fisica1-youn-e-freedman-exercicios-resolvidos

J
10.98: The velocity of the center of mass will change by ∆ v = , and the angular
velocity will change by ∆ω
=
J x x
I
( − cm )
⋅The change is velocity of the end of the bat will
J J ( x − xcm)
xcm
then be ∆vend
= ∆vcm
_ ∆ωxcm
= −
⋅ Setting
m Ι
∆ v
end
= 0 allows cancellation of J,
and gives Ι = ( x − xcm)
xcmm,
which when solved for x
is
−2
2
Ι (5.30×
10 kg ⋅ m )
x = + xcm
=
+ (0.600 m) = 0.710 m.
x m (0.600 m)(0.800 kg)
cm
cm
m
10.99: In Fig. (10.34(a)), if the vector
r , and hence the vector L are not horizontal but make an angle β with the horizontal, the
torque will still be horizontal (the torque must be perpendicular to the vertical weight).
The magnitude of the torque will be ω r cos β , and this torque will change the direction
of the horizontal component of the angular momentum, which has magnitude L cos β .
Thus, the situation of Fig. (10.36) is reproduced, but with L
expression found in Eq. (10.36) becomes
horiz
instead of L . Then, the
dφ
Ω = =
dt
d L
dt
L
horiz
=
L
τ
horiz
mgr cos β
=
Lcos
β
ωr
= ⋅
Ιω