fisica1-youn-e-freedman-exercicios-resolvidos

10.79: a) v =
10K
7m
=
( 10)( 0.800)( 1 2)( 400 N m)( 0.15 m)
7( 0.0590 kg)
2
= 9.34m s.
b) Twice the speed found in part (a), 18 .7 m s. c) If the ball is rolling without
slipping, the speed of a point at the bottom of the ball is zero. d) Rather than use the
intermediate calculation of the speed, the fraction of the initial energy that was converted
to gravitational potential energy is ( 0 .800)( 0.900) , so ( 0.720)( 1 2) kx 2 = mgh and solving
for h gives 5.60 m.
10.80: a)
b) R is the radius of the wheel (y varies from 0 to 2R) and T is the period of the
wheel’s rotation.
c) Differentiating,
v
v
x
y
2πR
⎡ ⎛ 2πt
⎞⎤
= ⎢1
− cos⎜
⎟
T
⎥
⎣ ⎝ T ⎠⎦
2πR
⎛ 2πt
⎞
= sin⎜
⎟
T ⎝ T ⎠
a
a
y
x
⎛ 2π
⎞
= ⎜ ⎟
⎝ T ⎠
2
⎛ 2π
⎞
= ⎜ ⎟
⎝ T ⎠
2
⎛ 2πt
R sin⎜
⎝ T
⎞
⎟
⎠
⎛ 2πt
⎞
R cos⎜
⎟.
⎝ T ⎠
⎛ 2πt
⎞
d) vx = v
y
= 0 when ⎜ ⎟ = 2π
or any multiple of 2π , so the times are integer
⎝ T ⎠
multiples of the period T. The acceleration components at these times are
2
4π
R
ax
= 0,
a
y
= .
2
T
2
2
2 2 ⎛ 2π
⎞
2 ⎛ 2πt
⎞ 2 ⎛ 2πt
⎞ 4π
R
e) ax
+ a
y
= ⎜ ⎟ R cos ⎜ ⎟ + sin ⎜ ⎟ = ,
2
⎝ T ⎠ ⎝ T ⎠ ⎝ T ⎠ T
independent of time. This is the magnitude of the radial acceleration for a point moving
2π on a circle of radius R with constant angular velocity
T
. For motion that consists of this
circular motion superimposed on motion with constant velocity ( a r = 0),
the acceleration
due to the circular motion will be the total acceleration.