fisica1-youn-e-freedman-exercicios-resolvidos

10.95: a) The initial angular momentum with respect to the pivot is mvr,
and the
2
final total moment of inertia is I + mr , so the final angular velocity is
2
ω = mvr ( mr + I ).
b) The kinetic energy after the collision is
1 2 2
K = ω ( mr + I ) = ( M + m)
gh,
or
2
ω =
2
( M + )
m gh
.
2
( mr + I )
2
⎛ m ⎞
c) Substitution of Ι = Μr into either of the result of part (a) gives ω = ⎜ ⎟( v r),
⎝ m + M ⎠
and into the result of part (b), ω = 2gh(1
r),
which are consistent with the forms for v.
10.96: The initial angular momentum is Ιω1 − mRv1
, with the minus sign indicating that
runner’s motion is opposite the motion of the part of the turntable under his feet. The
2
final angular momentum is ω
2
( Ι + mR ), so
Ιω1
− mRv1
ω2
=
2
Ι + mR
2
(80 kg ⋅ m )(0.200 rad s) − (55.0 kg)(3.00 m)(2.8 m s)
=
2
2
(80 kg ⋅ m ) + (55.0 kg)(3.00 m)
= −0.776 rad s,
where the minus sign indicates that the turntable has reversed its direction of motion (i.e.,
the man had the larger magnitude of angular momentum initially).
10.97: From Eq. (10.36),
ωr
Ω =
Ι ω
2
(50.0 kg)(9.80 m s )(0.040 m)
=
= 12.7 rad
2
(0.085 kg ⋅ m )((6.0m s) (0.33 m))
s,
or 13
rad
s to two figures, which is quite large.