Transcript Rotational Motion 2015x

Rotational Kinematics –
its just like linear kinematics except
stuff spins instead of going in a straight line…
Rotational Displacement, 
Consider a disk that rotates from A to B:
B

Angular displacement :
A
Measured in revolutions,
degrees, or radians.
1 rev = 360 0 = 2 rad
The best measure for rotation of
rigid bodies is the radian.
Definition of the Radian
One radian is the angle  subtended at
the center of a circle by an arc length s
equal to the radius R of the circle.
s

R
s
Example 1 A rope is wrapped many times
around a drum of radius 50 cm. How many
revolutions of the drum are required to raise a
bucket to a height of 20 m?
s
20 m
 
R 0.50 m
 = 40 rad
R
Now, 1 rev = 2 rad
 1 rev 
   40 rad  

 2 rad 
 = 6.37 rev
h = 20 m
Example 2 A bicycle tire has a radius of 25 cm. If
the wheel makes 400 rev, how far will the bike
have traveled?
 2 rad 
   400 rev  

 1 rev 
 = 2513 rad
s =  R = 2513 rad (0.25 m)
s = 628 m
Angular Velocity
Angular velocity,w, is the rate of change in
angular displacement. (radians per second.)
w

t
Angular velocity in rad/s.
Angular velocity can also be given as the
frequency of revolution, f (rev/s or rpm):
w  2f Angular frequency f (rev/s).
Example 3 A rope is wrapped many times around
a drum of radius 20 cm. What is the angular
velocity of the drum if it lifts the bucket to 10 m in 5
s?
s
10 m
 
R 0.20 m
w

t

 = 50 rad
R
50 rad
5s
h = 10 m
w = 10.0 rad/s
Those Oversized Tires!
If you drive a vehicle with oversize
tires, how is the speedometer
affected?
Those Oversized Tires!
• The use of oversized tires has an impact
on the speedometer reading. Your vehicle
is actually covering greater distance due to
the larger circumference of the oversized
tires. Your speedometer was calibrated for
the smaller tires sold on the vehicle so
your speedometer is reading less speed
and you are moving faster.
If your tires are 10% larger than the original tires
on your truck you travel 10% more than your truck
recognizes. This has two effects:
Your actual speed is 10% MORE than what your
speedometer is reflecting. (When your
speedometer reads 60, you are actually traveling
66) 10=11, 40=44, 100=110 etc.
Your actual mileage is 10% MORE than what
your odometer is reflecting. (When your odometer
reads 100 miles, you have actually traveled 110
miles.)
Example 4 In the previous example, what is the
frequency of revolution for the drum? Recall that w
= 10.0 rad/s.
w
w  2 f or f 
2
10.0 rad/s
f 
 1.59 rev/s
2 rad/rev
R
Or, since 60 s = 1 min:
rev  60 s 
rev
f  1.59

  95.5
s  1 min 
min
f = 95.5 rpm
h = 10 m
Angular Acceleration
Angular acceleration is the rate of change in
angular velocity. (Radians per sec per sec.)
w

t
2
Angular acceleration (rad/s )
The angular acceleration can also be found
from the change in frequency, as follows:
2 (f )

t
Since
w  2 f
Example 5 The block is lifted from rest until the
angular velocity of the drum is 16 rad/s after a time
of 4 s. What is the average angular acceleration?

w f  wo
t
0
or

wf
R
t
16 rad/s
rad

 4.00 2
4s
s
h = 20 m
 = 4.00 rad/s2
Angular and Linear Speed
From the definition of angular displacement:
s =  R Linear vs. angular displacement
s    R   
v


t  t   t

R

v=wR
Linear speed = angular speed x radius
Angular and Linear
Acceleration:
From the velocity relationship we have:
v = wR Linear vs. angular velocity
v  Δω∙R
v  R   Δω
v 
va 

  R
t  t   t 
a = R
Linear accel. = angular accel. x radius
Example 6:
Consider flat rotating disk:
wo = 0; wf = 20 rad/s
t=4s
What is final linear speed at
points A and B?
vAf = wAf R1 = (20 rad/s)(0.2 m);
vBf = wBf R2 = (20 rad/s)(0.4 m);
R2
R1
B
A
R1 = 20 cm
R2 = 40 cm
vAf = 4 m/s
vBf = 8 m/s
Acceleration Example 7
Consider flat rotating disk:
R1
A
B
wo = 0; wf = 20 rad/s
t=4s
What is the average angular
and linear acceleration at B?

w f  w0
t
20 rad/s

4s
a = R = (5 rad/s2)(0.4 m)
R2
R1 = 20 cm
R2 = 40 cm
 = 5.00 rad/s2
a = 2.00 m/s2
Angular vs. Linear Parameters
Recall the definition of linear
acceleration a from kinematics.
a
v f  v0
t
But, a = R and v = wR, so that we may write:
a
v f  v0
t
becomes
R 
Angular acceleration is the time
rate of change in angular velocity.
Rw f  Rw0
t

w f  w0
t
Trains ride on a pair of tracks.
For straight-line motion, both
tracks are the same lengths. But
which track is longer for a curve,
the one on the outside or the
one on the inside of the curve?
The outside track is longer, just
as a circle with a greater radius
has a longer circumference
A Comparison: Linear vs. Angular
w f  wo  t
v f  vo  at
sd  v0t  at
1
2
2asd  v  v
2
f
2
0
2
  w 0t   t
1
2
2
2  w  w
2
f
2
0
Linear Example: A car traveling initially at
20 m/s comes to a stop in a distance of
100 m. What was the acceleration?
V0 = 20m/s
d = 100 m
Vf = 0 m/s
100 m
Select Equation:
2as  v2f  v02
a=
0 - vo2
2s
vo = 20 m/s
vf = 0 m/s
-(20 m/s)2
=
2(100 m)
a = -2.00 m/s2
Example 8: A disk (R = 50 cm), rotating at 600
rev/min comes to a stop after making 50 rev.
What is the acceleration?
Select Equation:
2
2
2  w f  w0
wo = 600 rpm
wf = 0 rpm
rev  2 rad   1 min 
600


  62.8 rad/s
min  1 rev   60 s 
=
0 - wo2
2
-(62.8 rad/s)2
=
2(314 rad)
 = 50 rev
50 rev = 314 rad
 = -6.29 rad/s2
Problem Solving Strategy:
 Draw and label sketch of problem.
 Indicate + direction of rotation.
 List givens and state what is to be
found.
Given: ____, _____, _____ (,wo,wf,,t)
Find: ____, _____
 Select equation containing one and not
the other of the unknown quantities, and
solve for the unknown.
Example 9 A drum is rotating clockwise initially at 100
rpm and undergoes a constant counterclockwise
acceleration of 3 rad/s2 for 2 s. What is the angular
displacement?
Given: wo = -100 rpm; t = 2 s
 = +3
rad/s2

rev  1 min  2 rad 
100


  10.5 rad/s
min  60 s  1 rev 
  w ot  12  t 2  (10.5)(2)  12 (3)(2)2
 = -20.9 rad + 6 rad
 = -14.9 rad
Net displacement is clockwise (-)
R
Bet you can’t
Center of gravity or
Center of mass is the
point at the center of an
object’s weight
distribution, where the
force of gravity can be
considered to act.
If the Center of Mass is NOT supported,
an object will topple.
A Force that cause rotation
about an axis is
TORQUE
Pushing on a door is an example of a force
that causes rotation.
TORQUE
The pivot is in the hinges and
the force needed to cause it to
rotate is dependent on …
1) the amount of force
2) the angle of the force
3) the distance from the pivot
Torque = Forceperpendicular x lever arm
 = F r
Units = N.m
The same torque can be produced by a
large force with a short lever arm, or a
small force with a long lever arm.
Picture of lever arms from
page 151
Although the magnitudes of the applied forces are
the same, the torques are different. Only the
component of forces perpendicular to the lever arm
contribute to torque.
Let’s go back to the door
If the force is perpendicular
then the lever arm (r) is the
distance from the force to
the pivot.
If the force is along the
arm then, no rotation occurs
so no torque.
If you apply the force at an
angle, then the lever arm will
be smaller and the torque
less.
So how can we use this…
Say the bolt is too tight and
you can’t get it to turn, what
can you do?
Example 10: What torque is necessary to
lift a 10 N dumbbell by your muscles 0.80
m away?
F  = 10 N
r = 0.80m
 = F r
 = 10 N x 0.80 m
 = 8 Nm
Let’s review Equilibrium
The condition of equilibrium exists
where the resultant of all external
forces is zero.
We have used this
relationship to solve problem
where the objects were
stationary or at rest.
First Condition of Equilibrium: A body is in
translational equilibrium if and only if the
vector sum of the forces acting upon it is zero.
Σ Fx = 0
and
Σ Fy = 0
Second Condition of Equilibrium: An object
is in rotational equilibrium when the sum of
the forces and torques acting on it is zero.
Στ=0
By choosing the axis of rotation at the point of application of an
unknown force, problems may be simplified.
We will solve torque problems by using the idea
that the clockwise torque balances the
counterclockwise torque when an object is in
rotational equilibrium.
Clockwise Torque
Counterclockwise Torque
 = 0
clockwise = counterclockwise
Example 11: The 60 kg diver is standing 1.20 m
from the pivot. What force must be exerted by the
bolt 0.80 m away from the pivot? Assume the board
is supported at its center of mass.
r cw = 1.2 m
r ccw = 0.8 m
F cw = 600 N
Fccw = ??
0.8 m
counterclockwise = clockwise
Fr
ccw
= Fr
cw
F (0.80 m) = (600 N)(1.20 m)
F = 900 N
1.20 m
Example 12: What is the mass required to balance
this system?
Example 13: A uniform beam weighing 200 N is held
up by two supports A and B. Given the distances and
forces listed find the forces exerted by the
supports.
The dumbells below are all the same size, and
the forces applied all have the same
magnitude.Rank in order from largest to
smallest, the torques created (τ1 , τ2 or τ3 )
τ 1 > τ 2 = τ3
An equal force will lead to a
greater torque if applied at a
right angle. A torque applied at
90 degrees – θ creates the same
torque as 90 degrees + θ.
Comparing rotational
and linear motion
Angular Velocity
Angular velocity,w, is the rate of change in
angular displacement. (radians per second.)
w

t
Angular velocity in rad/s.
Angular velocity can also be given as the
frequency of revolution, f (rev/s or rpm):
w  2f Angular frequency f (rev/s).
Angular Acceleration
Angular acceleration is the rate of change in
angular velocity. (Radians per sec per sec.)
w

t
2
Angular acceleration (rad/s )
The angular acceleration can also be found
from the change in frequency, as follows:
2 (f )

t
Since
w  2 f
Force and Linear Acceleration
When an object is subject to a net
force, it undergoes an acceleration.
(Newton’s 2nd)
Torque and Angular Acceleration
When a rigid object is subject to a
net torque, it undergoes an angular
acceleration.
Inertia of Rotation
Consider Newton’s second law for the inertia of rotation
Linear Inertia, m = F/a
F = 20 N
m=
a = 4 m/s2
F = 20 N
R = 0.5 m
=2
rad/s2
20 N
= 5 kg
4 m/s2
Rotational Inertia, I
I=


(20 N)(0.5 m)
=
= 2.5 kg m2
2
2 rad/s
Force does for translation what
torque does for rotation:
  I
F  ma,
Moment of Inertia
This mass analog is called the moment of
inertia, I, of the object
I   mi ri
2
i
Is defined relative to rotation axis
SI units are kg m2
More About Moment of Inertia
• I depends on both the mass and its
distribution.
• If an object’s mass is distributed further
from the axis of rotation, the moment of
inertia will be larger.
Some dragsters are built so that the front
wheels are far ahead of the rear wheels.
The main
reason for this is
a. to streamline the dragster
b. to provide better traction.
c. to keep the front of the car down.
Dragsters baby!!
The front wheels are so far forward
to help prevent the front from
raising off the ground. It can
produce a counter torque to the
driving force of the wheels.
Concept Question: Cylinder Race
Two cylinders of the same size and mass
roll down an incline, starting from rest.
Cylinder A has most of its mass
concentrated at the rim, while cylinder B
has most of its mass concentrated at the
center. Which reaches the bottom first?
1) A
2) B
3) Both at the same time.
Concept Q. Ans.: Cylinder Race
Answer 2: Because the moment of inertia
of cylinder B is smaller, more of the
mechanical energy will go into the
translational kinetic energy hence B will
have a greater center of mass speed and
hence reach the bottom first.
Common Moments of Inertia
Common moments of inertia are on page 251.
Example 14 A circular hoop and a disk each have a
mass of 3 kg and a radius of 20 cm. Compare their
rotational inertias.
R
I  mR  (3 kg)(0.2 m)
2
I = 0.120 kg m2
R
I = ½mR 2
Disk
2
I = mR 2
Hoop
I  mR  (3 kg)(0.2 m)
1
2
2
1
2
2
I = 0.0600 kg m2
Reminders and info
• Should you take the AP test?
Important Analogies
For many problems involving rotation, there is an
analogy to be drawn from linear motion.
m
x

I R
f
4 kg
A resultant force F
produces negative
acceleration a for a
mass m.
F  ma
w w  50 rad/s
o
 = 40 N m
A resultant torque 
produces angular
acceleration  of disk
with rotational inertia I.
  I
Example 15
Treat the spindle as a solid cylinder.
a) What is the moment of Inertia of
the spindle?
b) If the tension in the rope is 10 N,
what is the angular acceleration of
the wheel?
c) What is the acceleration of the
bucket?
d) What is the mass of the bucket?
M
Solution
a) What is the moment of Inertia of
the spindle?
Given: M = 5 kg, R = 0.6 m
Moments of Inertia
2
MR , cylindrical shell
1
MR 2 , solid cylinder
2
2
MR 2 , solid sphere
5
2
MR 2 , spherical shell
3
1
ML2 , rod, about end
3
1
ML2 , rod, about middle
12
M
1
I  MR 2 = 0.9 kgm2
2
Solution
b) If the tension in the rope is 10 N,
what is ?
Given: I = 0.9 kg m2, T = 10 N, r = 0.6 m
Basic formula
  rF
  I
rT  I
  (0.6m)(10 N)/(0.9 kg∙m2)
 = 6.67 rad/s2
c) What is the acceleration of the
bucket?
Given: r=0.6 m,  = 6.67 rad/s
Basic formula
a  r
a = (6.67 rad/s2)(0.6 m)
M
a=4 m/s2
Solution
d) What is the mass of the bucket?
Given: T = 10 N, a = 4 m/s2
Basic formula
F  ma
Ma  Mg  T
T
M
g a
M = 1.72 kg
M
Concept Question: Cylinder
Race Different Masses
Two cylinders of the same size but different
masses roll down an incline, starting from
rest. Cylinder A has a greater mass. Which
reaches the bottom first?
1) A
2) B
3) Both at the same time.
Concept Q. Ans.: Cylinder Race
Different Masses
The initial mechanical energy is all potential
energy and hence proportional to mass.
When the cylinders reach the bottom of the
incline, both the mechanical energy
consists of translational and rotational
kinetic energy and both are proportional to
mass. So as long as mechanical energy is
constant, the final velocity is independent of
mass.So both arrive at the bottom at the
same time.
Other applications of Torque
• https://mail.google.com/mail/u/0/#inbox/14
a0b51b82af3d78?projector=1
Combined Rotation and
vcmTranslation
vcm
vcm
First consider a disk sliding
without friction. The velocity of
any part is equal to velocity vcm
of the center of mass.
w
Now consider a ball rolling without
slipping. The angular velocity w
about the point P is same as w for
disk, so that we write:
v
w
R
Or
v
R
v  wR
P
Two Kinds of Kinetic Energy
Kinetic Energy
of Translation:
Kinetic Energy
of Rotation:
K = ½mv2
w
v
R
K = ½Iw2
P
Total Kinetic Energy of a Rolling Object:
KT  mv  I w
1
2
2
KE of center-of-mass motion
1
2
2
KE due to rotation
Example 16
What is the kinetic energy of the Earth due to
the daily rotation?
Given: Mearth=5.98 x1024 kg, Rearth = 6.63 x106 m.
Basic formula
2
w
T
Solid sphere
2
I  MR 2
5
Basic formula
1 2
KE  Iw
2
First, find w
2
w
24  3600
1
2 2
KE  MR w
5
= 7.27 x10-5 rad/s
= 2.78 x1029 J
Summary – Rotational
Analogies
Quantity
Linear
Displacement Displacement
x
Inertia
Mass (kg)
Force
Newtons N
Rotational
Radians 
I (kgm2)
Torque N·m
Velocity
v
“ m/s ”
w
Rad/s
Acceleration
a
“ m/s2 ”

Rad/s2
Momentum
mv (kg m/s)
Iw (kgm2rad/s)
Analogous Formulas
Linear Motion
Rotational Motion
F = ma
K = ½mv2
Work = Fx
 = I
K = ½Iw2
Work = 
Power = Fv
Power = Iw
Fx = ½mvf2 - ½mvo2
 = ½Iwf2 - ½Iwo2
Example 17
A solid sphere rolls down a hill of height 40 m.
What is the velocity of the ball when it reaches
the bottom? (Note: We don’t know r or m!)
Basic formula
1 2 1 2
mgh  mv  Iw
2
2
For solid sphere
2 2
I  mr
5
Basic formula
v  wr
1 2 2 2 2
gh   v  r w 
2
5

1  2 2 2  2 2 gh
gh   v  v , v 
2
5 
7/5
v = 23.7 m/s
Angular Momentum
L  Iw
Rigid body
L  mvr
Point particle
Analogy between L and p
Angular Momentum
Linear momentum
L = Iw
p = mv
 = L/t
F = p/t
Conserved if no net Conserved if no net
outside torques
outside forces
Example 18
A 65-kg student sprints at 8.0
m/s and leaps onto a 110-kg
merry-go-round of radius 1.6 m.
Treating the merry-go-round as
a uniform cylinder, find the
resulting angular velocity.
Assume the student lands on the
merry-go-round while moving
tangentially.
Solution
Known: M, R, m, v0
Find: wF
First, find L0
Basic formula
L  mvr
L0  mvR
Next, find Itot
Solid cylinder
I Tot
1
2
 mR  MR
2
2
1
I  MR 2 Now, given I and L , find w
tot
0
2
Particle
Basic formula
I  MR 2
L  Iw
L0
= 2.71 rad/s
w
I Tot
An ice skater spins with her arms folded.
When she extends her arms outward her
angular
velocity
a. increases
b. decreases
a. remains the same
If the skater extends her arms her
radius becomes greater and she has a
greater momentum of inertia. A greater
momentum of inertia causes her to have
less ω due to conservation of
momentum. If her new I is greater, her
new ω must be smaller.
Summary of
Formulas:
K  Iw
1
2
I = mR2
Work  
2
  ½ Iw  ½ Iw
2
f
2
0
I ow o  I f w f
Power 

t
 w
Conservation:
Height?
mgho
Rotation?
½Iwo2
velocity?
½mvo2
=
mghf
Height?
½Iwf2
Rotation?
½mvf2
velocity?