mechanics of rotation and moment of inertia

MACHINE DYNAMICS 231

Lecture 6

Mechanics of Rotation & Moment of Inertia Calculating Moment of Inertia from First Principles

Centre of Mass, Radius of Gyration Parallel Axis Theorem, MoI of Thin Plates, Perpendicular Axis Theorem

Example Inertia Calculations: Discs, Cylinders, Cone

By

Brad Saracik

Lecturer &

Dr Ian Howard Associate Professor

Department of Mechanical Engineering Curtin University of Technology

MD231 Lecture 6: Kinetics of Rotation & Moment of Inertia 2

Kinetics of Rotation Overview In this topic, we study what causes rotation to occur. Rotational acceleration occurs when an unbalanced torque acts on a body. Newtons 2nd Law can be extended to relate applied forces and torques to the properties of a body and determine the resulting acceleration and rotational acceleration. Using the kinematics studied in other topics, you can then determine the complete resulting motion. Outcomes: After understanding this reading, completing the exercises and practicing problems of these types, the reader should be able to:

1. Define torque in terms of applied forces enabling the calculation of resulting torques.

2. Identify forces causing translation and torques causing rotation, separately and in combination, on a given body or FBD

3. Explain Moment of Inertia (rotational inertia) as a measure of the distribution of mass, conceptually and in equation form and that it is a constant property of rigid bodies.

4. Resolve torques in FBDs and accurately apply Newtons 2nd Law to determine the resulting rotational acceleration. Given a rotational acceleration, determine the required torque.

5. Calculate Moment of Inertia of uniform shapes using mathematical techniques. Apply these techniques to estimate the moment of inertia of machine components.

Reading & Problem Sets: Meriam: 6.1-6.5 pp408-459 Hibbeler:17.1-17.5 pp377-435


Rotational Concepts As seen in our study of kinematics, real objects dont just move through space, they can also rotate (change orientation). We have first studied Newtons Laws for linear motion (translation) and need to extend this for rotation to complete the full picture. The majority of engineering machines involve rotating parts. Although cars, trains and paddle steamers were designed to get us from A to B (i.e. translation), they achieve it by turning wheels (ie rotary motion or rotation). Electric motors, combustion engines and turbines all provide useful power in the form of rotation. Thus mechanical engineers need to have equivalent deep understanding of rotational concepts as for translation. While this may prove challenging at first, there are direct linear analogies for every rotational concept you come across. This is hardly surprising when you realise that rotation is movement about a given axis. The kinematics of rotation have been dealt with in detail. Recall the following table

Linear Dim Rotational Dim (angular) displacement s [L] [1]

(angular) velocity sdtdsv ,, [L/T] ,, dt

d [1/T]

(angular) acceleration sdt

sdvdtdva ,,,, 2

2

[L/T2]

,,,, 22

dtd

dtd [1/T

2]

const. acceleration 2atuts += 20 tt +=

const. acceleration asuv 222 += 2202 +=

circular motion rst = in radians

circular motion rvt = in rad/s ; rpm

circular motion rat = in rad/s2

Table 1: Table of Analogous 1D Formulae


It is again necessary to differentiate between rotation referring to: a) movement about a given axis b) change of orientation (i.e. a change in angle)

Newtons 2nd law, as previously studied, completely solves the first case. For example,

A particle rotating about an axis through O, requires a normal acceleration and therefore a normal Force (Fn).

Any unbalanced tangential force (Ft) will result in a tangential acceleration: the particles speed will change.

The instantaneous acceleration of a particle is completely determined by the linear Newtons 2nd Law, F=ma

For a rigid body rotating about O, compare the orientations evident in Figures 2 & 3.

As studied so far, Newtons 2nd Law does not explain the change in orientation evident in Figure 3. We need to extend Newtons laws to apply to this kind of rotation.

at

r

O Fn an

Ft

m

Fn

Fn

vt

at

r

O Fn an

Ft

m

Figure 3: Rigid Body Translation and Rotation

1

2

3

Fn

Fn

vt

Figure 1: Particle Travelling in a Circle

at

r

O Fn an

Ft

m

Figure 2: Rigid Body Circular Translation

1

2

3

Fn

Fn

vt


The Cause of Rotation: Newtons 2nd Law For a change in linear motion (acceleration) to occur, an unbalanced force must act on a body.

If the unbalanced force acts through the centre of mass ONLY translation occurs. (No net Torque)

If an unbalanced force exists, NOT acting through the centre of mass, translation AND rotation occurs. (Net Force & Torque)

If forces balance, however an unbalanced moment (or torque) exists, ONLY rotation occurs. (Net Torque)

Thus an unbalanced force causes linear acceleration according to Newtons 2nd Law. An unbalanced moment (or torque) causes angular acceleration. As we will see, the exact relationship can be derived from Newtons 2nd Law.

Torques and Moments Forces are interactions between bodies which (unopposed) cause changes in velocity. Torque is a vector concept used to describe the tendency of a force to cause rotation about a given axis. Torques do not exist in isolation, they only occur as a result of the location of forces. Note that in statics, torques cause bending, while in dynamics, torques cause rotation. Torques are also known as Moments; the two are synonymous.

The fundamental definition of Torque is a vector or cross product. T= F x r where (Eq 1)

T is the Torque vector due to a force about a given axis (Nm) F is the applied force vector (N) r is a vector from the given axis to the line of action of the applied force, called the moment arm (m)

Thus, the torque magnitude |T|=|F||r|.sin, where is the angle between F and r. The direction of T is given by the right hand rule and is the axis about which the torque causes rotation.


In the special case where =90, i.e. the applied force is perpendicular to the moment arm, |T|=|F||r|, i.e. T=Fr If r.sin is 0, the line of action of the force acts through the given point and no torque results.

Newtons 2nd Law for Rotation F=ma defines exactly how an unbalanced force causes any body to instantaneously accelerate in a line. Torque is the rotational affect of force on a body and causes rotational acceleration of the body (). Thus we seek an equation of the form T=C, where C is a constant property of the body (related to its mass, and the ease with which it can be rotated). We call this constant C, the Mass Moment of Inertia, and use the symbol I. Newtons 2nd Law for rotation is thus T=I (Eq 2)

T is the combined (resultant) torque vector acting on a body about a given axis (Nm = kgm2/s2)

is the resulting rotational acceleration vector of the body about that axis (1/s2)

I is a scalar constant, called the mass moment of inertia relating to that body about the given axis (N.m/s2 = kg.m2)

We will see later how mass moment of inertia is mathematically defined and how we can calculate it. Once the mass moment of inertia is known, solving problems with Newtons 2nd Law for rotation is similar to solving problems with Newtons 2nd Law for translation.


Worked Example using Moment of Inertia Problem: A cable is wrapped several times around a uniform solid cylinder having a moment of inertia about its axis of rotation of 0.090kg.m2 and an outer diameter of 12cm. The cable is pulled with a tension force of 9.0N. Assuming the cable does not stretch or slip, find its acceleration. What is the speed of the cable after travelling 2m from rest?

Solution: The drum does not translate due to reaction forces from the supports. i.e. Fdrum=0. Neglecting friction, the only torque present is due to the applied tension. i.e. Tdrum = Idrumdrum = T Since the Tension is perpendicular to the drum, T=FT.r = 9*(0.12 / 2) = 0.54Nm

T = Idrumdrum = T / Idrum = 0.54 / 0.090 = 6.0rad/s2

a=r = 0.06 * 6.0 = 0.36m/s2

Assuming the cable unwinding does not change the effective radius, the cable acceleration will be constant, thus v2=0+2as ; v=2*0.36*2 = 1.2m/s Alternative: 2m travelled means the drum has turned by =2m/ d * 2 radians. Then use = 6.0rad/s2. 2=2, v=r. Student Exercise: Draw a free body diagram for the drum. Student Exercise: After studying the remaining notes, return here with your new knowledge and calculate the mass of the cylinder in this example. Tip: How do you calculate the moment of inertia of a solid cylinder? (ans: 50kg)

d=0.12m FT=9.0N

Figure 4: Rotating drum


Moment of Inertia : Distribution of Mass Moment of Inertia is more than a mathematical concept given by a formula. Before deriving how we calculate moment of inertia, it is worthwhile using our intuition and existing knowledge to predict what the concept must look like:

Known Prediction / Intuition an objects mass is a measure

of its linear inertia mass moment of inertia will

be a measure of rotational inertia

an object of larger mass is more difficult to accelerate in a line than an object of smaller mass

an object with a larger mass moment of inertia will be more difficult to rotate than an object of smaller mass moment of inertia

A force causes translation regardless where it acts on a body. A torque depends on the distance from the axis where the force acts. A body accelerates linearly regardless how the mass is distributed

A bodys rotational acceleration will depend on the distribution (location) of its mass and the location of the forces magnitude of torques

In summary, the moment of inertia must be a measure of the distribution of mass within a body and inversely affect the resulting rotational acceleration when subjected to a given torque. For bodies of constant density, the mass is distributed according to the shape (area in 2D, volume in 3D) of the body.


Moment of Inertia Derivation A quite straight forward way of mathematically defining the mass moment of inertia uses the concept that a rigid body is a summation of smaller masses. Conceptually we can reduce any rigid body to a collection of point / particle masses. For a particle rotating about an axis, we know: F=m.a (3) a = r (4) T = Fxr (5)

Substituting into T=I => Ipoint mass= T / = F.r . (r/a) = m.r2 (Eq 6)

Figure 5: Point Mass Rotational Inertia

Thus over the entire rigid body, we only need to lump together the mass and distance (radius) for every point mass

Irigid body = Ipoint masses = =

N

i 1mi.ri2 (Eq 7)

Mathematically this is most accurately obtained as N approaches infinity and the above expression yields integration over the rigid bodys volume

Irigid body = =VolVolume

dVrdmr ... 22 since dm=.dV (Eq 8)

(If you are not sure about the last bit, recall av = mass/volume)

r

O m

a

r

O F=m.a

m

T=Fxr T=I.


Note that the value for the mass moment of inertia of a body depends on the axis about which it is to rotate. Technically every body has an infinite variety of mass moments of inertia, depending on where we define the axis of rotation (origin of r) to be. ie wrong: The mass moment of inertia for this object is x kg.m2 X right: The mass moment of inertia for this object about axis AA is x

Common Mass Moments of Inertia It is relatively easy to find tables such as those in Meriam Appendix D and the back inside page of Hibbeler showing the properties of common shapes, including the mass moment of inertia.

Figure 6: Example MMoI (Source http://hyperphysics.phy-astr.gsu.edu)

Some objects resemble common shapes and tables such as these can be used to determine the moment of inertia. However this is not always the case, therefore mechanical/mechatronic engineers (students) also need to be able to calculate these via integration.

Common Error to Avoid Concentrating the mass at the centre of an object and using MR2 will almost always yield the incorrect result (as demonstrated in the above table). Dont do it.


Worked Example MMoI Calculation Calculate the Mass Moment of Inertia of a uniform slender rod about an axis passing through its centre of mass as shown.

Solution Define the total mass of the rod M and the length of the rod L. Consider the small element dx, with mass dm shown in the Figure. dm = (M / L) * dx

About the axis shown, the moment of inertia can be calculated as

2/

2/

3222

3...

L

LlengthlengthVolumeaxis

xLMdxx

LMdmxdmrI

====

( )12

)(8*3*

233 MLLL

LMI axis == (about the axis shown only)

Student Exercise: show that an axis through the rods end yields a Moment of Inertia

3

2MLIend = . Explain why the moment of inertia must be larger about an axis at the end than in the centre of the rod

Moment of Inertia via Kinetic Energy You may see elsewhere that Mass Moments of Inertia is commonly derived via the concept of kinetic energy. Recall for linear motion that the Kinetic Energy of a particle is given by

K.E linear = mv2. The rotational analogy for velocity v is angular velocity . Since v = .r, K.E rotation = (m.r2) 2 = Iparticle 2. So Iparticle =mr2. For the general case, K.E rotation = I2

where Ibody = Iparticles = =

N

i 1mi.ri2 ==

VolVolume

dVrdmr ... 22

Axis

L/2

x

dm

L/2

dx

Figure 7: MMoI for a slender rod


Composite Bodies If an object is constructed of a combination of parts, the total mass is the sum of the mass of the parts i.e. M=m1+m2+m3. Similarly the inertia of parts can be summed to determine the total Moment of Inertia of a composite body (I = I1 + I2 + I3). The mass or inertia which is missing from a hole can similarly be subtracted from that of a solid part. The distance to the axis of rotation must be correct for each component added or subtracted as shown below.

Figure 8: Adding or Subtracting Composite Bodies

Radius of Gyration There is one distance (radius) from any axis for which the whole mass of a rigid body could be concentrated and result in the same moment of inertia about that axis. We call this the radius of gyration (k) and it is given by 2mkI = or mIk = (Eq 9)

Note that in linear systems, we often treat a rigid body as if all its mass was concentrated at its centre of mass. This is the rotational equivalent, since equation Eq 9 resembles the moment of inertia of a particle given by I = mr2, when k replaces r as the radius of gyration. Tabled formulae or values for Radius of Gyration are generally given about an axis of rotation through the centre of mass. Worked Example: Find the radius of gyration about the axis through the centre of mass of the slender rod shown in Figure 7.

Solution: From the previous solution, 12

2MLI z = , therefore

=== 122L

MIk zz 12

L


Worked Example: Falling Load Accelerating a Drum A 60kg, 800mm wide drum has a radius of gyration about its axis of rotation of 0.25m. A 20kg load is suspended from a massless rope wrapped around the drum. Assuming no slip, determine the drums angular acceleration.

Solution The drums acceleration depends on the tension in the rope and will be accelerated by the mass m1, i.e. the drum will accelerate clockwise. Free body diagrams are required for the two bodies

FBD Drum: TO=IO. T.r = IO. (1) FBD Load: m1.g - T = m1a (2) IO is given by the mass and radius of gyration, IO = mkO2 = 60*0.252 = 3.75kg.m2 No stretching/slip a = .r. (3) T = m1(g - r). Substitute in (1) m1(g-r).r = IO.. so (IO+m1.r2)=m1.g.r. On Earth: = (20*9.81*0.4)/(3.75+20*0.42)=78.48/6.95 =11.3 rad/s2

r=0.4m

m1=20kg

m=60kg k=0.25m

Figure 9: Load accelerating drum T

m1.g

a

T R

O

mg


Centre of Mass How do you expect the boomerang to move under application of the different forces in the following free body diagrams?

Figure 10: Boomerang FBDs

Recall that: an object will rotate as well as move if the resultant force does not act through its centre of mass. On Earth (in a uniform gravitational field), the centre of mass coincides with the centre of gravity, so we can find it by finding the point at which the object balances. Mathematically, the distance to

the centre of mass is given by M

dmy

mym

yM

dmx

mxm

xi

iicom

i

iicom

====

.;

.

(10)

Moments of Inertia are often given for axes running through the centre of mass of an object. We will show why these are most useful. Note the use of the following notation. In general the axis used for a given moment of inertia will be designated by the symbol I with two subscript letters denoting the axis (Ixx, Iyy, Izz). One subscript is used to demonstrate when the Moments of Inertia is about an axis going through the centre of mass (Ix, Iy, Iz) Thus in the slender rod example Iaxis could be designated Iz since it went through the centre of mass, while Iend could be designated Izz.

F

F

F


Parallel Axis Theorem The parallel axis theorem gives us a relationship between the moment of inertia about an axis at the centre of mass and any other axis parallel to it. Parallel Axis Theorem: IAA=IA + Md2 (Eq 11)

Proof:

= dmrI AAA .2 & = dmrI BBB .2

Geometry: rB2=a2 + y2 and rA2=(a+d)2 + y2 therefore rA2 = rB2 + 2.a.d + d2 thus

++== dmddmaddmrdmrI BAAA 222 .2.

++= dmadmdII BBAA .22 (Eq 12)

Since the final term in equation Eq 12 is 0 when BB is an axis through the centre of mass, we have the Parallel Axis Theorem. Equation 10 shows why the final term in equation Eq 12 is 0. It also shows that the theorem is valid only between two axes when one of them is through the centre of mass.

Relationship with the Second Moment of Area You will almost certainly come across the second moment of area during your studies, since it is used to determine how cross-sections respond to bending. Second Moment of Area =

Area

dAr .2 .

Be careful not to confuse this or its radius of gyration, with the mass moment of inertia and its radius of gyration.

x

y

z

A

A B

B

d a

rB rA

dm

Figure 11: Mass element about parallel


Moment of Inertia of Thin Plates Consider a thin homogeneous plate of thickness t and density About the axis AA we know

= dmrI AA .2

For the element dA, we know dm = .t.dA thus = dArtI AA .2

We just noted that this integral is the second moment of area, therefore FOR THIN PLATES ONLY: IAA(mass) = .t.IAA(area) (Eq 13)

Perpendicular Axis Theorem The perpendicular axis theorem states that for a planar object (thin plate), the moment of inertia about an axis perpendicular to the plate is equal to the sum of the moments of inertia about two perpendicular axes going through the same point in the plane of the object. i.e. FOR PLANAR OBJECTS ONLY: Iz = Ix + Iy (Eq 14) Proof: for a point mass, dm at x,y,z Ix = dm.y2 Iy = dm.x2 Iz = dm.r2 Since r2 = y2+ x2,

Iz = dm(y2+x2)= Ix + Iy This theorem is useful for 3d objects which can be broken down into planar sections and then integrated (summed) along its length.

A

A

dA

r

Figure 12: Thin Plate


Moment of Inertia using Tables of Common Shapes Use of tabled moment of inertias, sometimes in combination with the parallel and/or perpendicular axis theorems, often enable estimation of moment of inertia for rotating machinery components. This is often useful as a rapid check of values obtained by other means (e.g. catalogs, computer aided engineering models etc).

Worked Example: MMoI of Composite Body Two bars welded together as shown in Figure 13 will be fitted as part of a new lightweight bucket design. Each 1m bar is 5kg.

i. Find the Centre of Mass ii. Find the moment of inertia about an axis

through the pivot. iii. Find the Radius of Gyration from the pivot Solution: The body consists of two slender rods, which may be found in tables such as Figure 6. Label the section connected to the pivot rod 1 and the other rod 2. Use Equation 10 to locate the centre of mass. Due to symmetry, the centre of mass is above the pivot at a height halfway between the centre of the equal mass sections. i.e. yCOM = (0.5 + 1)/2=0.75m

Rod 1: I1 = MoI of slender rod about one end I1= ML2/3 = 1.67kgm2 Rod 2: I2 = MoI of slender rod about centre + parallel axis theorem, i.e. I2= ML2/12 + Md2 = 0.42 + 5 = 5.42kgm2 Thus, Ipivot = I1 + I2 = 7.1 kgm2 The radius of gyration about the pivot: k =I/m = (7.1/10) = 0.84m

Student Exercise: Show the CoM and radius of gyration about the pivot on Figure 13. Find the radius of gyration about the CoM and sketch a circle showing this radius. (0.38m, hint: parallel axis)

pivot

1m

1m

Figure 13


The Mass Moment of Inertia about a perpendicular axis through the centre of mass for a rectangular plate is:

IZ Rect Plate = ( )

12

22 baM + (Eq 15)

Worked Example: MMoI via Subtraction Use the previous equation to determine the MoI and radius of gyration about an axis through the pivot at O for the 5mm thick uniform plate shown, with density =2700kg/m3. Solution First recognise that the Moment of Inertia of this plate can be found by subtracting a circular section (I2) from a square section (I1): I = I1 I2 Both sections have the same centre of mass (C). Thus, I1C = m1(a2+b2)/12 ; m1 = tab = 26.5kg ; I1C = 8.64kgm2 I2C = m2*(/2)2/2 ; m2 = t2/4 = 0.954kg ; I2C = 0.01kgm2 IC = I1C I2C = 8.53 kgm2, where C is for an axis through the CoM (note that I2c is insignificant) IOO = IC + md2 ; d = 1.4/2 = 0.99m ; IOO = 33.6kgm2 kOO=(33.6/25.5)=1.15m

Student Exercise: Show the CoM and radius of gyration about the pivot on Figure 14. Find the radius of gyration about the CoM and sketch a circle showing this radius. (0.58m) Note (from the exercises) that mass further away from the axis of rotation contributes more to the Moment of Inertia. With practice, this realisation enables you to intuitively estimate the Radius of Gyration based on the distribution of mass for any object. This enables rapid checking that your answer seems reasonable.

=300mm

1.4m 1.4m O

Figure 14


Moment of Inertia via Numerical Integration Using the previous results, a deep understanding of integral calculus and some practice, you should now be able to calculate the moments of inertia for 2D and 3D objects. This is useful where the moment of Inertia cannot be found using tables and may be programmed in computer aided engineering software.

Worked Example: Mass Moment of Inertia of a Thin Disc Calculate the moment of inertia about the x, y and z axis through the centre of a thin homogeneous disc.

Solution For a thin disc, Iz = Ix + Iy Symmetry Ix = Iy = Iz / 2 So we only need to calculate the moment of inertia about any one axis to know all 3.

For these notes, I have chosen to integrate about the z axis. You should be able to obtain the result yourselves integrating about the x or y axis. See that a disc consists of rings of growing radius. Each ring has circumference 2.r, therefore dm = 2.r.dr.

So 24

2.2.4

0

4

0

32 RrdrrdmrIzR

Rr

rArea

=

===

=

=. Note that the mass of the

disc is given by the area times the density, i.e. M = R2

Thus 22

24 MRRIz ==

Perpendicular Axis Theorem Ix = Iy = Iz / 2 = MR2 / 4 Since we now know the inertial properties of a disc, we can use these results in any 3-dimensional object built up of discs, in much the same way as the disc was built up of rings.

y

x dm

dr

Figure 15: Inertia of a Disc


3D Example: Mass Moment of Inertia of a Cylinder Problem: Calculate the Mass Moment of Inertia for a homogeneous cylinder of mass M, radius R and length L about axes through its centroid.

Solution From Symmetry we see that Iz = Iy. Since we know the inertial properties of a thin disc, use these as dm as shown above. Since the cylinder is homogeneous note that dm = (M / L)dx.

To solve for Ix, note that the x axis runs through the centroid of each thin disc. For each incremental dm, dIx = 0.5dm.R2 =(0.5M R2 / L)dx

Thus [ ] === 2

2

22

2

2

22

L

L

L

L

xL

MRdxL

MRIx2

2MR . This is an interesting, though

unsurprising) result. About the axis of rotation, the moment of inertia of a cylinder is independent of its length. To solve for Iz, we will need to use the parallel axis theorem since the discs are away from the centroid axis. For each incremental dm, dIZ = 0.25dm.R2 + dm.x2 = dm(0.25.R2 + x2)

Thus =

+=

+=

+=

124344

322

2

322

2

22 LLR

LMxxR

LMdxxR

LMIz

L

L

L

L

124

22 MLMR+

Note how easy it now is to find Izz about the end of the cylinder.

x

x

x

dx

L

R R

dm

z z y


3D Example: Mass Moment of Inertia of a Circular Cone Calculate the Mass Moment of Inertia for a homogeneous circular cone of mass M, base radius R and height H about the XX and YY axes shown.

Figure 16: Moment of Inertia of a Uniform 3D Cone

Solution

Well use the same technique as the previous example. At a distance along the x-axis, the cross section is a disc of radius r = (R / H).x and mass dm. dm can be calculated, by knowing the volume of a cone is V=R2H/3 and the volume of each disc is given by r2.dx. Therefore dm=(M. dV / V) = 3M.x2dx / H3.

XX again runs through the centroid of each disc, so for each

incremental dm, dIx = 0.5dm.r2 Thus =

==

HH xH

MRdxxH

MRIx0

5

5

2

0

45

2

523

23

103 2MR

Iz again requires the parallel axis theorem. For each incremental dm, dIZ = 0.25dm.r2 + dm.x2. Thus

=

+=+=

HHH xHR

HMdxx

HMdxx

HMRIz

0

5

2

2

30

43

0

45

2

51

433

43

+ 2

2

453 HRM

X

X x

X

dx

H

R R

Y

Y


Summary: Linear and Rotational Analogies Linear Dim Rotational Dim Conversion

(angular) displacement s [L] [1] rst =

(angular) velocity sdtdsv ,, [L/T]

,,dtd

[1/T] rvt =

(angular) acceleration sdt

sdvdtdva ,,,, 2

2

[L/T2] ,,,, 2

2

dtd

dtd

[1/T2] rat =

const. acceleration 2atuts += 20 tt +=

const. acceleration asuv 222 += 2202 +=

force / torque F N T Nm T = F x r

inertia (mass/rotational) m kg I kgm

2 I = Volume

dmr .2

Newtons 2nd Law F = ma IT =

Work sFsdFW av .. ==

Joule avTdTW ==

. Joule

Kinetic Energy .m.v2 Joule .I.2 Joule

Instantaneous Power FvdtdWP == Watt T

dtdWP == Watt

Kinetics of RotationOverviewRotational ConceptsThe Cause of Rotation: Newtons 2nd LawTorques and MomentsNewtons 2nd Law for RotationWorked Example using Moment of Inertia

Moment of Inertia : Distribution of MassMoment of Inertia DerivationCommon Mass Moments of InertiaCommon Error to AvoidWorked Example MMoI CalculationMoment of Inertia via Kinetic EnergyComposite BodiesRadius of GyrationWorked Example: Falling Load Accelerating a Drum

Centre of MassParallel Axis TheoremRelationship with the Second Moment of AreaMoment of Inertia of Thin PlatesPerpendicular Axis TheoremMoment of Inertia using Tables of Common ShapesWorked Example: MMoI of Composite BodyWorked Example: MMoI via Subtraction

Moment of Inertia via Numerical IntegrationWorked Example: Mass Moment of Inertia of a Thin Disc3D Example: Mass Moment of Inertia of a Cylinder3D Example: Mass Moment of Inertia of a Circular Cone

Summary: Linear and Rotational Analogies

mechanics of rotation and moment of inertia

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