fisica1-youn-e-freedman-exercicios-resolvidos

10.72: (a) Στ = Iα
and a = Rα
T
PR =
a
T
1
2
2
MR α =
2P
= =
M
2
Distance the cable moves: x = 1
at
2
1
2 2
50 m = ( 50 m/s ) t
2
v = v + at = 0 +
0
1
2
200 N
4.00 kg
MR
⎛
⎜
⎝
2
aT
R
⎞
⎟
⎠
= 50 m/s
→ t = 1.41s.
2
( 50 m/s )( 1.41s) = 70.5 m s
2
(b) For a hoop, I = MR , which is twice as large as before, so α and aT
would be
2 2
half as large. Therefore the time would be longer. For the speed, v = v0
+ 2ax,
in which
x is the same, so v would be smaller since a is smaller
2
10.73: Find the speed v the marble needs at the edge of the pit to make it to the level
ground on the other side. The marble must travel 36 m horizontally while falling
vertically 20 m.
Use the vertical motion to find the time. Take
v
0 y
y − y
= 0, a
0
y
= v
Then x − x
2
= 9.80 m/s , y − y
0 y
0
t +
1
2
= v
a t
0x
y
2
gives t = 2.02 s
t gives v
0x
0
= 20 m, t = ?
= 17.82 m/s.
+ y to be downward.
Use conservation of energy, where point 1 is at the starting point and point 2 is at the
edge of the pit, where v = 17.82 m/s. Take y = 0 at point 2, so y
2
= 0 and y1
= h.
K + U = K + U
1
mgh =
1
1
2
mv
2
2
+
1
2
2
2
Iω
Rolling without slipping means
mgh =
7
10
2
7v
h =
10g
mv
2
7(17.82 m/s)
=
= 23 m
2
10(9.80 m/s )
1 2 1 2
b) I ω = mv , Independent of r.
2
5
ω = v r =
ω =
2 2 1 2 1 2
. I mr , so I mv
5
2 5
c) All is the same, except there is no rotational kinetic energy term in
mgh = 1
mv
2
2
2
v
h = = 16 m, 0.7 times smaller than the answer in part ( a).
2 g
K : K = mv
1
2
2