Download AP Rot Mech

Rotational Mechanics
AP Physics C: Mechanics
Enough with the particles…
• Do you ever get tired of being
treated like a particle?
• We can not continue to lump all
objects together and pretend
that they undergo the same
motion when acted upon by the
same force…
• We will now study the rotation
of rigid bodies.
What do we already know?
Objects rotate
about their center
of mass.
Ideas of circular
motion and
centripetal force.
For rotational
motion it is
useful to
consider
tangential and
radial
components
instead of x and y
What is a rigid body?
• An extended object whose size
and shape do not change as it
moves.
• This size and shape can not be
neglected when modeling its
motion.
• Objects that are held together by
“massless rods” of molecular
bonds.
• Some objects can be modeled as
rigid during parts of their motion.
Three types of motion of
a rigid body:
There is an rotational analogy to every
concept of linear motion.
• We have looked at the basics when studying circular
motion:
s  r
ds
vt 
dt
d

dt
dr 
vt 
 r
dt
dv t
at 
dt
d
  
dt
dr 
at 
 r
dt

Tangential Components:
vt
r
ω
Rotational Velocity
Tangential Velocity
vt

r
vt  r
Rotational Acceleration
at

 r
Tangential Acceleration
at  r
Radial (centripetal)
Component
Radial Velocity
vt
vr  0
ar
Radial Acceleration
ω
2
t
v
ar 
 r
r


2
ar
ar   r
2
r
Rotational Kinematic
Equations:
v  v0  at
1 2
d  at  v0t  d0
2
2
v  v0  2ad
2

  0  t
1 2
  t   0t  0
2
2
  0  2
2
Sign Conventions:
Counter-clockwise is positive direction for ω
and vt. So positive α can be speeding up in the
ccw direction or slowing down in the cw
direction.
A Rotating Crankshaft
A car engine is idling at 500rpm. When the light
turns green, the crankshaft rotation speeds up
at a constant rate to 2500rpm over an interval
of 3 seconds. How many revolutions does the
crankshaft make during this time interval?
This is a rotating rigid body with constant angular acceleration
Imagine painting a dot on the
crankshaft. IF the dot is at θ=0
and t=0, at a later time, the
dot will be at:
1 2
  t   0t
2
A Rotating Crankshaft
A car engine is idling at 500rpm. When the light
turns green, the crankshaft rotation speeds up
at a constant rate to 2500rpm over an interval
of 3 seconds. How many revolutions does the
crankshaft make during this time interval?
500rev 1min 2rad 
0 


 52.4rad /s
min 60sec  rev 
2500rev

 5 0  262rad /s
min
  0

t
A Rotating Crankshaft
A car engine is idling at 500rpm. When the light
turns green, the crankshaft rotation speeds up at
a constant rate to 2500rpm over an interval of 3
seconds. How many revolutions does the
crankshaft make during this time interval?
262  52.4 rad /s
  0


 69.9rad /s2
t
3s
1 2
1
  t   0t  69.932  52.43  472rad
2
2
 1rev 
  472rad
 75revs
2rad 
The Center of Mass
A 500g ball and a 2kg ball are connected by a
massless 50cm long rod.
Where is the center of mass?
What is the speed of each ball if they rotate
about the center of mass at 40rpm?
The Center of Mass
A 500g ball and a 2kg ball are connected by a
massless 50cm long rod.
Where is the center of mass?
x cm
2kg0  0.5kg0.5

0.10m
2.5kg
The Center of Mass
A 500g ball and a 2kg ball are connected by a
massless 50cm long rod.
What is the speed of each ball if they rotate
about the center of mass at 40rpm?
x cm  0.10m
rev 1min 2rad 
  40


 4.16rad /s
min  60s  1rev 
v t1  r1  0.10m4.16rad /s  0.42m/s
v t 2  r2  0.40m4.16rad /s  1.68m/s
What can we measure?
• What can we measure and analyze with a bike
in order to further understand rotational
mechanics?
Other Rotational Analogs?
Rotational Energy?
• Why does the Sun rise in the morning? Why do
magnets stick together? Because everybody
says so. Everybody.
• - Michael Scott
Rotational Energy?

Rotational Energy
Why must there be such a thing?
All of the atoms in the object are moving so they
must have kinetic energy!
Translational Kinetic Energy
1 2
K t  mv
2
Can we use our analogies to find an
expression for Rotational Kinetic Energy?
Rotational KE for a particle
traveling in a circle:
1 2
K t  mv
2

Kt  Kr
v r

m ?

mI
1
2
K r  mr
2
1 2 2
K r  mr 
2
1
2
2
K r  mr 
2
1 2
K r  I
2
Moment of Inertia
• Has nothing to do with a moment in time. The word
comes from the Latin momentum which means motion.
• The rotational analog to mass.
• Describes the distribution of mass relative to the axis of
rotation.
• Is different for each shape and orientation.
• An object with a large mass is hard to accelerate, an
object with a large moment of inertia is difficult to rotate.
Moment of Inertia
(Rotational Laziness)
• Inertia is the resistance to changes in motion
• Moment of inertia is the resistance to changes in
rotation.
Moment of Inertia
Consider an irregular shape that is rotating:
The object’s rotational energy
is the sum of the kinetic
energies of each particle
1
1
2
2
K r  m1r1  m2 r2  ...
2
2
1
Kr 
2


2
2
m
r

 ii
Moment of Inertia can be calculated as the sum of
the contributions from each particle in an object
I  m r
2
i i
Moment of Inertia
Calculating moment
of inertia can be
very difficult for odd
shapes. I for many
shapes has been
tabulated and
printed in
handbooks for
scientists and
engineers.
Note about the axis…
If the rotation axis is not through the center of mass, then rotation
may cause the center of mass to move up or down in a
gravitational field. The gravitational potential energy of the object
will change as it spins.
With no friction
in the axle or
other dissipative
forces, the
mechanical
energy can be
described as:
E mech
1 2
 K r  U g  I  Mgy cm
2
The three masses, held together by lightweight plastic rods, rotate
about an axle passing through the right–angle corner. At what
angular velocity does the triangle have 100mJ of rotational energy?
150g
1 2
K r  I
2
6cm
ω
axle
300g
250g
8cm

I  m r

2
i i


I  0.15kg0.06m  0.25kg0.08m  0.3kg0
2

2
3
I  2.14 10 kgm
2
2

The three masses, held together by lightweight plastic rods, rotate
about an axle passing through the right–angle corner. At what
angular velocity does the triangle have 100mJ of rotational energy?
I  2.14 103 kgm 2
150g
6cm
ω
axle

250g
300g
8cm


2K r

I
2100 10
 J
1 2
K r  I
2
3
3
2.14 10 kgm
2
 9.67rad /s
 60s 1rev 
  9.67rad /s

 92rpm
1min  2 
A 1m long, 0.2kg rod is hinged at one end and connected
to a wall. It is held out horizontally, then released. What is
the speed of the tip of the rod as it hits the wall?
Conservation of
Energy:
K
r
 U g i  K r  U g f
1 2
Mgy

I

cm
2
L  1 1
2  2
Mg   ML 
2  2 3




Ei  Ef
g L 2
 
2 6
A 1m long, 0.2kg rod is hinged at one end and connected
to a wall. It is held out horizontally, then released. What is
the speed of the tip of the rod as it hits the wall?
Conservation of
Energy:
g L 2
 
2 6

vt

r

3g

L
r L
3g
L
 vt
L
vt  3gL  5.4m/s
Calculating Moment of
Inertia:
Like finding inertia, we can not simply place the
object on a scale to find its moment of inertia.
We must go through the calculation.
Recall: Moment of Inertia can be calculated as the sum
of the contributions from each particle in an object
as Δm approaches zero it can be replaced with the
differential dm.

I  r mi
2
i
I
 r dm
2
Calculating Moment of Inertia: Tips I 
 r dm
2
Break the object into elements that you will sum together. Do this
in a way that keeps the same distance from the axis for all particles
in each element.
You will sum the elements over a range of distances so you must

find an expression to substitute dm with a differential dx, dy, or dz.
Densities are helpful but not necessarily the only way to solve:
M dm
 
L dx
M dm
 
A dA
M dm
 
V dV
For a complex object made up of parts with known moments of
inertia, sum the terms of each to find part to find the moment of
inertia of the object:


I object  I1  I 2  I 3  ...
Find the moment of inertia of a circular disk of
radius R and mass M that rotates on an axis
passing through its center.
I
R
 r dm
2
0
M dm
 
A dA

R

A  R 2
dA  2rdr
M
I  r
2rdr
2
R
0
2
Find the moment of inertia of a circular disk of
radius R and mass M that rotates on an axis
passing through its center.
R
M
I r
2rdr
2
R
0
2


2M
I 2
R
R
3
r
 dr
0
R
2M r 
I  2  
R  4 0
4
MR
I
2
2
The four T’s in the diagram are made from
identical rods. Rank in order, from largest to
smallest, the moments of inertia for rotation
about each dashed line.
Which has the most mass distributed farthest
from the axis???
Ia>Id>Ib>Ic
Parallel-Axis Theorem
We have been calculating the moment of inertia
for rotational axes that run through the center of
mass. This theorem helps us if we wish to use an
off-center axis but know where it is in relation to
a parallel, on-center axis.
I  I cm  Md
2
Parallel-Axis Theorem
Find the moment of inertia of a thin rod with
mass M and length L about an axis 1/3rd of the
length from one end.
I  I cm  Md
Icm
1
2
 ML
12
2
1/3rd
cm
d=L/6
Moment of inertia through the center
of mass of a thin rod from table.
2
2
2  
2 

ML2 
L  ML
ML
ML
I  
 

 M   
6   12   36  9
 12 
Rotational Force?
Why must there be such a thing?
Because all net forces do not cause rotation!
How can rotational force be maximized?
Apply a force at the
proper location.
How do we calculate it?
Torque
• Torque is to rotational motion as force is to linear
motion.
• Torque is given the Greek symbol capital tau (τ)
Torque
• Is greater with a greater
force
 F
• Is greater if the force is
applied farther from the axis
or rotation.
 r
• Is greater if the angle of
application of the force is
perpendicular to the radial
line.



 sin 
  Frsin 
 Fr
Torque

 Fr
Torque and Lever arm or
Moment Arm
How much Torque does
Luis Provide?
Rank the Torques
τe>τa=τd>τb>τc
Analog to Newton’s
nd
2
Law
F  0
  0
No Change in motion
(no acceleration)
No Change in
rotation
EQUILIBRIUM

Torque due to gravity
 grav  Mgx cm
Center of mass is
relative to the
axis of rotation
Torque due to gravity
The gravitational torque is found by
treating the object as if all its mass were
concentrated at the center of mass.
What is the torque on
the 500kg steel beam?
 grav  Mgx cm  500kg9.8m/s
2
0.8  3920Nm
Torque
Equilibrium Lab
Challenge:
• Please DO NOT touch or alter
any set-up.
• Report the unknown mass in
each. I will collect one per
group for score based on
amount correct.
• 7 Minutes per table remaining
time may be used to
reconsider past tables but
not revisit them while other
groups are working.
• Bonus goes to group with
most correct.
Rotational Dynamics
A net centripetal force will cause an
object’s path to change direction.
2
t
v
2
 Fc  ma c  m r  m r
A net tangential force will cause a rotating
object to speed up or slow down.
F  ma
t
t
 mr
at
ac
Rotational Dynamics
Only component of force tangent to a
circle causes rotation change:
F  ma
t
t
 mr
  F  r  F r
t
   F r  mrr  mr   I
2
t

Surprised?
at
ac
Analog to Newton’s
F  ma
Unbalanced
Forces
nd
2
Law
  I
Unbalanced
Torques
A net torque causes an angular
acceleration!

Rank the angular
accelerations!
αb > αa > α c = αd = αe
Angular Momentum (L)
Linear Momentum
p  mv
p  Ft

mI
v 
F  


Angular Momentum for a particle in
circular motion:


2
2 v 
L  I mr  mr   mrv
r  
Angular Momentum
L  I
L  t


L  F  r t
L  p  r

Angular Momentum
Newton’s 2nd Law:
dL
net 
dt
dp
Fnet 
dt

A net torque causes a
particles angular
momentum to change.
The rate of change of a
systems angular
momentum is the net
torque on the system.
Conservation of Angular
Momentum:
The angular momentum of an isolated (no external
torques) system is conserved.
Li  Lf

A note on direction:
Why is it easier to balance a bike when the
wheels are spinning than when they are not?
An object with a large linear momentum is difficult to
slow down or to knock it off of its straight line path.
An object with a large angular momentum is difficult to
slow down or to change the direction of its axis
 F  r
L  p  r
General Direction of
Cross Product: The RHR
1. Point the fingers of your right hand in the direction of vector A.
2. Close your fingers into the direction of vector B.
3. Your thumb points in the direction of vector C.
Gyroscope!
Use of a gyroscope
Precession
Gyrobowl!