Transcript Rotational Energy and Momentum

Work and Power for Rotation
Work = Fd = FRq
t = FR
q
Work = tq
tq
Work
Power =
= t
t
s
q
w=
t
F
F
s = Rq
Power = t w
Power = Torque x average angular velocity
Example 1: The rotating disk has
a radius of 40 cm and a mass of
6 kg. Find the work and power if
the 2-kg mass is lifted 20 m in 4 s.
Work = tq = FR q
s
20 m
q=
=
= 50 rad
R
0.4 m
q
2 kg
6 kg
Power =
Work
= 392 J
t
4s
F
F=W
s = 20 m
F = mg = (2 kg)(9.8 m/s2); F = 19.6 N
Work = (19.6 N)(0.4 m)(50 rad)
s
Work = 392 J
Power = 98 W
The Work-Energy Theorem
Recall for linear motion that the work done is equal
to the change in linear kinetic energy:
Fx = ½mv  ½mv
2
f
2
0
Using angular analogies, we find the rotational work
is equal to the change in rotational kinetic energy:
tq = ½ Iw  ½ Iw
2
f
2
0
Two Kinds of Kinetic Energy
Kinetic Energy
of Translation:
Kinetic Energy
of Rotation:
K=
w
½mv2
v
R
P
K = ½Iw2
Total Kinetic Energy of a Rolling Object:
KT = mv  I w
1
2
2
1
2
2
Translation or Rotation?
If you are to solve for a linear parameter, you
must convert all angular terms to linear terms:
s
q=
R
v
w=
R
I = (?)mR 2
If you are to solve for an angular parameter, you
must convert all linear terms to angular terms:
s =qR
v = wR
Example 2: A circular hoop and a circular
disk, each of the same mass and radius,
roll at a linear speed v. Compare the
kinetic energies.
w
w
Two kinds of energy:
KT = ½mv2
v
Kr = ½Iw2
Total energy: E =
½mv2
+
½Iw2
2


v
2
2
Disk: E = ½mv  ½ ½mR  2 
R 
2


v
2
2
Hoop: E = ½mv  ½ mR  2 
R 




v
w=
R
E = ¾mv2
E = mv2
v
Conservation of Energy
The total energy is still conserved for
systems in rotation and translation.
However, rotation must now be considered.
Begin: (U + Kt + KR)o = End: (U + Kt + KR)f
Height?
mgho
Rotation?
½Iwo2
velocity?
½mvo2
=
mghf
Height?
½Iwf2
Rotation?
½mvf2
velocity?
Example 3: Find the velocity of the 2-kg mass
just before it strikes the floor.
R = 50 cm
mgho
mghf
=
½Iwo2
½Iwf2
½mvf2
½mvo2
mgh0 = 12 mv 2  12 I w 2
(2)(9.8)(10) = (2)v  (6)v
2
2 kg
h = 10 m
I = 12 MR 2
2


v
2
2
1
1 1
mgh0 = 2 mv  2 ( 2 MR )  2 
R 
1
2
6 kg
1
4
2
2.5v2 = 196 m2/s2
v = 8.85 m/s
Example 4: A hoop and a disk roll from the
top of an incline. What are their speeds at
the bottom if the initial height is 20 m?
mgho = ½mv2 + ½Iw2
Hoop: I = mR2
2


v
2
2
mgh0 = ½ mv  ½(mR )  2 
R 
20 m
mgho = ½mv2 + ½mv2; mgho = mv2
v = gh0 = (9.8 m/s2 )(20 m)
Hoop:
Disk: I = ½mR2; mgho = ½mv2 + ½Iw2
2

v 
2
2
mgh0 = ½mv  ½(½mR )  2 
R 
v = 14 m/s
v=
4
3
gh0
v = 16.2 m/s
Angular Momentum Defined
Consider a particle m
moving with velocity v
in a circle of radius r.
Define angular momentum L:
L = mvr
Substituting v= wr, gives:
L = m(wr) r = mr2w
For extended rotating body:
L = (Smr2) w
v = wr
m
w
m1
axis
m
4
m3
m2
Object rotating at constant w.
Since I = Smr2, we have:
L = Iw
Angular Momentum
Example
Rank the following from largest to smallest
angular momentum.
answer
Example 5: Find the angular
momentum of a thin 4-kg rod of
length 2 m if it rotates about its
midpoint at a speed of 300 rpm.
1
For rod: I = 12mL2 =
1
(4
12
kg)(2 m)2
L=2m
m = 4 kg
I = 1.33 kg m2
rev  2 rad  1 min 

w =  300


 = 31.4 rad/s
min  1 rev  60 s 

L = Iw = (1.33 kg m2)(31.4 rad/s)2
L = 1315 kg m2/s
Impulse and Momentum
Recall for linear motion the linear impulse is equal to
the change in linear momentum:
F t = mv f  mv0
Using angular analogies, we find angular impulse to
be equal to the change in angular momentum:
t t = Iw f  Iw 0
Example 6: A sharp force of 200 N is applied to
the edge of a wheel free to rotate. The force acts
for 0.002 s. What is the final angular velocity?
I = mR2 = (2 kg)(0.4 m)2
I = 0.32 kg m2
Applied torque t = FR
 t = 0.002 s w = 0 rad/s
w o
R
R = 0.40 m
F
2 kg
F = 200 N
Impulse = change in angular momentum
0
t t = Iwf  Iwo
FR t = Iwf
wf = 0.5 rad/s
Conservation of Momentum
If no net external torques act on a system then the
system’s angular momentum, L, remains constant.
Speed and radius can change just as long as angular
momentum is constant.
0
Ifwf = Iowo
Ifwf  Iowo = t t
Io = 2 kg m2; wo = 600 rpm
I 0w 0 (2 kg  m )(600 rpm)
wf =
=
If
6 kg  m 2
If = 6 kg m2; wo = ?
2
wf = 200 rpm
Example
The figure below shows two masses held
together by a thread on a rod that is rotating
about its center with angular velocity, ω. If the
thread breaks, what happens to the system's (a)
angular momentum and (b) angular speed.
(Increase, decrease or remains the same)
Examples of the Conservation of
Angular Momentum
Natalia Kanounnikova World Record Spin - 308
RPM
Diving
18 meter high dive
Amazing Fouette Turns on Pointe
Playground Physics
Summary – Rotational Analogies
Quantity
Linear
Rotational
Displacement
Displacement x
Radians q
Inertia
Mass (kg)
I (kgm2)
Force
Newtons N
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
t = I
K = ½Iw2
Work = tq
Power = Fv
Power = Iw
Fx = ½mvf2 - ½mvo2
tq =
½Iwf2 - ½Iwo2
Summary of Formulas:
K = Iw
1
2
Work = tq
2
tq = ½ Iw  ½ Iw
2
f
Height?
mgho
Rotation?
½Iwo2
velocity?
½mvo2
2
0
=
I = SmR2
I ow o = I f w f
Power =
tq
t
= tw
mghf
Height?
½Iwf2
Rotation?
½mvf2
velocity?
CONCLUSION: Angular Motion