fisica1-youn-e-freedman-exercicios-resolvidos

10.94: a), g)
r r r r
b) Using the vector product form for the angular momentum, v = −v
and = − , 2
so
r r r r
m
,
2
× v2
= m
1
× v1 r
so the angular momenta are the same. c) Let ω = ωĵ.
Then,
r r r
v
( ˆ ˆ
1
= ω×
1
= ω zi
− xk) , and
r r r
( ) ˆ 2
L = m × v = mω − xR i + ( x
2 + y ) ˆj
+ ( xR)
ˆ).
1 1 1
k
1 2 1
r
r
r
2 2 2
2 r 2 2
With x + y = R , the magnitude of L is 2mω R , and L ⋅ ω = mω
R , and so
2 2
m ω R 1
π
cosθ =
2
= , andθ
= . This is true for L r 2
6
2
as well, so the total angular
(2mωR
)( ω )
momentum makes an angle of 6
π
with the +y-axis.
1
d) From the intermediate
2
calculation of part (c), L y 1
= mω R = mvR,
so the total y-component of angular
momentum is L = 2mvR.
e) L is constant, so the net y-component of torque is zero. f)
y
y
Each particle moves in a circle of radius R with speed v, and so is subject to an inward
force of magnitude mv 2 R.
The lever arm of this force is R, so the torque on each has
2
magnitude mv . These forces are directed in opposite directions for the two particles,
and the position vectors are opposite each other, so the torques have the same
2
magnitude and direction, and the net torque has magnitude 2mv .