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Unformatted text preview: 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...
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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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