Transcript Rotational Motion Power Point

Rotational Motion
6-1 Angular Position, Velocity, & Acceleration
Angular Displacement (q)
Angular Quantities
Measured in radians (rad)
this unit has no dimensions
Counterclockwise is positive
1rev  2rad  360
o
6-1 Angular Position, Velocity, & Acceleration
Angular Velocity (w)
Defined in the same way as velocity
Average velocity is displacement divided by
time
q
w
t
SI unit
rad
s
s
1
Instantaneous Angular Velocity
w
lim
t  0
q
t
6-1 Angular Position, Velocity, & Acceleration
Angular Acceleration (a)
Defined like acceleration
Angular acceleration is change in angular
velocity divide by time
w
a 
t
SI unit
rad
s2
s
2
Instantaneous Angular
Acceleration
a
lim
t  0
w
t
6-1 Angular Position, Velocity, & Acceleration
6.2 Rotational Kinematics
All the equations used in kinematics (for
conditions of constant acceleration)
x  x0  (v0  v)t
w  w0  at
1
q  q 0  2 (w0  w )t
x  x0  v0t  at
q  q 0  w 0t  at
v  v0  at
1
2
1
2
2
v  v  2a ( x  x0 )
2
2
0
1
2
2
w  w  2a (q  q 0 )
2
2
0
Are exactly the same, but with the new
quantities
6.2 Rotational Kinematics
6.3 Connections Between Linear & Rotational
For an object moving in a circular path
At any given time, the
object is moving with a
2r
v t rw
t
linear velocity tangent
to the arc
The speed would be circumference divide by
the time for one revolution
Since w would be 2/t
6.3 Connections Between Linear & Rotational
Similarly we can determine that
So in a situation where a spinning
object causes linear motion
at  ra
Circumference of the
tire
Will equal the
displacement
6.3 Connections Between Linear & Rotational
6.4 Rotational K & the Moment of Inertia
In a spinning object, each particle has kinetic
energy
Lets assume that we have a rod of uniform
mass rotating about its end
w
v
r
6.4 Rotational K & the Moment of Inertia
Now, just consider a piece of the rod at the
very end
To calculate the kinetic energy of the mass
KK K
 (m
mr
mv
(rw)w)
11 1
22 2
2 2 22
w
v
Converting
To rotational
r
quantities
6.4 Rotational K & the Moment of Inertia
So K depends on angular velocity
And it depends on the distribution of mass
This mass distribution is called Moment of
Inertia
2
2
2
1 1
KK
 2(mr
w
2 Iw)w
v
r
6.4 Rotational K & the Moment of Inertia
Moment of Inertia varies with shape, mass,
and axis of rotation (You will need these to
solve
problems)
6.4 Rotational K & the Moment of Inertia
6.5 Conservation of Energy
We are adding a new type of Kinetic Energy
into our existing Energy Equation
K L0  U g 0  U s0  K L  U g  U s
Adding Rotational Kinetic Energy
K r0  K L0  U g 0  U s0  K r  K L  U g  U s
Where Kr is defined as
K r  Iw
1
2
2
6.5 Conservation of Energy
So if a solid ball is rolling down a slope
The equation would become
Using the moment of inertia of a solid sphere
And the relationship between v and w
227v2 2 221 21 11 2 2 22
1112U
1 2K
K r0  K Lmgy

U

K

K

U

U
U

K
mgy
mgy


mgy

(
mr
mr
mv

I
w
w
)
w
mv


mv
mv
mv
mgy

mr
(
)

mv
g
s
r
L
g
s
g
r
L
r
2 0 10
2 2 22
0
025 5
0 55
6.5 Conservation of Energy
Example: A yo-yo is released from rest and
allowed to drop as the top end of the string is
kept stationary. The mass of the yo-yo is
0.056kg its radius is 1.5 cm. Assume it acts as
a solid disc rotating around its center. What is
the angular speed of the yo-yo after it has
dropped 0.50m?
K r0  K L0  U g 0  U s0  K r  K L  U g  U s
6.5 Conservation of Energy
Example: A yo-yo is released from rest and allowed to drop as the top end of the
string is kept stationary. The mass of the yo-yo is 0.056kg its radius is 1.5 cm.
Assume it acts as a solid disc rotating around its center. What is the angular speed
of the yo-yo after it has dropped 0.50m?
What quantities remain in the equation below?
Expanded equation?
Moment of inertia?
Solve (remember v=rw)
Cancel and insert values
22 2rad
2 221121 1221 2 22 22
3322
1111U
122K
K r0  K mgy

U
K

K

U

U
U

K
w

17
mgy
(
9
mgy
.
8


)(.

gy

5
(
mr
mr
)
mr
mr
I
w
w
w
r
(.
)
w
015
w

m
mv
mr
)
mv
(
w
r
mv
w
w
)
mgy
mr
w
Lmgy
g
s
r
L
g
s
g
r
L
s 22 2 2
44 4022 044
0
02
6.5 Conservation of Energy
6.6 Torque
Rotational Dynamics – the causes of rotational
motion
How do we make an object spin
Apply a force away from the pivot
6.6 Torque
The ability to spin increases with force and the
distance from the pivot point
If the force is parallel to the distance, no
rotation occurs
6.6 Torque
It is the perpendicular component of force that
causes rotation
This quantity is called Torque (t)
And is measure in Nm
t  Fr sin q
6.6 Torque
As with angular quantities
Counterclockwise torque is positive
Clockwise torques are defined as negative
r
r
F
F
6.6 Torque
6.7 Torque and Angular Acceleration
If an unbalanced torque is applied to an object
Consider a point where the force is applied
The acceleration of that point is
F
a
m
6.7 Torque and Angular Acceleration
Since
a  ra
We can write this equation as
Combining we get
a
a
r
Now, math tricks, multiply by r/r
The of the fraction is torque, and the bottom is
inertia
Usually written as
 rF FFt FFr
a a taam
Ia
2
mr
 r rmr
mI mr
6.7 Torque and Angular Acceleration
Applying Newton’s second law to rotational
motion we get
t  Ia
6.7 Torque and Angular Acceleration
6.8 Static Equilibrium
Static Equilibrium
How do we calculate the forces needed to
support a bridge?
The bridge is at static (not moving) equilibrium
6.8 Static Equilibrium
Using a diagram of the bridge
P1
P2
W
Two conditions for equilibrium
1. The sum of the forces equals zero
P1  P2  W  0
6.8 Static Equilibrium
F  0
2.Sum of the torques equals zero
t  0
For torque we need a pivot point – the actual
point does not matter because the object is
not rotating
Use a pivot point that eliminates a variable
6.8 Static Equilibrium
The equation becomes
P1
P2
F  0
W
P2 r2 sin q 2  Wrw sin q w  0
6.8 Static Equilibrium
Try the problem with some variable
P1
P2
W
The bridge is 100 m long with two pylons that
are 10 m from each end. The mass of the
bridge is 10,000 kg. What is the force on each
pylon?
6.8 Static Equilibrium
The bridge is 100 m long with two pylons that are 10 m from each end. The mass
of the bridge is 10,000 kg. What is the force on each pylon?
P1
P2
W
First Condition of Equilibrium
P1 PP12P(210000
 98000
)(9.8
N)  0
6.8 Static Equilibrium
The bridge is 100 m long with two pylons that are 10 m from each end. The mass
of the bridge is 10,000 kg. What is the force on each pylon?
P1
P2
W
Second Condition of Equilibrium
r

Wr

0
P
sinP
q

Wr
sin
q w)  00
P22r(280
)P2

(
98000
49000
)(
N
40
ww
22
6.8 Static Equilibrium
The bridge is 100 m long with two pylons that are 10 m from each end. The mass
of the bridge is 10,000 kg. What is the force on each pylon?
P1
P2
W
Combine equations
P1 P
1 49000
P
1 P2 49000
 98000
 98000
N NN
6.8 Static Equilibrium
A 75 kg man climbs a ladder that is at 60o to
the horizontal. The ladder has mass of 40 kg,
is 20 m long and the coefficient of friction
between the ground and ladder is 0.20. How
far up the ladder can the man climb before it
begins to slip? (Assume that there is no friction
at the top of the ladder)
6.8 Static Equilibrium
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass
of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder
is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume
that there is no friction at the top of the ladder)
Diagram?
Dude
Wall
ladder
q=60o
6.8 Static Equilibrium
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass
of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder
is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume
that there is no friction at the top of the ladder)
N2
Force Diagram?
Dude
Wall
ladder
N1
WD
WL
q=60o
f
6.8 Static Equilibrium
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass
of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder
is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume
that there is no friction at the top of the ladder)
Force Equations?
N2
Fy  N1  WL  WD
Wall
Fx  f  N 2
N1
ladder
WD
WL
q=60o
f
6.8 Static Equilibrium
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass
of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder
is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume
that there is no friction at the top of the ladder)
Pivot Point for Torque?
N2
Wall
ladder
N1
WD
WL
q=60o
Fy  N1  WL  WD
f
6.8 Static Equilibrium
Fx  f  N 2
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass
of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder
is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume
that there is no friction at the top of the ladder)
N2
Distances?
20m
10m
ladder
N1
r
Wall
WD
WL
q=60o
Fy  N1  WL  WD
f
6.8 Static Equilibrium
Fx  f  N 2
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass
of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder
is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume
that there is no friction at the top of the ladder)
N2
Angles?
20m
10m
q=30o
ladder
N1
q=30o
r
Wall
WD
q=30o
WL
o
q=60 q=60
Fy  N1  WL  WD
f
o
6.8 Static Equilibrium
Fx  f  N 2
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass
of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder
is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume
f
f
L L
L
D D
D
1 1
that there is no friction at the top of the ladder)
t  fr sin q  W r sin q  W r sin q  N r sin q1
N2
Torque Equation?
20m
10m
q=30o
ladder
N1
q=30o
q=30o
q=60o
f
r
Wall
WD
WL
Fy  N1  WL  WD
6.8 Static Equilibrium
Fx  f  N 2
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass
of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder
is
0.20.
before
begins
( 20
) sin(How
60
( 40the
.8)(10
sin(
30man
) L(climb
75)(9D
.D
8) DDitsin(
30
)Dto slip?1 ((Assume
f ) far up
f)(9ladder
L)can
L the
1201) sin(301)
1
that there is no friction at the top of the ladder)
f t  fr sin
r sin
q .5W
sinNq 0NN r sin q
17
 .q3 f W
1960
 367
r rr10
N2
Values?
20m
10m
q=30o
ladder
N1
q=30o
q=30o
q=60o
f
r
Wall
WD
WL
Fy  N1  WL  WD
6.8 Static Equilibrium
Fx  f  N 2
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass
of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder
is
0.20.
before
begins
( 20
) sin(How
60
( 40the
.8)(10
sin(
30man
) L(climb
75D)(D
9D
.D
8) DDitsin(
30
)Dto slip?1 ((Assume
f ) far up
f)(9ladder
L)can
L the
1201) sin(301)
1
that there is no friction at the top of the ladder)
f t  fr1717
sin
r sin
q.5..r55rW
17
.3..3qf3ff 1960
W
1960
1960
367
367
367
r10
rr11270
10
(sin
1127
Nq )00NN
0 r sin q
N2
Value for N1?
0 N
N11 392
1127
 735
20m
10m
q=30o
ladder
N1
q=30o
q=30o
q=60o
f
r
Wall
WD
WL
Fy  N1  WL  WD
6.8 Static Equilibrium
Fx  f  N 2
A 75 kg man climbs a ladder that is at 60o to the horizontal. The ladder has mass
of 40 kg, is 20 m long and the coefficient of friction between the ground and ladder
is 0.20. How far up the ladder can the man climb before it begins to slip? (Assume
that there is no friction at the top ofDthe ladder) DD D
17.17
33899
(225
.3 f .41960
)1960
 1960
r367
367
14367
..557rm
r.5r11270
11270
11270
00 0
N2
Value for f?
0 N
N11 392
1127
 735
20m
f f f(.2225
)(1127
N.14 )
N1
10m
q=30o
ladder
q=30o
q=30o
r
Wall
WD
WL
q=60o
f
6.8 Static Equilibrium
Fx  f  N 2
6.9 Center of Mass and Balance
An object is balanced, if the net torque is zero
If the rod above is uniform, then when the
Wall
pivot is in the middle, half the mass is on one
side and half on the other
1
2
mr  12 mr
This balance point is called the center of mass
6.9 Center of Mass and Balance
6.10 Conservation of Angular Momentum
When an object spins, it has angular
momentum
v
Defined as
L  Iw
The conservation of angular momentum is
treated just like the conservation of linear
momentum
Conservation of L
6.10 Conservation of Angular Momentum
So the equation for conservation becomes
I A0 w A0  I B0 w
L0B0LI Aw A  I Bw B
6.10 Conservation of Angular Momentum
Example: A small 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
v0=2.4m/s in a circle of radius r0=0.80m. The
string is then pulled slowly through the hole so
that the radius is reduced to r=0.48m. What is
the speed, v, of the mass now?
6.10 Conservation of Angular Momentum
Example: A small 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 v0=2.4m/s in a circle of radius r0=0.80m.
The string is then pulled slowly through the hole so that the radius is reduced to
r=0.48m. What is the speed, v, of the mass now?
Using conservation of momentum (only one
object)
w
I(.ww
Lrv
vL
4rmr
(.mr
8)(
.v4)
48w)v
rIr2w
22 v0
0000 r000
m22 v2
s r
Mass on the end of a string so
Masses cancel and we can convert to v
Enter numbers and solve
6.10 Conservation of Angular Momentum
Example: A record player with a mass of 1.5
kg is spinning at 33.3 rev/min. The radius of
the turntable is 15cm. A bug with a mass of
0.25 kg lands 10cm from the center of the
record. What is the new velocity of the record
player?
6.10 Conservation of Angular Momentum
Example: A record player with a mass of 1.5 kg is spinning at 33.3 rev/min. The
radius of the turntable is 15cm. A bug with a mass of 0.25 kg lands 10cm from the
center of the record. What is the new velocity of the record player?
22
2
2
2 rad
rev 2
rad
1
1
1
1
v
I
w

I
w

I
w
w

3
w
.
04
3
.
04

29
.
0
L

L
(
1
.
5
)(.
15
)
(
3
.
49
)

(
1
.
5
)(.
15
)
w

10B) w B
m
r
w

m
r
w

m(.BB25
.
0589

0169
w
wrB)(.
.
0589
.
0589

.
0169

.
0194

.
0025
R
R
R
R
B
0
s
s
R
min
R R 0 R0 0 22
R R R
Bw
2
2
w
w
w
The record player is a solid disk, the bug is a
point mass
Convert rev/min to rad/s
33.3
rev
min
(
2rad
rev
)(
1min
60 s
)  3.49
rad
s
Substitute and solve
6.10 Conservation of Angular Momentum