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Unformatted text preview: Vector nature of rotation
r
ω
r
ri CHAPTER 10 θ increasing ccw dθ θ CONSERVATION OF ANGULAR MOMENTUM
• Vector nature of rotation
o the cross product
o the right hand rule Because rotation can be clockwise or counter clockwise, • Angular momentum The directions are given by the righthandrule. • Conservation of angular momentum
o everyday examples
o the gyroscope*
o Kepler’s 2nd Law* angular velocity and angular acceleration are actually
vector quantities, with both direction and magnitude . ccw * You study using the notes provided.
cw r
F φ r
r l = r sin φ RMM02VD2.MOV ω= dθ
and v = rω
dt Torque is also a vector. Expressed mathematically, the
torque is defined as rrr
τ = r × F, r
which is a vector (or cross) product. The direction of τ
r
is given by the righthandrule and the magnitude of τ is
given by
RMM02VD3.MOV dω d 2θ
α=
=
and a t = rα
dt dt 2 r
r
rr
τ = r F sin φ = l F . r
Thus, in the case shown, the direction of τ is out of the page.
Note: the value of the torque depends on the rotation
axis and the direction of the force. r
B θ r
A We have already seen that the scalar or dot product of
two vectors is defined as:
rr
rr
A • B = A B cos θ
r
r
If A ⇒ ( A x , A y , A z ) and B ⇒ (B x , B y , B z ) , then
rr
A • B = A xB x + A yB y + A zB z.
~ see revision notes on website ~
Example: r
r
Work done (W) = F • ∆x However, the cross (vector) product is very different ...
for example: the result is a vector. r
B ˆ
n
θ r
A The vector or cross product of two vectors is defined as:
rrr
rr
ˆ
C = A × B = A B sin θ n
r
ˆ
where n is a unit vector that is perpendicular to both A
r
and B and in a direction given by the righthandrule.
~ See website ~ ijˆ
Since the unit vectors ˆ , ˆ , k are orthogonal we have:
ˆ×ˆ=k
ˆ × ˆ = −k
ˆ
ijˆ
ji ˆ×k=ˆ
jˆi
ˆij
k×ˆ=ˆ ˆj
k × ˆ = −ˆ
i
ˆ × k = −ˆ
iˆ
j and ˆ × ˆ = ˆ × ˆ = k × k = 0,
iijjˆˆ ˆ
i
− ve
VTM07AN2.MOV ˆ
k The Right hand rule for
the cross product
rrr
C= A×B + ve ˆ
j ~ Here’s an easy way to
remember it
~ r
B ˆ
j
θ r
r
i
j
ij
Question 1: If A = 2ˆ + 3ˆ and B = −4ˆ + ˆ , (a) what is
rr
r
r
A × B , and (b) what is the angle between A and B? r
A r
i
j
(a) Given: A = 2ˆ + 3ˆ
r
ij
and B = −4ˆ + ˆ ,
ˆ
i rrr
i
j
ij
then C = A × B = (2ˆ + 3ˆ ) × ( −4ˆ + ˆ )
= −8( ˆ × ˆ ) + 2( ˆ × ˆ ) − 12( ˆ × ˆ ) + 3( ˆ × ˆ )
ii
ij
ji
jj
0
0
ˆ
ˆ
ˆ
= 2 k − 12( − k ) = 10 k . rr
rr
ˆ
(b) We know A × B = A B sin θ k ,
r
but A = 22 + 32 = 13
r
and B = ( −4)2 + 12 = 17 rr
10
ˆ
= 0.672 .
So A × B = 14.87 sin θ k , i.e., sin θ =
14.87
The solution θ = 42.2o is clearly not appropriate, but we
remember that sin α ≡ sin(180 − α ).
∴ θ = 180o − 42.2o = 137.8o. Another important concept in rotation
about an axis is angular momentum ... y ˆ
j
r
v r
r
O φ
m
r
r
( p = mv )
ˆ
i
x r
L is outward The angular momentum about O is defined as:
rrr
L = r × p,
r
r
where p = mv is the linear momentum of the object.
The magnitude of the angular momentum is;
L = mvr sin φ = pr sin φ ,
r
and the direction of L is given by the righthandrule and
r
r
is perpendicular to r and v . Note: a particle does not have to be rotating to have
angular momentum.
Dimension: [M][L]2
[ T] Units: kg ⋅ m2 /s (or J.s) Question 2: An object of mass 3kg is moving with a
r
ˆ
velocity v = (3.0m/s)i along the line z = 0, y = 5.3m. (a)
What is the angular momentum of the object relative to
the origin at the point x = 0, y = 5.3m? (b) What is the
angular momentum of the object relative to the origin at
r
i
the point x = 12 m, y = 5.3m? (c) If a force F = ( −3N)ˆ
is applied to the object, what is the torque of this force
relative to the origin? ˆ
j
y (0, 5.3m) r
5.3m r1 r
r2 (12m, 5.3m)
r
v = (3m/s) ˆ
i
ˆ
k
x O Take a closer look at a
similar scenario ... ˆ
j Origin ˆ
i d ˆ
i r
rr
(a) Angular momentum about O is L1 = r1 × p,
r
j
where the position vector is: r1 = (5.3ˆ ) m and the
r
r
momentum is p = mv .
r
ˆ
∴ L = 5.3ˆ × 9ˆ = ( −47.7 k ) kg ⋅ m2 /s .
ji
1 (Note the direction) r
i
j
(b) For the position vector r2 = (12ˆ + 5.3ˆ ) m,
r
rr
ˆ
L2 = r2 × p = (12ˆ + 5.3ˆ ) × 9ˆ = ( −47.7 k ) kg ⋅ m2 /s,
i
j
i
i.e., the same ... very interesting!
rrr
(c) The torque about O is: τ = r × F
ˆ
= (12ˆ + 5.3ˆ ) × ( −3ˆ ) = ( +15.9 k ) N ⋅ m.
i
j
i (Note the direction) r
r1 r
r r2
p
θ1 r
p r
r3 θ2 r
p
θ3 The angular momentum of the object about the origin is
r
rr
ˆ
Li = ri × p = ri p sin θi k (outward)
r
∴ Li = Li = ri p sin θi
r
But ri sin θi = d , which is constant for all pairs of ri , θi,
r
providing the direction of p is unchanged!
r
So, the angular momentum Li is constant provided there
r
are no external forces to change the direction of p, i.e., no torques.
~ Rotation analog of the conservation
of linear momentum ~ ˆ
j r
ri Look at the special case of
motion in a circle ... angular
momentum is conserved .... r
vi
∆mi r
L ˆ
i
O
r
r r
p ... although a centripetal force acts on the object, it is
Consider a planar object rotating about O. If ∆mi is a
mass element, its angular momentum about the rotation
axis is:
But v i = riω . directed through the rotation axis; it produces zero
torque and so there’s no change in angular momentum! r
r
r
ˆ
Li = ri × ( ∆mi ) v i = ∆mi ri v i k
v
ˆ
∴ Li = ( ∆mi ri2 )ω k . So, the total angular momentum is
r
r
ˆ
ˆ
L = ∑i Li = ∑i ( ∆mi ri2 )ω k = Iω k
ˆr
But ω k = ω .
r
r
∴ L = Iω ,
r
r
i.e., in the direction of ω . cf: linear momentum p is
r
parallel to v . r
r
L = Iω
No net torque ...
ω
I
∴1= 2
I2 ω1 RMM09VD1.MOV If there is a net external torque acting on an object then,
from Newton’s 2nd Law of rotation, we find
r
r
r
r
r
dω d (Iω ) dL
τ = Iα = I
=
=,
dt
dt
dt DISCUSSION PROBLEM [10.1]: i.e., the rate of change of angular momentum of the
object is equal to the net external torque. (Compare with
r
r dp
Newton’s 2nd Law F = .) Note: if there is no net
dt
r
r
dL
= 0 , as before.
external torque, i.e., τ ⇒ 0, then
dt
A rotating cloud of gas begins to collapse under If a torque is applied for time ∆t , then
r
r
(Units of L ⇒ J.s)
τ.∆t = ∆L . r
r
Remember: F.∆t = ∆p, the impulse equation? gravitational forces within. As it collapses, what
happens to its:
• Moment of inertia?
• Angular momentum? Let’s look some situations
involving angular momentum ... • Angular velocity? ω1
l l ω2 ⇒
I2 I1 This is a conservation of angular momentum problem ...
I1ω1 = I2ω 2 ... but we need to make some assumptions ...
• body is a cylinder ( r ≈ 0.2 m, Mb ≈ 50 kg ). Question 3: An ice skater starts a spin with arms
outstretched, rotating at 1.5rev/s. Estimate her rotational
speed (in rev/s) when she brings her arms against her
body. • each arm is a thin rod ( l ≈ 1m, Ma ≈ 5kg).
1
1
With arms “out” I1 ⇒ MbR12 + 2 × Ma (2l)2
2
12
1
1
= × 50 × (0.2)2 + 2 × × 5 × 22 = 4.33 kg ⋅ m2.
2
12
1
With arms “in” I2 ⇒ (Mb + 2Ma )R 22
2
1
= × (50 + 10) × (0.22)2 = 1.45 kg ⋅ m2 .
2
Iω
4.33 × 1.5
∴ ω2 = 1 1 =
= 4.5rev/s.
I2
1.45 (a) The angular momenta
ωo ωo ωo
2r are
r 2r ωo
r L1 = I1ω o and L2 = I2ω o,
but they are in opposite directions. Since there are
no external torques angular momentum is conserved. Question 4: Two disks, with identical masses but The initial angular momentum is different radii (r and 2r) are spinning on frictionless L1 + ( −L2 ) = I1ω o − I2ω o = (I1 − I2 )ω o,
and the final angular momentum is bearings with the same angular speed ω o but in opposite (I1 + I2 )ω . directions. The two disks are brought slowly together
and the resulting frictional force between their surfaces
eventually brings them to a common angular velocity.
(a) What is the magnitude of the final angular speed in
terms of ω o ? (b) What is the change in the rotational
kinetic energy? 1
1
But I1 = M(2 r )2 = 2Mr 2 and I2 = Mr 2.
2
2
1
1
∴ 2 − Mr 2ω o = 2 + Mr 2ω , 2
2
i.e., 32
5
Mr ω o = Mr 2ω
2
2
3
∴ ω = ω o.
5 The total initial rotational kinetic energy:
1
1
1
Ki = I1ω o2 + I2ω o2 = (I1 + I2 )ω o2
2
2
2
The total final rotational kinetic energy:
1
K f = (I1 + I2 )ω 2.
2
∴ ∆K K f − Ki ω 2 − ω o2
=
=
,
K
Ki
ω o2 3
but ω = ω o, so (ω 2 − ω o2 ) < 0, i.e., ∆K is always < 0.
5
∆K
∴
= −0.64,
K i.e., 64% of the kinetic energy is “lost”.
• What’s happened to it?
Note: this is the analog of the inelastic collision we
studied in chapter 8. Question 5: (a) Assuming the Earth to be a
homogeneous sphere of radius R and mass M, show that
the period of rotation about its axis is
4 πM 2
T=
R,
L
where L is the angular momentum of the Earth about its
axis. (b) Suppose the radius changes by a small amount
∆R . Show that the fractional change in period is
∆T
∆R
≈2
.
T
R (c) By how many kilometers would the Earth have to
expand for the period to change by 0.25 days/year so
that leap years would no longer be necessary? (a) If the angular velocity is ω we have
• angular momentum L = Iω ,
2
• moment of inertia I = MR 2.
5
2π
The period of rotation is T =
ω ω = 2 π 2 πI 4 π M 2
R.
=
=
L
L 5L I QED. DISCUSSION PROBLEM [10.2]:
A student sits on a stool that is free to rotate. She is
handed a wheel that is spinning in the horizontal plane as
shown below. (b) Let R ⇒ R + ∆R so T ⇒ T + ∆T , then
4 πM ∆T dT
T
≈
= 2
R = 2 2 R . 5L R ∆R dR
∴ ∆T
∆R
≈2
.
T
R Assuming she (and the stool) are stationary to begin (c) ∆R ≈ R ∆T
∆T 0.25 day 0.25 day
=
=
and we need
.
2T
T
1 year
365 day ∴ ∆R = 6.37 × 103 × 0.25
= 2.18 km.
2 × 365 with, what happens if she inverts the wheel (through
180o) ? We have a perfectly inelastic collision and no external
torques. So, angular momentum is conserved at the
collision. Question 6: A uniform rod of length 1.20m and mass
0.80kg is pivoted at its top. The rod, initially at rest, is
struck by a lump of putty of mass 0.30kg at a point x cos θ
d θ x x
x(1 − cos θ) 0.96m below the pivot point. If the putty sticks to the d rod and the maximum deflection of the rod is 60o from θ 2 +
d×
cm the vertical, what was the speed of the putty at impact? d (1 − cos θ)
2 Also, mechanical energy is conserved, i.e.,
0.96m
1.20m 0.30kg v M = 0.80kg Ki + Ui = K f + Uf .
But K f = Ui = 0, so Ki = Uf . The rotational kinetic
1
energy of the rod and putty at impact is Ki = Iω 2,
2 where I is the moment of inertia of the rod plus putty.
An increase in potential energy occurs for both the rod
and the putty. The increase in potential energy at the top of the swing is
d
Uf = Mg (1 − cos θ) + mgx (1 − cos θ) ,
2
where M is the mass of the rod and m is the mass of
putty. Conservation of mechanical energy gives
12
d
Iω = g M + mx (1 − cos θ).
2 2
Also, at impact angular momentum is conserved, i.e., the
angular momentum of the putty before the impact equals
the angular momentum of the rod and putty after the
impact.
mvx 0.288 v
∴ mvx = Iω ⇒ ω =
=
,
I
I 1
where I = Md 2 + mx 2 3 0.80 × (1.20)2
=
+ 0.30 × (0.96)2 = 0.660 kg ⋅ m2.
3 ∴ω = 0.288 v
= 0.436 v ,
0.660 which gives ω in terms of v, the velocity of the putty. Hence, substituting for ω gives
1
× 0.660 × (0.436 v )2
2 0.80 × 1.20
= 9.81
+ 0.30 × 0.96 (1 − cos 60o ). 2
∴ 0.0627 v 2 = 9.81(0.480 + 0.288) × Hence v = 1
= 3.767.
2 3.767
= 7.75m/s.
0.0627 Physics of the Gyroscope The Gyroscope The axle of a wheel is supported by a pivot at O and held A spinning wheel exhibits a remarkable property: horizontal by a hand, say. The weight of the wheel
r
produces a torque τ in the horizontal plane, so, if the end
r
τ r
τ d rr
∆L = τ∆t
r
L ω
O
r
F ω r
mg r
L
∆φ gyro.mov To download a copy of this movie go to the “Useful notes
...” link on the homepage and look under Chapter 10. r
∆L r
L of the axle is released the wheel falls. If the wheel is
given a very large angular velocity ω before the end is
released, it does not fall; rather, the axle rotates slowly in
the horizontal plane about O. Why?
The action is explained by Newton’s 2nd Law for
rotation. r
τ
rr
∆L = τ∆t
r
L r
L
∆φ r
∆L r
τ r
L rr
∆L = τ∆t
r
L r
r
The wheel has large angular momentum L = Iω along its
spin axis. When the end is released the weight of the
r
wheel applies a torque τ on the wheel. Therefore,
according to Newton’s 2nd Law for rotation, there will ∴ωp = r
L
∆φ r
∆L r
L ∆φ mgd mgd
=
=
.
∆t
L
Iω be a change in angular momentum, given by
r
rr
r ∆L
τ=
⇒ ∆L = τ∆ t ,
∆t
r
in the direction of τ, i.e., in the horizontal plane. All the If the angular momentum of the spinning wheel is large, time the wheel is spinning, the “free” end will rotate the axle would point downward resulting in a large about the pivot support at O. This motion is called vertical component of angular angular momentum; but precession. If the angular velocity of the precession is there is no torque in the vertical direction! So, the end ω p, then (working with magnitudes), does not fall. ∆L = τ∆t = ( mgd )∆t ,
and ∆L = L∆φ = ( mgd )∆t . the precession can be very slow.
Note that if the “free” end were to fall when released, Question 7: A bicycle wheel of radius 28cm has an axle
that is 10cm long. The total mass of the wheel and tire is
0.95kg and it is rotating at 15rev/s. If one end of the
axle is supported by a pivot, (a) what is the angular
momentum of the spinnng wheel? (Assume the wheel is (a) The angular momentum of the spinning wheel is L = Iω,
1
1
where I = MR 2 = × 0.95 × (0.28)2 = 0.0372 kg ⋅ m2 .
2
2 ∴ L = 0.0372 × (15 × 2π ) = 3.51J ⋅ s. a hoop.) (b) What is the angular velocity of precession,
and (c) how long does it take the axle to rotate through 360o about the supporting pivot? (b) From before ω p =
= mgd
L 0.95 × 9.81 × 0.05
= 0.133rad/s.
3.51 (c) Time for one revolution is
2π
2π
T=
=
= 47.2s.
ω p 0.133 Kepler’s 2nd Law: the Law of equal areas
Kepler’s 2nd Law of planetary motion states that if a line
is drawn from a planet to the Sun, the areas swept by the
line in the same time intervals are equal throughout the
orbit. r
vdt
dA dt
dA
dt dA Sun r
r vp dA dt The shaded area is onehalf the area of the parallelogram,
1r r
1r
1r r
r
∴ dA = r × vdt =
r × mv dt =
r × p dt .
2
2m
2m
i.e., Since the mass of the planet is constant
dA L
=
= constant .
dt 2 m dA L
=
,
dt 2 m rr
where L = r × p is the magnitude of the angular momentum of the planet about the Sun. The gravitational
r
force between the Sun and planet is along r so it has no
torque about the Sun. Thus, L is conserved, i.e., constant. Aphelion v
a ra rp Perihelion Also, since L is constant, rv sin φ is constant, where φ is
r
r
the angle between r and v . At aphelion and perihelion, φ = 90o. ∴ ra v a = rp v p. ...
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This note was uploaded on 07/13/2011 for the course PHY 2048 taught by Professor Guzman during the Spring '08 term at FAU.
 Spring '08
 Guzman
 Physics, Angular Momentum, Momentum

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