118 6. TOPIC AK-6
platform before it stops. Also, dene impact impulses on the cylinder from both the step and the inclined plane.
Using Carnot’s theorem, verify the obtained expressions for the angular velocities of the cylinder after the impacts
with both the step and the inclined plane.
B
s
D
C
A
F
h
N
K
E
Figure 6.31.
6.3 SOLUTION
Due to the sudden stop of the platform, a translational motion of the cylinder instantly converts to the rotational
motion about the side D of the step BD, e.g. the cylinder experiences the hit.
We will apply a change of the angular momentum (kinetic moment) of the mechanical system during the impact. As
an axis of moments we will consider the xed horizontal axis provided along the side D of the step BD (Figure 6.32):
L
II
D
– L
I
D
= M
D
(S
⃗
i
E
).
In Figure 6.32 the positions I and II corresponding to the beginning and the end of the impact on the side D of the
step BD coincide.
e sum of the moments of the external impact impulses applied to the cylinder about axis D:
M
D
(S
⃗
i
E
) = 0.
is means that the impact impulse S
⃗
D
crosses the axis D. erefore:
L
II
D
= L
I
D
.
e kinetic moment of the cylinder about the axis D at the beginning of the impact:
L
I
D
= mv
C
I
(r – h),
where v
C
I
= v is the velocity of the center of the gravity of the cylinder at the beginning of the impact, which equals
to the velocity of the platform before a sudden stop.
119
e kinetic moment of the cylinder about the axis D at the end of the impact:
L
II
D
=
D
ω
II
,
where
D
is the moment of the inertia of the cylinder about axis D and ω
II
is the angular velocity of the cylinder at
the end of the impact. en:
L
II
D
= (
C
+ mr
2
)ω
II
=
mr
2
+ mr
2
ω
II
=
3
mr
2
ω
II
.
2 2
As: L
II
D
= L
I
D
, then:
3
mr
2
ω
II
= mv(r – h).
2
From here:
ω
II
=
2(r – h)v
. (6.1)
3r
2
Let’s verify the obtained expression (6.1) for the angular velocity of the cylinder using the Carnot’s theorem for the
case when an ideal elastic support is applied to the mechanical system:
T
I
– T
II
= T
*
,
where T
I
is the kinetic energy of the particles system at the beginning of the impact; T
II
is the kinetic energy of
the particles system at the end of the impact; and T
*
is the kinetic energy corresponding to the lost velocities of the
system’s particles.
e kinetic energy of the cylinder before it hits the step:
T
I
=
1
mv
2
.
2
e kinetic energy of the cylinder after it hits the step:
T
II
=
1
D
ω
II
.
2
B
D
C
A
x
y
s
Dx
s
Dy
v
CI
v
CII
ω
II
90°
I
II
Figure 6.32.
Next, we determine the kinetic energy of the cylinder which corresponds to the lost velocities of its particles Δv
⃗
i
I
=
v
⃗
i
I
– v
⃗
i
II
(Figure 6.33):
2
6.3 SOLUTION
120 6. TOPIC AK-6
T
*
=
1
m
i
(v
⃗
i
I
– v
⃗
i
II
)
2
2
=
1
m
i
(v
⃗
i
I
– v
⃗
i
II
)
(v
⃗
i
I
– v
⃗
i
II
)
2
=
1
m
i
(v
⃗
i
I
v
⃗
i
I
– 2v
⃗
i
I
v
⃗
i
II
+ v
⃗
i
II
v
⃗
i
II
)
2
=
1
m
i
(v
⃗
i
I
– 2v
⃗
i
I
v
⃗
i
II
cos γ
i
+ v
⃗
i
II
).
2
Here v
i
I
= v is the velocity of the particle M
i
before the impact; and
v
i
II
= ω
I
DM
i
is the velocity of the particle M
i
after the impact.
en:
T
*
=
1
m
i
v
2
2
– m
i
vω
II
DM
i
cos γ
i
+
1
m
i
ω
II
(DM
i
)
2
2
=
1
v
2
m
i
2
– vω
II
m
i
DM
i
cos γ
i
+
1
ω
II
m
i
(DM
i
)
2
=
1
mv
2
– vω
II
m
i
y
i
+
1
D
ω
II
.
2
2
2
However,
m
i
y
i
= my
C
= m(r – h).
erefore,
T
*
=
1
mv
2
– mvω
II
(r – h) +
1
D
ω
II
.
2
2
Hence,
1
mv
2
–
1
D
ω
II
=
1
mv
2
– mvω
II
(r – h) +
1
D
ω
II
.
2
2
2
2
or
D
ω
II
= mvω
II
(r – h),
e.g.,
3
mr
2
ω
II
= mvω
II
(r – h),
2
and from here:
ω
II
2(r – h)v
.
3r
2
2
2
2
2
2
22
2
2
121
B
D
C
A
x
y
c
Δv
i
v
iI
v
iII
ω
II
90°
I
II
y
γ
i
γ
i
Figure 6.33.
Next, we apply the kinetic energy work principle for the mechanical system, which corresponds to the lift of the
cylinder on the step BD from the position II to the position III (Figure 6.34):
T
III
– T
II
= A
i
E
.
Because the cylinder rotates about axis D, then:
1
D
ω
III
–
1
D
ω
II
= – Gh
2
2
or
3
mr
2
ω
III
–
3
mr
2
ω
II
= –mgh,
2
2
and from here:
ω
III
=
ω
II
–
4gh
. (6.2)
3r
2
B
D
C
A
x
v
CIII
v
CII
ω
II
90°
h
II
C
III
90°
ω
III
Figure 6.34.
Next, we apply the kinetic energy work principle for the mechanical system, which corresponds to the rolling of the
cylinder on the segment DE from the position III to the position IV (Figure 6.35):
T
IV
– T
III
= A
i
E
.
As in this displacement A
i
E
= 0, then
T
IV
= T
III
,
2
2 2
2
2
6.3 SOLUTION
122 6. TOPIC AK-6
which means:
v
C
IV
= v
C
III
and ω
IV
= ω
III
. (6.3)
When the cylinder touches the inclined plane, the instantaneous axis of rotation of the cylinder (the instantaneous
center of velocities) suddenly moves from point E to the point F, e.g., the cylinder experiences the hit.
B
D
C
E
v
CIII
v
CIV
ω
IV
III
C
ω
III
G
IV
Figure 6.35.
Next, we apply the kinetic energy work principle for the mechanical system at the impact. e axis of the moments
will be a xed horizontal axis, which corresponds to the generatrix F of the cylinder (IV and V positions of the cyl-
inder corresponding to prior and after the impact, coincide) (Figure 6.36):
L
V
F
– L
IV
F
= M
F
(S
⃗
i
E
).
As the impact impulse S
⃗
F
applied to the cylinder crosses the axis F: M
F
(S
⃗
i
E
) = 0.
erefore,
L
V
F
= L
IV
F
.
To calculate the kinetic moments of the cylinder L
IV
F
and L
V
F
we will apply the theorem of the kinetic moment of
the system for the general case of its motion.
e kinetic moment of the cylinder about axis F prior to the impact:
L
IV
F
= mv
C
IV
r cos α +
C
ω
IV
,
where v
C
IV
= ω
IV
EC, as the point E will be the instantaneous center of velocities.
en,
L
IV
F
= mω
IV
r
2
cos α +
mr
2
ω
IV
= mω
IV
r
2
cos α +
1
.
2 2
e kinetic moment of the cylinder about axis F after the impact:
L
V
F
= mv
C
V
r +
C
ω
V
,
where v
C
V
= ω
V
∙ CF, because the point F is the instantaneous center of velocities.
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