Homework 6 Chapt 8 & 9
Accelerating along a Racetrack
Since acceleration is a vector quantity, an object moving at constant speed along a curved path
has nonzero acceleration because the direction of its velocity
v
harpoonrightnosp
is changing, even though the
magnitude of its velocity (the speed) is constant. Moreover, if the speed is constant, the object's
acceleration is always perpendicular to the velocity vector
v
harpoonrightnosp
at each point along the curved path
and is directed toward the center of curvature of the path.
The magnitude of the acceleration of an object that moves with constant speed v along a circular
path of radius r is given by
.
8.35.
Model:
Use the particle model for the ball in circular motion.
Visualize:
Solve: (a)
The mass moves in a
horizontal
circle of radius
20 cm.
r
=
The acceleration
a
arrowrightnosp
and the net force vector point
to the center of the circle,
not
along the string. The only two forces are the string tension
,
T
arrowrightnosp
which does point along the
string, and the gravitational force
G
.
F
arrowrightnosp
These are shown in the freebody diagram. Newton’s second law for circular
motion is
G
cos
cos
0 N
z
F
T
F
T
mg
θ
θ
=

=

=
∑
2
sin
r
r
mv
F
T
ma
r
θ
=
=
=
∑
From the
z
equation,
(
)
(
)
2
0.500 kg
9.8 m/s
5.00 N
cos
cos11.54
mg
T
θ
=
=
=
°
(b)
We can find the tangential speed from the
r
equation:
sin
0.63 m/s
rT
v
m
θ
=
=
The angular speed is
0.63 ms
3.15 rad/s
30 rpm
0.20 m
v
r
ω
=
=
=
=
8.46.
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 Spring '10
 ZHOU
 Force, Mass, Momentum, m/s

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