fisica1-youn-e-freedman-exercicios-resolvidos

8.108: The impulse applied to the cake is J = µ k1
mgt = mv , where m is the mass of the
cake and v is its speed after the impulse is applied. The distance d that the cake moves
1 2
during this time is then d = µ gt . While sliding on the table, the cake must lose its
1 2
kinetic energy to friction, or mg ( r − d ) = mv .
gives
1 2
µ
r − d =
k1
2
g
µ k 2
t
2
2
k1
µ Simplification and substitution for v
k2
, substituting for d in terms of
2
2
t gives
2
1 2⎛
µ ⎞
k1
1 2 µ
k1
r = gt µ
k1
= ( µ
k1
+ µ
k2
),
2
⎜ +
µ
⎟ gt
⎝
k2 ⎠ 2 µ
k2
which gives t = 0.59 s.
8.109: a) Noting than dm = dx avoids the intermediate variable ρ . Then,
L
M
x
1 L M L
= x dx
M
∫ .
0
L 2
cm
=
b) In this case, the mass M may be found in terms of ρ and L, specifically by using
2
dm = ρ Adx = αAdx
to find that M = α A ∫ xdx = αAL
2 . Then,
x
cm
2
=
2
αAL
∫
0
L
2 2
αAx
dx =
αAL
2
3
L
3
2L
= .
3
8.110: By symmetry, x = 0 . Using plane polar coordinates leads to an easier
a 3
( y = )
4 πa
3
cm
2
integration, and using the Theorem of Pappus 2π
cm( )
2
method of Problem 8.109 involves Cartesian coordinates.
For the x-coordinate,
π is easiest of all, but the
2
dm = ρ t a
2 − x dx , which is an even function of x, so
2 2
∫ x dx = 0.
For the y-coordinate, dm = ρt2 a − y dy , and the range of integration is
from 0 to a, so
2ρt
a
2 2
ycm = ∫ y a − y , dy.
M 0
2
2 2
Making the substitutions M = 1 ρπa
t,
u = a − y , du = −2
y , and
y
cm
2
1
3
− 2 o − 4 ⎡ ⎤ 4a
2
2
= u du u
2 2
2 ⎥ =
πa
∫ = ⎢ .
a
3πa
⎣ ⎦ 2
3π
0
a