fisica1-youn-e-freedman-exercicios-resolvidos

10.21: From Eq. (10.11), the fraction of the total kinetic energy that is rotational is
where
2
2
( 1 2)
Icmω
1
=
2
2
( 1 2) Mv ( 1 2) I ω 1+
( M I )
cm
2 2
+ v / ω
cm
cm
cm
1
=
1+
2
MR
Icm
v cm
= Rω
for an object that is rolling without slipping has been used.
2
2
a) I = (1 2) MR ,so the above ratio is1 3. b)I (2 5)MR , so the above ratio is
7.
cm
=
2
2
c) I = 2 3MR
, so the ratio is 2 5. d) I = 5 8MR , so the ratio is 5 13.
,
f
10.22: a) The acceleration down the slope is a = g sin θ − ,
M
the torque about the
center of the shell is
a 2 a 2
τ = Rf = Iα = I = MR
2 = MRa,
R 3 R 3
so f
= 2
a.
M 3
Solving these relations a for f and simultaneously gives 5 a = g sin θ , or
3
3 3
2
2
a = g sinθ
= (9.80m s )sin38.0°
= 3.62m s ,
5 5
2 2
2
f = Ma = (2.00kg)(3.62m s ) = 4.83 N.
3 3
The normal force is Mg cos θ , and since f ≤ µ n s
,
2
f
3
Ma 2 a 2
5 3 g sin θ 2
µ
s
≥ = = = = tan θ = 0.313.
n Mg cosθ
3 g cos θ 3 g cosθ
5
b) a = 3.62m
2
s
since it does not depend on the mass. The frictional force, however,
is twice as large, 9.65 N, since it does depend on the mass. The minimum value of
s
µ also
does not change.