Dynamics of Rotational Motion
1021
S
ET
U
P
:
The force diagram for the cylinder is given in Figure 10.62b.
E
XECUTE
:
z
z
I
τ
α
=
∑
,
z
Fl
TR
I
α
−
=
where
0.12 m
l
=
and
0.25 m
R
=
a
R
α
=
so
/
z
a R
α
=
/
Fl
TR
Ia R
=
+
Figure 10.62b
2
2
0.25 m
(2.9 kg
m
)(0.80 m/s
)
530 N
1200 N
0.12 m
(0.25 m)(0.12 m)
R
Ia
F
T
l
Rl
⋅
⎛
⎞
⎛
⎞
=
+
=
+
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
E
VALUATE
:
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.
I
DENTIFY
:
Apply
ext
cm
m
=
∑
F
a
!
!
and
cm
z
z
I
τ
α
=
∑
to the roll.
S
ET
U
P
:
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.
E
XECUTE
:
(a)
Balancing vertical forces,
rod
cos
,
F
f
w
F
θ
=
+
+
and balancing horizontal forces
rod
k
sin
.
With
,
F
n
f
n
θ
μ
=
=
these equations become
rod
k
cos
,
F
n
F
w
θ
μ
=
+
+
rod
sin
.
F
n
θ
=
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
w
F
F
θ
μ
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
(
)
, and
F
f R
f
n
μ
−
=
may be found by insertion of the value found for
rod
F
into either of the above
relations;
i.e.,
k
rod
sin
33.2 N.
f
F
μ
θ
=
=
Then,
2
2
2
(40.0 N
31.54 N)(18.0
10
m)
4.71 rad/s
.
(0.260 kg
m )
I
τ
α
−
−
×
=
=
=
⋅
E
VALUATE
:
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
α
for a given
F
.
10.64.
I
DENTIFY
:
Apply
z
z
I
τ
α
=
∑
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)
S
ET
U
P
:
Apply
z
z
I
τ
α
=
∑
to the rotation of the flywheel about the axis. The freebody diagram for the
flywheel is given in Figure 10.64a.
E
XECUTE
:
The forces
n
and
Mg
act
at the axis so have zero torque.
z
TR
τ
=
∑
z
TR
I
α
=
Figure 10.64a
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1022
Chapter 10
S
ET
U
P
:
Apply
m
=
∑
F
a
!
!
to the translational motion of the block. The freebody diagram for the block is given
in Figure 10.64b.
E
XECUTE
:
y
y
F
ma
=
∑
cos36.9
0
n
mg
−
° =
cos36.9
n
mg
=
°
k
k
k
cos36.9
f
n
mg
μ
μ
=
=
°
Figure 10.64b
x
x
F
ma
=
∑
k
sin36.9
cos36.9
mg
T
mg
ma
μ
° −
−
° =
k
(sin36.9
cos36.9 )
mg
T
ma
μ
° −
° −
=
But we also know that
block
wheel
,
a
R
α
=
so
/
.
a R
α
=
Using this in the
z
z
I
τ
α
=
∑
equation gives
/
TR
Ia R
=
and
2
( /
) .
T
I
R
a
=
Use this to replace
T
in the
x
x
F
ma
=
∑
equation:
2
k
(sin36.9
cos36.9 )
( /
)
mg
I
R
a
ma
μ
° −
° −
=
k
2
(sin36.9
cos36.9 )
/
mg
a
m
I
R
μ
° −
°
=
+
2
2
2
2
(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
⋅
=
=
E
VALUATE
:
If the string is cut the block will slide down the incline with
2
k
sin36.9
cos36.9
3.92 m/s
.
a
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 Spring '06
 Buchler
 Physics, Force, Friction, rα

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