fisica1-youn-e-freedman-exercicios-resolvidos

10.84: (a)
Στ
= Iα
1 1 2
mg cosθ = mL α
2 3
2
3 cosθ
⎛ 3 ⎞ (9.80 m s )cos60°
α = g = ⎜ ⎟
= 0.92 rad
2 L ⎝ 2 ⎠ 8.00 m
s
2
(b) As the bridge lowers, θ changes,
not valid.
soα
is
not constant. Therefore Eq. (9.17) is
(c) Conservation of energy:
mg
L
2
PE = KE
i
1 ⎛ 1 2 ⎞
sin θ = ⎜ mL ⎟ω
2 ⎝ 3 ⎠
ω =
=
f
→ mgh =
3g
sinθ
L
1
2
2
3(9.8 m s )sin 60°
8.00 m
2
Iω
2
= 1.78 rad
s
10.85: The speed of the ball just before it hits the bar is v = 2 gy = 15.34 m s.
Use conservation of angular momentum to find the angular velocity ω of the bar just
after the collision. Take the axis at the center of the bar.
2
L
1
= mvr = ( 5.00 kg)( 15.34m s)( 2.00 m) = 153.4 kg ⋅ m
Immediately after the collsion the bar and both balls are rotating together.
L I ω
= 2 tot
1 2 2 1
2
2
2
tot
= Ml + 2mr
= ( 8.00 kg)( 4.00 m) + 2( 5.00 kg)( 2.00 m) = 50.67 kg ⋅ m
I
12
12
2
L
2
= L1
= 153.4 kg ⋅ m
ω = L2 Itot
= 3.027 rad s
Just after the collision the second ball has linear speed
v = rw = 2 .00 m 3.027rad s = 6.055m and is moving upward.
1 2
mv
2
( )( ) s
= mgy gives y = 1.87 m for the height the second ball goes.