Transcript Rotational Kinematics (Part I from chapter 10)

Finish Momentum
Start Spinning Around
March 22, 2006
Watsup?
• Quiz on Friday
– Last part of energy conservation through
today.
• Watch for still another WebAssign
• Full calendar for the remainder of the
semester is on the website near the end
of the last set of PowerPoint slides.
• Next exam is next week!
F external  m1a1  m2a 2
F external
d2
d 2 m1r1  m2r2
 2 (m1r1  m2r2 )  M 2 (
)
dt
dt
M
where
M  m1  m2
This would look like F  ma if we define
the center of mass vector R as
R CM
m1r1  m2r2

M
and
F  Ma cm
We therefore define the center of mass as
1 

R CM 
  m1ri 
M i

M   mi
i
Center of Mass, Coordinates

The coordinates of the center of mass are
xCM 

m x
i i
i
M
yCM 
m y
i
i
M
i
zCM 
m z
i i
i
M
where M is the total mass of the system
Center of Mass, position

The center of mass can be located by its
position vector, rCM
rCM 

m r
i i
i
M
ri is the position of the i th particle, defined by
ri  xi ˆi  yi ˆj  zi kˆ
Center of Mass, Example



Both masses are on the
x-axis
The center of mass is
on the x-axis
The center of mass is
closer to the particle
with the larger mass
Center of Mass, Extended Object


Think of the extended object as a system
containing a large number of particles
The particle distribution is small, so the mass
can be considered a continuous mass
distribution
Center of Mass, Extended Object,
Coordinates

The coordinates of the center of mass of a
uniform object are
xCM
1
1

x dm yCM 

M
M
1
zCM 
z dm

M
 y dm
Center of Mass, Extended Object,
Position

The position of the center of mass can also
be found by:
rCM

1

r dm

M
The center of mass of any symmetrical object
lies on an axis of symmetry and on any plane
of symmetry
Density r
mass
r
unit - volume
m  rV
dm  rdv
Usually, r is a constant but NOT ALWAYS!
Center of Mass, Example


An extended object can
be considered a
distribution of small
mass elements,
Dm=rdxdydz
The center of mass is
located at position rCM
For a flat sheet, we define a mass per unit
area … s
mass
s
unit volum e
dm  sda
Linear Density m
m  mL
dm  mdx
Center of Mass, Rod



Find the center of mass
of a rod of mass M and
length L
The location is on the
x-axis (or
yCM = zCM = 0)
xCM = L / 2
A golf club consists of a shaft connected to a club
head. The golf club can be modeled as a uniform rod
of length L and mass m1 extending radially from the
surface of a sphere of radius R and mass m2. Find the
location of the club’s center of mass, measured from
the center of the club head.
R
L
m2
m1
Motion of a System of Particles



Assume the total mass, M, of the system
remains constant
We can describe the motion of the system
in terms of the velocity and acceleration of
the center of mass of the system
We can also describe the momentum of
the system and Newton’s Second Law for
the system
Velocity and Momentum of a System
of Particles

The velocity of the center of mass of a system of
particles is
mv
vCM

dr
 CM 
dt
i
i
i
M
The momentum can be expressed as
Mv CM 
m v
i
i


i

p
i
 p tot
i
The total linear momentum of the system equals the
total mass multiplied by the velocity of the center of
mass
Acceleration of the Center of Mass

The acceleration of the center of mass can
be found by differentiating the velocity with
respect to time
aCM
dv CM
1


dt
M
m a
i i
i
Newton’s Second Law for a System of
Particles


Since the only forces are external, the net external
force equals the total mass of the system multiplied
by the acceleration of the center of mass:
SFext = M aCM
The center of mass of a system of particles of
combined mass M moves like an equivalent particle
of mass M would move under the influence of the
net external force on the system
Momentum of a System of Particles


The total linear momentum of a system of
particles is conserved if no net external force
is acting on the system
MvCM = ptot = constant when SFext = 0
Motion of the Center of Mass,
Example



A projectile is fired into the
air and suddenly explodes
With no explosion, the
projectile would follow the
dotted line
After the explosion, the
center of mass of the
fragments still follows the
dotted line, the same
parabolic path the projectile
would have followed with
no explosion
Chapter 10
Rotation of a Rigid Object
about a Fixed Axis
Rigid Object

A rigid object is one that is nondeformable


The relative locations of all particles making up
the object remain constant
All real objects are deformable to some extent, but
the rigid object model is very useful in many
situations where the deformation is negligible
Angular Position



Axis of rotation is the
center of the disc
Choose a fixed
reference line
Point P is at a fixed
distance r from the
origin
Angular Position, 2




Point P will rotate about the origin in a circle of
radius r
Every particle on the disc undergoes circular motion
about the origin, O
Polar coordinates are convenient to use to represent
the position of P (or any other point)
P is located at (r, q) where r is the distance from the
origin to P and q is the measured counterclockwise
from the reference line
Angular Position, 3



As the particle moves,
the only coordinate
that changes is q
As the particle moves
through q, it moves
though an arc length
s.
The arc length and r
are related:

s=qr
Radian

This can also be expressed as

q is a pure number, but commonly is given

the artificial unit, radian
One radian is the angle subtended by an
arc length equal to the radius of the arc
Conversions

Comparing degrees and radians
1 rad =

= 57.3°
Converting from degrees to radians
θ [rad] =
[degrees]
Angular Position, final

We can associate the angle q with the entire rigid
object as well as with an individual particle


Remember every particle on the object rotates through
the same angle
The angular position of the rigid object is the
angle q between the reference line on the object
and the fixed reference line in space

The fixed reference line in space is often the x-axis
Angular Displacement

The angular displacement is
defined as the angle the
object rotates through
during some time interval
Dq  q f  q i

This is the angle that the
reference line of length r
sweeps out
Average Angular Speed

The average angular speed, ω, of a rotating
rigid object is the ratio of the angular
displacement to the time interval
q f  qi
Dq


t f  ti
Dt
Instantaneous Angular Speed

The instantaneous angular speed is defined
as the limit of the average speed as the time
interval approaches zero

lim
Dt 0
Dq dq

Dt
dt
Angular Speed, final

Units of angular speed are radians/sec



rad/s or s-1 since radians have no dimensions
Angular speed will be positive if θ is
increasing (counterclockwise)
Angular speed will be negative if θ is
decreasing (clockwise)
Average Angular Acceleration

The average angular acceleration, a,
of an object is defined as the ratio of the
change in the angular speed to the time it
takes for the object to undergo the change:
 f  i
D
a

t f  ti
Dt
Instantaneous Angular Acceleration

The instantaneous angular acceleration is
defined as the limit of the average angular
acceleration as the time goes to 0
a
lim
Dt 0
D d

Dt
dt
Angular Acceleration



Units of angular acceleration are rad/s² or s-2
since radians have no dimensions
Angular acceleration will be positive if an
object rotating counterclockwise is speeding
up
Angular acceleration will also be positive if an
object rotating clockwise is slowing down
Angular Motion, General Notes

When a rigid object rotates about a fixed axis
in a given time interval, every portion on the
object rotates through the same angle in a
given time interval and has the same angular
speed and the same angular acceleration
 So
q, , a all characterize the motion of the entire
rigid object as well as the individual particles in
the object
Directions


Strictly speaking, the
speed and acceleration
(, a) are the
magnitudes of the
velocity and
acceleration vectors
The directions are
actually given by the
right-hand rule
Using this convention,  is a VECTOR!
Hints for Problem-Solving

Similar to the techniques used in linear
motion problems


With constant angular acceleration, the
techniques are much like those with constant
linear acceleration
There are some differences to keep in mind

For rotational motion, define a rotational axis
The choice is arbitrary
 Once you make the choice, it must be maintained


The object keeps returning to its original
orientation, so you can find the number of
revolutions made by the body
Rotational Kinematics

Under constant angular acceleration, we
can describe the motion of the rigid object
using a set of kinematic equations


These are similar to the kinematic equations for
linear motion
The rotational equations have the same
mathematical form as the linear equations
Rotational Kinematic Equations
 f  i  a t
1 2
q f  q i  i t  a t
2
 2f  i2  2a (q f  qi )
1
q f  q i   i   f ) t
2
The derivations are similar to what we did
with kinematics. Example:
d
 constant  a
dt
d  adt
   adt  at  C
C  0
dq
   0  at 
dt
1 2
q  q 0   0 t  at
2
Comparison Between Rotational and
Linear Equations
Relationship Between Angular and
Linear Quantities

Displacements

Speeds

s qr
v  r
Accelerations
a  ar


Every point on the
rotating object has the
same angular motion
Every point on the
rotating object does not
have the same linear
motion
A dentist's drill starts from rest. After 3.20 s of
constant angular acceleration, it turns at a rate of 2.51
 104 rev/min. (a) Find the drill's angular acceleration.
(b) Determine the angle (in radians) through which the
drill rotates during this period.
An airliner arrives at the terminal, and the engines are shut off.
The rotor of one of the engines has an initial clockwise
angular speed of 2 000 rad/s. The engine's rotation slows with
an angular acceleration of magnitude 80.0 rad/s2. (a)
Determine the angular speed after 10.0 s. (b) How long does it
take the rotor to come to rest?
A centrifuge in a medical laboratory rotates at an angular
speed of 3 600 rev/min. When switched off, it rotates 50.0
times before coming to rest. Find the constant angular
acceleration of the centrifuge.
Speed Comparison

The linear velocity is
always tangent to the
circular path


called the tangential
velocity
The magnitude is
defined by the
tangential speed
ds
dq
v
r
 r
dt
dt
Acceleration Comparison

The tangential
acceleration is the
derivative of the
tangential velocity
dv
d
at 
r
 ra
dt
dt
Speed and Acceleration Note



All points on the rigid object will have the same
angular speed, but not the same tangential
speed
All points on the rigid object will have the same
angular acceleration, but not the same tangential
acceleration
The tangential quantities depend on r, and r is not
the same for all points on the object
Centripetal Acceleration

An object traveling in a circle, even though it
moves with a constant speed, will have an
acceleration

Therefore, each point on a rotating rigid object will
experience a centripetal acceleration
v2
aC 
 r 2
r
Resultant Acceleration


The tangential component of the acceleration
is due to changing speed
The centripetal component of the
acceleration is due to changing direction
Rotational Motion Example


For a compact disc
player to read a CD,
the angular speed must
vary to keep the
tangential speed
constant (vt = r)
At the inner sections,
the angular speed is
faster than at the outer
sections