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Dynamics 2e 1517
Problem 7.53
One way to measure the mass moment of inertia of any body
relative to its mass center
G
is to horizontally suspend it by a
string at one end
A
, and to support it by a scale
C
at the other end
B
. The location of points
A
and
B
is not important as long as we
know the distance
d
between them and the mass
m
of the body.
When the body is statically supported as described, we take note
of the reading on the scale. We then cut the string at
A
and note
the new reading on the scale immediately after the string is cut.
Knowing the mass
m
of the body, the distance
d
, and the reading
on the scale immediately before
Nb
and after
Na
the string is cut,
determine IGfor the body.
1.27831
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of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Dynamics 2e 1519
Problem 7.54
The composite rigid body lies in the vertical plane and consists of the
uniform block
D
and the L-shaped bar
ABC
. The block is rigidly
attached to the L-shaped bar, which is uniform. Each segment of the
L-shaped bar has length
LD0:75
m and mass
mAB DmBC D2kg
,
the mass of the block is
mDD5kg
, and the width and height of the
block are hD0:3 m.
If the system is released from rest in the position shown, determine
the initial angular acceleration of the composite body and the reaction
at the pin Aon the composite body.
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Force Laws. All forces have been accounted for on the FBD.
Kinematic Equations. Since the system is released from rest, the angular velocity of the composite body
is zero. Since the body is in fixed axis rotation about
A
, we can relate the acceleration components of the
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Dynamics 2e 1521
Problem 7.55
The composite rigid body lies in the vertical plane and consists of the
uniform block
D
and the L-shaped bar
ABC
. The block is rigidly
attached to the L-shaped bar, which is uniform. Each segment of the
L-shaped bar has length
LD0:75
m and mass
mAB DmBC D2kg
,
the mass of the block is
mDD5kg
, and the width and height of the
block are hD0:3 m.
If the system is rotating clockwise with the angular speed
!bD
10 rad=s
in the position shown, determine the reaction at the pin
A
on
the composite body.
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permission of McGraw-Hill, is prohibited.
Force Laws. All forces have been accounted for on the FBD.
Kinematic Equations.
Since the body is in fixed axis rotation about
A
, we can relate the acceleration
components of the mass center of each part to the angular velocity and angular acceleration of the composite
body using
L
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Dynamics 2e 1523
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permission of McGraw-Hill, is prohibited.
Problem 7.56
The composite rigid body lies in the vertical plane and consists of the
uniform block
D
and the L-shaped bar
ABC
. The block is rigidly
attached to the L-shaped bar, which is uniform. Each segment of the
L-shaped bar has length
LD0:75
m and mass
mAB DmBC D2kg
,
the mass of the block is
mDD5kg
, and the width and height of the
block are hD0:3 m.
If the system is rotating clockwise with the angular speed
!bD
10 rad=s
in the position shown, determine the internal reaction at the
rigid joint Bon the bar BC of the composite body.
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1525
Force Laws. All forces have been accounted for on the FBD.
Kinematic Equations.
Since the body is in fixed axis rotation about
A
, we can relate the acceleration
components of the mass center of each part to the angular velocity and angular acceleration of the composite
body using
L
2˛bL!2
2!2
or, breaking out the needed components
2˛b; aF y DL˛bCL
2!2
2!2
Computation.
Substituting the expressions for the mass moments of inertia and the kinematic equations
into the Newton-Euler equations, we obtain
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