Unformatted text preview: 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...
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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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