It basically states that the Moment of Inertia ( Ip) around any axis "P" is
equal to the known moment of inertia (Icm) about some center of mass plus
M ( the total mass of the system) times the square of "d" ( the distance
between the two parallel axes)
Using the prior example let’s use the parallel axis theorem to calculate the
moment of inertia when it is rotating around one end and 2m from a fixed
axis.
I p  I cm  Md 2
I p  (24)  (6)( 2) 2 
48 kgm2
m1
m2
d = 2m
m1
I p  I cm  Md 2
m2
4m
I p  (24)  (6)( 4) 
2
120 kgm2
(not drawn to scale)
The earlier equation, I =Smr2, worked fine for what is called
POINT masses. But what about more continuous masses like
disks, rods, or sphere where the mass is extended over a
volume or area. In this case, calculus is needed.
dr
I rod @ cm
ML2

12
I rod @ end
ML2

3
R
r
2pr
I disk@ cm
MR 2

2
Consult the table for a list of Moments of
Inertia to view common expressions for “I”
for various shapes and rotational
situations.
Solid disk flipping
Hoop rotating
Solid Sphere
Spinning Rod
Solid disk rotating
Hoop flipping
Hallow Sphere
Rotating rod
A playground merry-go-round of radius 2.00 m has a
moment of inertia I = 275 kgm2 and is rotating about a
frictionless vertical axle. As a child of mass 25.0 kg
stands at a distance of 1.00 m from the axle, the
system (merry-go-round and child) rotates at the rate
of 14.0 rev/min. The child then proceeds to walk
toward the edge of the merry-go-round. What is the
angular speed of the system when the child reaches
the edge?
Solution (details given in class):
1.17 rad/s
C  AB sin 

The above formula for the cross product is useful when the magnitudes
of the two vectors and the angle between them are known.

If you only know the components of the two vectors:
 
A  B  ( Ay Bz  Az By )iˆ  ( Az Bx  Ax Bz ) ˆj  ( Ax By  Ay Bx )kˆ
  
C  A B
C x  Ay Bz  Az B y
C y  Az Bx  Ax Bz
C z  Ax B y  Ay Bx
Components of cross
product vector

Units: N·m
clockwise rotation is
into the page
(negative)
counter-clockwise
rotation is out of the
page (positive)
Into page

Torque is a vector quantity that measures the tendency of a force to rotate an
object about an axis. The magnitude of the torque produced by a force is
defined as
  rF sin   Fperpr

r = distance between the pivot
point and the point of
application of the force.

F = the magnitude of the
force.

 = the angle between the
force and a line extending thru
the pivot and the point of
application.

Fperp = F sin() = the component
of the force perpendicular to the
line connecting the pivot and the
point of application.

  r F




What is the best way to loosen a bolt using a wrench?
Notice that where you exert the force matters!
Which of the following forces is most likely to loosen the bolt
(assuming they all have the same magnitude):
Bolt
Answer: Force C, because the tendency of a force
to cause a rotation about some point depends on
its magnitude and the perpendicular distance l
between the line of action of the force and that
lC point
Point about which bolt rotates
Line of action of force A
Line of action of force B
A
(lA = 0)
Line of action of force C
lB
B
C
A)
B)
C)
D)
Figure A
Figure B
Figure C
Figure D
A)
B)
C)
D)
increasing force increases torque.
decreasing force decreases torque.
increasing lever arm increases torque.
decreasing lever arm decreases torque.

z
 Ia z
N2L for a rigid body in rotational form
In equilibrium the net torque is zero
ONLY external
torques affect the
rigid body’s rotation!
Problem-Solving Strategy
IDENTIFY the relevant concepts:

Sz=Iaz
SET UP the problem using the following steps:
1.
Sketch
2.
Free-body diagram
3.
Coordinate axes for each body and indicate a positive sense of
rotation for each rotating body.
EXECUTE the solution as follows:

Plug and chug
EVALUATE your answer:

Check that the algebraic signs of your results make sense.

We can apply Newton’s 1st law two different ways to a rigid
body when it is in equilibrium

For translational equilibrium:


Linear acceleration

a 0
For rotational equilibrium:


F  0

  0
Angular acceleration

a 0
A)
B)
C)
D)
It would remain the same.
It would increase by a factor of 1.5.
It would increase by a factor of 2.
It would increase by a factor of 4.
A)
B)
C)
D)
0 Nm
400 Nm
800 Nm
2000 Nm
An 8.00-m, 200-N uniform ladder rests against a
smooth wall. The coefficient of static friction
between the ladder and the ground is 0.600,
and the ladder makes a 50.0° angle with the
ground. How far up the ladder can an 800-N
person climb before the ladder begins to slip?
Solution (details in class):
6.15 m
Consider the model at right of a person
bending forward to lift a 200-N object.
The spine and upper body are
represented as a uniform horizontal rod
of weight 350 N, pivoted at the base of
the spine. The erector spinalis muscle,
attached at a point 2/3 of the way up
the spine, maintains the position of the
back. The angle between the spine
and this muscle is 12.0°. Find the
tension in the back muscle and the
compressional force in the spine.
l
2l/3
l/2
y
+
Solution (details given in class):
T = 2.71  103 N = 2.71 kN
Rx = 2.65 kN
(Note that when object is in equilibrium, you have freedom to choose the
point about which to calculate the torques.)
x
An Atwood’s machine consists of
blocks of masses m1 = 10.0 kg and m2 =
20.0 kg attached by a cord running over
a pulley as shown. The pulley is a solid
cylinder with mass M = 8.00 kg and
radius r = 0.200 m. The block of mass
T1
T
2
m2 is allowed to drop, and the cord
turns the pulley without slipping. (a)
Why must the tension T2 be greater
than the tension T1? (b) What is the
acceleration of the system, assuming
the pulley axis is frictionless? (c) Find
the tensions T1 and T2.
Solution (details given in class):
(b) 2.88 m/s2
(c) T1 = 127 N, T2 = 138 N
Cat on a ladder
Two workmen are carrying a 6 meter ladder as shown in the
sketch below. The ladder has a mass of 15 kilograms. A cat,
with a mass of 5 kg, jumps on the ladder and hangs on, 1
meter from the end of the ladder. How much force does
each workman have to exert to hold the ladder up?