hernandez (ejh742) – oldhomework 23 – Turner – (58120)
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001
10.0 points
A disk has mass 8 kg and outer radius 20 cm
with a radial mass distribution (which may
not be uniform) so that its moment of iner
tia is
2
3
m R
2
.
It disk is given a hard kick
(impulse) along a horizontal surface at time
t
0
.
The kicking force acts along a horizontal
line through the disk’s center, so the disk ac
quires a linear velocity 1
.
2 m
/
s but no initial
angular velocity.
The coefficient of friction
between the disk and the surface is 0
.
05
.
The kinetic friction force between the sur
face and the disk slows down its linear motion
while at the same time making the disk spin
on its axis at an accelerating rate. Eventually,
the disk’s rotation catches up with its linear
motion, and the disk begins to roll at time
t
rolling
without slipping on the surface.
20 cm
8 kg
1
.
2 m
/
s
μ
= 0
.
05
Once the disk rolls without slipping, what
is its linear speed? The acceleration of gravity
is 9
.
8 m
/
s
2
.
Correct answer: 0
.
72 m
/
s.
Explanation:
Let :
R
= 20 cm = 0
.
2 m
,
v
0
= 1
.
2 m
/
s
,
m
= 8 kg
,
and
μ
= 0
.
05
.
From the perspective of the surface, let the
speed of the center of the disk be
v
surface
.
summationdisplay
F
surface
=
f
m a
=
μ m g
a
=
μ g ,
so
v
surface
=
v
0
−
a t
=
v
0
−
μ g t .
After pure rolling begins at
t
rolling
there
is no longer any frictional force and conse
quently no acceleration. From the perspective
of the center of the disk, let the tangential ve
locity of the rim of the disk be
v
disk
and the
angular velocity be
ω ,
so
summationdisplay
τ
=
I α
α
=
τ
I
=
f R
I
=
μ m g R
2
3
m R
2
=
3
2
μ g
R
,
the time dependence of
ω
is
ω
=
α t
=
3
2
μ g
R
t ,
and
v
disk
=
R ω
=
3
2
μ g t .
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 Spring '08
 Turner
 Physics, Angular Momentum, Friction, Mass, Work, Moment Of Inertia, Velocity, Physical quantities

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