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Dynamics of Rotational Motion
1021
SET UP:
The force diagram for the cylinder is given in Figure 10.62b.
EXECUTE:
z
z
I
τ
α
=
∑
,
z
Fl
TR
I
−=
where
0.12 m
l
=
and
0.25 m
R
=
aR
=
so
/
z
=
/
Fl
TR
Ia R
=
+
Figure 10.62b
22
0.25 m
(2.9 kg m )(0.80 m/s )
530 N
1200 N
0.12 m
(0.25 m)(0.12 m)
RI
a
FT
lR
l
⋅
⎛⎞
⎛
⎞
=+
=
+
=
⎜⎟
⎜
⎟
⎝⎠
⎝
⎠
EVALUATE:
The tension in the rope is greater than the weight of the crate since the crate accelerates upward. If
F
were applied to the rim of the cylinder (
l
=
0.25 m), it would have the value
567 N.
F
=
This is greater than
T
because it must accelerate the cylinder as well as the crate. And
F
is larger than this because it is applied closer to
the axis than
R
so has a smaller moment arm and must be larger to give the same torque.
10.63.
IDENTIFY:
Apply
ext
cm
m
=
∑
F
a
!
!
and
cm
z
z
I
=
∑
to the roll.
SET UP:
At the point of contact, the wall exerts a friction force
f
directed downward and a normal force
n
directed to the right. This is a situation where the net force on the roll is zero, but the net torque is
not
zero.
EXECUTE:
(a)
Balancing vertical forces,
rod
cos
,
Ff
w
F
θ
=++
and balancing horizontal forces
rod
k
sin
.
With
,
Fn
f
n
θμ
==
these equations become
rod
k
cos
,
F
w
=
++
rod
sin
.
=
Eliminating
n
and
solving for
rod
F
gives
2
rod
k
(16.0 kg) (9.80 m/s )
(40.0 N)
266 N.
cos
sin
cos 30
(0.25)sin30
wF
F
θμ θ
=
−°
−
°
(b)
With respect to the center of the roll, the rod and the normal force exert zero torque. The magnitude of the net
torque is
k
()
,
a
n
d
FfR
f
n
μ
may be found by insertion of the value found for
rod
F
into either of the above
relations;
i.e.,
kr
o
d
sin
33.2 N.
fF
μθ
Then,
2
2
2
(40.0 N
31.54 N)(18.0 10 m)
4.71 rad/s .
(0.260 kg m )
I
−
−×
=
⋅
EVALUATE:
If the applied force
F
is increased,
rod
F
increases and this causes
n
and
f
to increase. The angle
φ
changes as the amount of paper unrolls and this affects
for a given
F
.
10.64.
IDENTIFY:
Apply
z
z
I
=
∑
to the flywheel and
m
=
∑
F
a
!
!
to the block. The target variables are the tension in
the string and the acceleration of the block.
(a)
SET UP:
Apply
z
z
I
=
∑
to the rotation of the flywheel about the axis. The freebody diagram for the
flywheel is given in Figure 10.64a.
EXECUTE:
The forces
n
and
Mg
act
at the axis so have zero torque.
z
TR
=
∑
z
TR
I
=
Figure 10.64a
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View Full Document1022
Chapter 10
SET UP:
Apply
m
=
∑
F
a
!
!
to the translational motion of the block. The freebody diagram for the block is given
in Figure 10.64b.
EXECUTE:
y
y
Fm
a
=
∑
cos36.9
0
nm
g
−
°=
cos36.9
g
=
°
kk
k
cos36.9
fn
m
g
μμ
=
=°
Figure 10.64b
x
x
a
=
∑
k
sin36.9
cos36.9
mg
T
mg
ma
μ
°− −
k
(sin36.9
cos36.9 )
mg
T
ma
°−
° − =
But we also know that
block
wheel
,
aR
α
=
so
/ .
=
Using this in the
z
z
I
τ
=
∑
equation gives
/
TR
Ia R
=
and
2
(/ ).
TI
R
a
=
Use this to replace
T
in the
x
x
a
=
∑
equation:
2
k
(sin36.9
cos36.9 )
( /
)
mg
I R a
ma
° −
=
k
2
(sin36.9
cos36.9 )
/
mg
a
mIR
°
=
+
2
2
22
(5.00 kg)(9.80 m/s )(sin36.9
(0.25)cos36.9 )
1.12 m/s
5.00 kg
0.500 kg m /(0.200 m)
a
°
==
+⋅
(b)
2
2
2
0.500 kg m
(1.12 m/s ) 14.0 N
(0.200 m)
T
⋅
EVALUATE:
If the string is cut the block will slide down the incline with
2
k
sin36.9
cos36.9
3.92 m/s .
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Force

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