# Center of Gravity Moments of Inertia Parallel Axis Theorem by sanmelody

VIEWS: 22 PAGES: 3

• pg 1
```									     MEEN 363 – Exam 5 Summary Sheet (Exam 5 is closed book and notes, no ”cheat” sheets) c Paul Stiverson, TAMU
You are responsible for all of these equations, none of these will be provided on the exam.

1     Center of Gravity
1. Deﬁne a reference system for the assembly at some point.
2. Identify (label with a number or letter) each component that makes up the assembly, if the assembly consists of a single
body you can safely skip this step.
3. Deﬁne the mass and center of gravity for each element relative to the deﬁned reference system.
4. Calculate the mass center of the assembly using the following equation, note that holes cut in a plate can be treated as a
subtraction of mass.
¯       ¯
xi mi   ¯       ¯
yi mi
X=            ; Y =
mi             mi

5. If the assembly is symmetric you can simplify this process by using symmetry principles.

2     Moments of Inertia
1. Identify the point at which you want to ﬁnd the moment of inertia, which will be referred to as the point of interest.
2. Identify (label with a number or letter) each component that makes up the assembly, if the assembly consists of a single
body you can safely skip this step.
3. Deﬁne the moment of inertia, I, for each component of the assembly at its center of gravity.
(a) The moment of inertia of a slender rod at the center of gravity is:

ml2
Ig =
12
Where m and l are the mass and length of the rod, respectively.
(b) The moment of inertia for a circular disk is at its center of gravity is:

mr2
Ig =
2
Where m and r are the mass and radius of the disk, respectively.
4. Use the parallel axis theorem (§3) to ﬁnd the moment of inertia for each component at the point of interest.
5. Sum the moments of inertia for each component, it is acceptable to subtract inertias for things like holes cut in plates.
2
6. The radius of gyration at some point P is deﬁned as KP , and can be found using the following IP = mKP .
7. Be mindful of units when working problems that require ﬁnding the moment of inertia or the center of gravity. Remember
that we do not use lbm in the study of dynamics.

3     Parallel Axis Theorem
If you need to ﬁnd the moment of inertia about an axes when you only know it for another axis you can use the parallel-axis
theorem. . . so long as the the axes are parallel. For instance if you know the moment about the z axis at the center of gravity
you can ﬁnd the moment about the z axis of that body at any point on or oﬀ the body. The parallel-axis formula is shown in
equation 1.

Io − Ig = m|bog |2                                                      (1)
Where point g is the center of mass, point o is the point of interest, and bog is the vector between points o and g.
Equation 1 can be rearranged to each of the following:

Io = Ig + m|bog |2

1 of 3
or
Ig = Io − m|bog |2
Which are used, respectively, for moving from the center of mass to a point of interest, and moving from a point of interest
to the center of gravity. Remember that you cannot move from any point to another point unless you go through the center of
gravity.

4      Torsion Problems
Problems that involve rotation are similar in most respects to problems you have already seen, only instead of linear forces you
will be dealing with moments, which are forces that impart rotation to a body. All the same elements exist in these rotational
problems as did in previous systems, they just look diﬀerent.
1. As usual, deﬁne an assumed positive direction of motion (rotation in this case) of all bodies (pendulums or disks).
2. If the problem has multiple degrees of freedom deﬁne relative motions (θ1 > θ2 > 0).

3. Find the equation of motion by applying                ¨
Mo = Io θ. This is applied to each disk. Note that to ﬁnd torsional stiﬀness
you use the following:
πr4 G
Kθ =
2L
Where L is the length of the shaft, r is the radius of the shaft, and G is a material property of the shaft.
4. Rearrange the equation of motion into standard form, unknowns on the left, knowns on the right. . . make sure you divide
¨
by the term in front of the acceleration (θ).

5      Energy Method of Solving Torsion Problems
1. Identify all the elements that store kinetic energy, T . Recall that kinetic energy is never negative. To ﬁnd the velocity of
the center of mass you may need to apply kinematic strategies developed for exam 4.
1    ˙
(a) Bodies which are rotating about a ﬁxed point have a kinetic energy of: T = 2 Io θ2
˙           ˙
(b) Bodies that are translating have a kinetic energy of: T = 1 mQ2 ; where Q is the velocity of the center of mass of the
2
body.
˙        ˙
(c) Bodies that are rotating and translating (planar motion) have kinetic energy: T = 1 Ig θ2 + 1 mQ2 .
2        2

2. Identify all the elements that store potential energy, V .
1   2
(a) Springs never have negative potential energy, and are related by V =          2 kδ ;   where δ is the deﬂection of the spring
(linear or angular).
(b) Potential energy due to weight is measured relative to the datum, which you must deﬁne: V = mgh; where h is the
vertical distance from the datum.
(c) It is unwise to assume small displacement angles for pendulums when using the energy approach to ﬁnd the equation
of motion. (Assume small angles after the equation of motion is found)
3. Deﬁne elements that do not conserve energy. To obtain equation of motion using energy integral substitution, drop these
non-conservative elements. Non-conservative elements can be included in terms of work done.
(a) Forces: WF = F dr (if co-linear); where dr is the displacement of the point where the force is acting. Assign
appropriate sign for whether the work is a loss or a gain.
(b) Torques/moments: WT = T dθ; where dθ is the rotational displacement of the body that the torque is acting on.
Again, apply appropriate sign.
˙
Q2
¨
4. If energy integral substitution is to be used, apply Q =      d
, where Q is a general coordinate.
dQ   2

5. To obtain equation of motion with non-conservative elements: Wgain − Wloss = ∆ (T + V ) where Wgain is the sum of non-
conservative work of elements that input power into the system and Wloss is the sum of non-conservative work elements
that remove power from the system. Remember that this approach is limited to single degree of freedom systems.

2 of 3
6     Base Acceleration – Support Reactions
1. Deﬁne a reference system.

2. Deﬁne the acceleration of the base in vector form (magnitude and direction), aO .

3. Deﬁne the position vector from the base to the center of mass of the body, bOG . The unit vectors need to match those
deﬁned for aO .

4. Apply vector cross-product operation m bOG × aO . It is important that the right hand rule applies for the reference
system.

5. The total result is then:           ¨
MO = IO θ + m bOG × aO
Z

6. To ﬁnd support reactions: apply                   ¨    ˙
fr = mar = m r − rθ2 , and                              ¨    ˙˙
fθ = maθ = m rθ + 2rθ showing the proper
reference axis or unit vectors in the free body diagram. You can apply           ¨
fX = mXG , and                ¨
fY = mYG , but the algebra is
˙
a bit more involved. Then apply energy integral substitution with deﬁned initial conditions to get expressions for θ2 . The
¨ term comes from
θ                       MO .
7. For compound pendulum problems, the acceleration of the base is zero. Apply the same strategy as in the previous step to
˙
obtain support reactions. If the system is conservative, θ2 can be found by applying conservation of energy. Again, do not
apply small angle assumption for weight potential. Linearization (small angle assumption: sin θ ≈ θ and cos θ ≈ 1) should
not be applied until after the derivative of the energy function is obtained.
˙   ¨
8. Be able to determine equilibrium position such that Q = Q = 0 is satisﬁed. Be able to determine the equation of motion