fisica1-youn-e-freedman-exercicios-resolvidos

10.76: (a) 1 2 1
mgh = mv + Iω
2 (1)
2 2
2 ⎛ 1 2 ⎞
I = I
rim
+ I
spokes
= M
rR
+ 6⎜
msR
⎟
⎝ 3 ⎠
Uniform density means: m = λ πR and m = λR.
No slipping means that ω = v R.
r
2
s
r
+ ms
= 2πRλ
+ 6Rλ
= 2Rλ
+
Also, = m
( π 3)
m substituting into (1) gives
2Rλ
1
2
( π + 3) gh = ( 2Rλ)( π + 3)( Rω)
ω =
( π + 3)
gh
2
R ( π + 2)
=
and v = Rω = 26.0 m s
2
1 ⎡
+ 2
2
⎢ πRλR
⎣
⎛ 1
+ 6⎜
πRR
⎝ 3
2
( π + 3)( 9.80 m s )( 58.0 m)
2
( 0.210 m) ( π + 2)
2
2
⎞⎤
⎟⎥ω
⎠⎦
2
= 124 rad
s
(b) Doubling the density would have no effect because it does not appear in the answer.
1
ω α ,
R
so doubling the diameter would double the radius which would reduce
ω by half, but v = Rω would be unchanged.
10.77: a) The front wheel is turning at ω = 1.00 rev s = 2π
rad s.
υ = rω = ( 0.330 m)(2π
rad s) = 2.07 s
b) ω = v r = ( 2.07 m s) (0.655 m = 3.16 rad s = 0.503 rev s
c) ω = v r = ( 2.07 m s) (0.220 m) = 9.41rad s = 1.50 rev s
10.78: a) The kinetic energy of the ball when it leaves the tract (when it is still rolling
2
without slipping) is (7 10)
mv and this must be the work done by gravity, W = mgh, so
v = 10gh 7. The ball is in the air for a time t = 2y
g,
so x = vt = 20hy
7.
b) The answer does not depend on g, so the result should be the same on the moon.
c) The presence of rolling friction would decrease the distance.
2
d) For the dollar coin, modeled as a uniform disc, K = (3 4) mv , and so x = 8hy
3.