Parallel axis theorem and their use on Moment Of Inertia
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you can learn how to use parallel axis theorem to calculate the moment of inertia about centroidal axis and its transfer rule.
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![–h
2
h
X
X
y
–h
2
b
h
X
X'
X
X'
y
Statement and Derivation of parallel axis theorem
Parallel axis theorem states that the moment of inertia of a plane area about any axis parallel to the centroidal axis of that area
is equal to the sum of moment of inertia about a parallel centroidal axis of that plane area and the product of the area and
square of the distance between the two axes.
[Ix'x' = Ixx +Ah
2
]
Proof:
Consider a lamina having area 'A', X – X be the centroidal axis of this lamina and x'- x' be the another axis which is parallel to the
centroidal x – x axis and at a distance of h from centroidal axis . Consider an elemental area dA at a distance 'y' from the
centroidal axis x – x.
Ixx = M.O.I of the lamina about x - x axis (centroidal axis)
we know, IXX = y
2
dA
Here,
IX'X' = moment of inertia of the lamina about x'-x' axis
so,
IX'X' = (h+y)
2
dA [(h+y) is the distance of elemental area from x' – x' axis.]
=(h
2
+2hy +y
2
) dA.
=h
2
dA +2h.ydA+ y
2
dA
= Ah
2
+2h× 0 + IXX
[Ix'x' = Ixx +Ah
2
]
Hence proved the theorem
Q. Determine moment of inertia about centroidal xx and yy axes of the plane figure shown in figure below.
Axis दिएको छैन भने जता मान्िा पनन हुन्छ !
तर सके सम्म object first quadrant मा पने गरर मान्नु !
Solution:
first divide the whole figure into known geometrical figure and taking left bottom corner as a origin.
For figure (i) A1 = 0.14 × 1.5 = 0.21 m
2
x1 =
0.14
2 = 0.07m
y1 =
1.5
2 = 0.75 m
20cm
1.5m
14cm
1.2m
1.17m1.5m
0.91m
0.13m
1.06m0.14m
20cm
1
2
3
14cm
(y.dA = 1
st
moment of area =
centroid = 0
i.e ydA=
-h/2
h/2
y.xdy = 0 )](https://image.slidesharecdn.com/parellelaxistheoremandtheiruse-200908034853/85/Parallel-axis-theorem-and-their-use-on-Moment-Of-Inertia-1-320.jpg)
![y1
y
h
x = 0.39m
1.17m1.5m
0.91m
0.13m
1.06m0.14m
20cm
1
2
3
14cm
For figure (2)
A2 = 1.06 × 0.13 = 0.1378m
2
x2 =
0.14 +
1.06
2 = 0.67 m
y2 =
0.13
2 = 0.065m
For figure (3)
A3 =
1
2 × 0.91 × 1.17 = 0.532m
2
x3 =
0.14 +
0.91
3 = 0.443m
y3 =
0.13 +
1.17
3 = 0.52m
-
X =
A1x1 + A2 x2 + A3x3
(A1 + A2 + A3) =
0.21 × 0.07 + 0.1378 × 0.67 + 0.532 × 0.443
(0.21 + 0.1378 + 0.532)
= 0.39m
-
Y =
A1y1 + A2 y2 + A3y3
(A1 + A2 + A3) =
0.21 × 0.75 + 0.1387 × 0.065 + 0.532 × 0.52
(0.21 + 0.1378 + 0.532)
= 0.503m
M.O.I about centriodal x - x axis
Ixx = [ ]Ixx rectangle1 + [ ]Ixx rectangle2 + [ ]Ixx rectangle3
M.O.I about centriodal axis is calculating by using parallel axis theorem.
(Note: i.e. M.O.I about centroidal axis of individual figure is transferred in to the centroidal axis of whole figure.)
Ixx =
b1h1
3
12 + A1 (y1 – –y)
2
+
b2h2
3
12 + A2 (y2 – –y)
2
+
b3h3
3
36 + A3 (y3 – –y)
2
=
0.14 × 1.5
3
12 + 0.21 (0.75 – 0.503)
2
+
1.06 × 0.13
3
12 + 0.1378 (0.06575 – 0.503)
2
+
0.91 × 1.17
3
36 + 0.532 (0.53 – 0.503)
2
= 0.0521 + 0.0266 + 0.0408
[Ixx= 0.11957 m
4
]
Iyy = [ ]Iyy rectangle1 + [ ]Iyy rectangle2 + [ ]Iyy rectangle3
=
h1b1
3
12 + A1 (x1 – –x)
2
+
h2b2
3
12 + A2 (x2 – –x)
2
+
h3b3
3
36 + A3 (x3 – –x)
2
=
1.5 × 0.14
3
12 + 0.21 (0.07 – 0.39)
2
+
0.13 × 1.06
3
12 + 0.1378 (0.67 – 0.39)
2
+
1.17 × 0.91
3
36 + 0.532 (0.443 – 0.39)
2
= 0.02184 + 0.0237 + 0.0276
= 0.07317 m
4](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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The document defines several terms:
1. Centroid - The point at which the total area of a plane is concentrated. It is where the average position of the total weight of the plane would balance.
2. Radius of gyration - The distance from the axis of rotation to the centroid. It is calculated as the square root of the moment of inertia divided by the total area.
3. Area moment of inertia - The product of the plane area and the square of the perpendicular distance to the axis of reference. It is a measure of the resistance offered by a plane figure to bending or twisting about an axis.
L10-T7 MOS Jul 2019-1.pptx .
The document discusses the concepts of centroid, moment of inertia, and radius of gyration. It provides examples of calculating the centroid and moment of inertia of simple geometric shapes like rectangles, triangles, and semicircles using the first and second moments of area. Key formulas introduced include the parallel axis theorem and polar moment of inertia theorem. Methods like direct integration are demonstrated for finding the moment of inertia of areas about different axes. Later examples show applying these concepts to solve tutorial problems involving composite shapes.
Prof. V. V. Nalawade, Notes CGMI with practice numerical
Centre of gravity is a point where the whole weight of the body is assumed to act. i.e., it is a point where entire distribution of gravitational force is supposed to be concentrated
It is generally denoted “G” for all three dimensional rigid bodies.
e.g. Sphere, table , vehicle, dam, human etc
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Centroid is a point where the whole area of a plane lamina is assumed to act.
It is a point where the entire length, area & volume is supposed to be concentrated.
It is a geometrical centre of a figure.
It is used for two dimensional figures.
e.g. rectangle, circle, triangle, semicircle
Lesson 16 length of an arc
1. This document discusses methods for calculating the length of an arc of a curve and the surface area of revolution. It provides formulas for finding arc length and surface area when curves are defined by rectangular coordinates, parametric equations, or polar coordinates.
2. Several examples are given of applying the formulas to find the arc length of curves and the surface area when graphs are revolved about axes. This includes revolving curves like y=x^3, y=x^2, and xy=2 about the x-axis and y-axis.
3. The key formulas presented are that arc length can be found using an integral of the form ∫√(dx/dy)^2 + 1 dy or
Moment of inertia revision
This document discusses moment of inertia, which is a measure of an object's resistance to changes in rotation. It begins by defining moment of inertia as the second moment of force or mass of an object. It then provides formulas for calculating the moment of inertia of common shapes like rectangles, circles, and hollow sections. For rectangles, the moment of inertia depends on the cube of the distance of the axis from the object's sides. For circles, the moment of inertia is proportional to the diameter to the fourth power. The document also presents theorems for calculating moment of inertia about different axes, such as perpendicular axes and parallel axes. Sample problems are worked through to demonstrate calculating moment of inertia for rectangular and circular sections.
Prof. V. V. Nalawade, UNIT-3 Centroid, Centre off Gravity and Moment of Inertia
The document discusses concepts related to center of gravity and moment of inertia. It defines center of gravity as the point where the entire weight of a body acts and centroid as the point where the entire area of a plane figure acts. It provides formulas for calculating the centroid of composite figures and discusses the parallel axis theorem and perpendicular axis theorem for calculating moment of inertia about different axes. The document also defines radius of gyration and provides formulas for calculating moment of inertia and radius of gyration of common plane figures.
Properties of surfaces-Centre of gravity and Moment of Inertia
The document discusses properties of surfaces, including centre of gravity and moment of inertia. It defines key terms like centre of gravity, centroid, area moment of inertia, radius of gyration, and mass moment of inertia. Methods for calculating these properties are presented for basic shapes like rectangles, triangles, circles, and composite shapes. Theorems like the perpendicular axis theorem and parallel axis theorem are also covered. Examples are provided for determining the moment of inertia of various plane figures and structures.
Centre of Gravity
1) The document discusses concepts related to centroid and moment of inertia including: the centroid is the point where the total area of a plane figure is assumed to be concentrated; formulas are provided for finding the centroid of basic shapes; the difference between centroid and center of gravity is explained; properties and methods for finding the centroid are described such as using moments.
2) Formulas are given for moment of inertia including how it is calculated about different axes and the parallel axis theorem.
3) Example problems are provided to demonstrate calculating the centroid and moment of inertia for various shapes.
8 arc length and area of surfaces x
The document discusses calculating the length of a curve defined by a function f(x). It explains that the length can be approximated by partitioning the domain into small intervals, drawing line segments between the corresponding points on the curve, and taking the sum of the lengths. This sum approaches the true curve length as the interval size approaches 0. Using the Mean Value Theorem, the sum can be written as an integral involving the derivative of f, known as the arc length formula. While difficult generally, the integral can be solved for some basic functions like polynomials.
Applications of Differential Calculus in real life
Applications of Differentiation in economics, optimization, kinematics, tangent and normal lines, related rates, linear approximation or small changes
conference_poster_5_UCSB
The document summarizes the brachistochrone problem from calculus of variations. It introduces the brachistochrone curve as the curve of fastest descent under gravity between two points. The problem is then solved using tools from calculus of variations, arriving at the Euler-Lagrange equation. This equation shows that the brachistochrone curve between two points is a cycloid. Additionally, the document discusses that the cycloid is a tautochronic curve, meaning an object will take the same amount of time to slide from any point on it to the lowest point.
Solution set 3
The document discusses quantum mechanical concepts including:
1) The time derivative of the momentum expectation value satisfies an equation involving the potential gradient.
2) For an infinite potential well, the kinetic energy expectation value is proportional to n^2/a^2 and the potential energy expectation value vanishes.
3) Eigenfunctions of an eigenvalue problem under certain boundary conditions correspond to positive eigenvalues that are sums of squares of integer multiples of pi.
Stoke’s theorem
1) Stokes' theorem relates a surface integral over a surface S to a line integral around the boundary curve of S. It states that the line integral of a vector field F around a closed curve C that forms the boundary of a surface S is equal to the surface integral of the curl of F over the surface S.
2) In Example 1, Stokes' theorem is used to evaluate a line integral around an elliptical curve C by calculating the corresponding surface integral over the elliptical region S bounded by C.
3) In Example 2, Stokes' theorem is again used, this time to evaluate a line integral around a circular curve C by calculating the surface integral over the part of a sphere bounded by C.
math vysh.pptx
The document discusses several integral theorems including the divergence theorem, Gauss's divergence theorem, and Stokes' theorem. It provides statements and examples of applying each theorem. The divergence theorem relates the volume integral of the divergence of a vector field over a volume to the surface integral of the vector field over the bounding surface. Gauss's divergence theorem is a similar relationship between a volume integral and surface integral. Stokes' theorem relates a surface integral over an oriented surface S to a line integral around the boundary curve of S, allowing conversion between different types of integrals.
Developing Expert Voices
The student reflects on completing a math project for their calculus course as a way to study for an upcoming exam. They acknowledge that they procrastinated significantly but were able to cover a broad range of calculus concepts through multi-step word problems selected from different units. While the assignment did not dramatically increase their knowledge, it helped reinforce some details and connections between topics. The student resolves to select deadlines more wisely and stop procrastinating for future projects.
Lesson 12 centroid of an area
Here are the steps to find the centroid of each given plane region:
1. Region bounded by y = 10x - x^2, x-axis, x = 2, x = 5:
- Set up the integral to find the area A: ∫2^5 (10x - x^2) dx
- Evaluate the integral: A = 96
- Set up the integrals to find the x- and y-moments: ∫2^5 x(10x - x^2) dx and ∫2^5 (10x - x^2)x dx
- Evaluate the integrals: Mx = 192, My = 960
- Use the formulas for centroid: C
(5) Ellipse (Theory). Module-3pdf
An ellipse is the locus of a point which moves in such a way that its distance form a fixed point is in constant ratio to its distance from a fixed line. The fixed point is called the focus and fixed line is called the directrix and the constant ratio is called the eccentricity of a ellipse denoted by (e).
In other word, we can say an ellipse is the locus of a point which moves in a plane so that the sum of it distances from fixed points is constant.
2.1 Standard Form of the equation of ellipse
Let the distance between two fixed points S and S' be 2ae and let C be the mid point of SS.
Taking CS as x- axis, C as origin.
Let P(h,k) be the moving point Let SP+ SP = 2a (fixed distance) then
(ii) Major & Minor axis : The straight line AA is called major axis and BB is called minor axis. The major and minor axis taken together are called the principal axes and its length will be given by
Length of major axis 2a Length of minor axis 2b
(iii) Centre : The point which bisect each chord of an ellipse is called centre (0,0) denoted by 'C'.
(iv) Directrix : ZM and Z M are two directrix and their equation are x= a/e and x = – a/e.
(v) Focus : S (ae, 0) and S (–ae,0) are two foci of an ellipse.
(vi) Latus Rectum : Such chord which passes through either focus and perpendicular to the major axis is called its latus rectum.
Length of Latus Rectum :
If L is (ae, 𝑙 ) then 2𝑙 is the length of
SP+S'P=
{(h ae)2 k 2} +
= 2a
Latus Rectum.
Length of Latus rectum is given by
2b2
.
h2(1– e2) + k2 = a2(1– e2)
Hence Locus of P(h, k) is given by. x2(1– e2) + y2 = a2(1– e2)
2
a
(vii) Relation between constant a, b, and e
a 2 b2
b2 = a2(1– e2) e2 =
a 2
x2
a 2 +
y
a 2 (1 e 2 ) = 1
e =
a 2
Result :
Major Axis
(a) Centre C is the point of intersection of the axes of an ellipse. Also C is the mid point of AA.
(b) Another form of standard equation of ellipse
x 2 y2
a 2 b2
1 when a < b.
Directrix Minor Axis Directrix x = -a/e x = a/e
Let us assume that a2(1– e2 )= b2
The standard equation will be given by
x2 y2
a2 b2
2.1.1 Various parameter related with standard ellipse :
In this case major axis is BB= 2b which is along y- axis and minor axis is AA= 2a along x- axis. Focus S(0,be) and S(0,–be) and directrix are y = b/e and y = –b/e.
2.2 General equation of the ellipse
The general equation of an ellipse whose focus is (h,k) and the directrix is the line ax + by + c = 0 and the eccentricity will be e. Then let P(x1,y1) be any point on the ellipse which moves such that SP = ePM
Let the equation of the ellipse x
y2
a > b
(x –h)2 + (y –k)2 =
e 2 (ax1 by1 c) 2
a 2 b2
1 1 a 2 b2
(i) Vertices of an ellipse : The point of which ellipse cut the axis x-axis at (a,0) & (– a, 0) and y- axis at (0, b) & (0, – b) is called the vertices of an ellipse.
Hence the locus of (x1,y1) will be given by (a2 + b
6161103 10.6 inertia for an area
Moment of inertia and product of inertia are properties that depend on the orientation of the axes they are calculated around. The product of inertia, Ixy, of an area can be positive, negative, or zero depending on the location and orientation of the x and y axes. Ixy is calculated by taking the double integral of xy over the entire area or using the parallel axis theorem which relates Ixy to the product of inertia about the centroidal axes, Ix'y', plus the cross product of the distance from the centroidal axes to the x and y axes.
Similar to Parallel axis theorem and their use on Moment Of Inertia (19)
Prof. V. V. Nalawade, Notes CGMI with practice numerical
Prof. V. V. Nalawade, Notes CGMI with practice numerical
Prof. V. V. Nalawade, UNIT-3 Centroid, Centre off Gravity and Moment of Inertia
Prof. V. V. Nalawade, UNIT-3 Centroid, Centre off Gravity and Moment of Inertia
Properties of surfaces-Centre of gravity and Moment of Inertia
Properties of surfaces-Centre of gravity and Moment of Inertia
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The Python for beginners. This is an advance computer language.
Literature Review Basics and Understanding Reference Management.pptx
Literature Review Basics and Understanding Reference Management.pptx
International Conference on NLP, Artificial Intelligence, Machine Learning an...
International Conference on NLP, Artificial Intelligence, Machine Learning an...
Eric Nizeyimana's document 2006 from gicumbi to ttc nyamata handball play
Eric Nizeyimana's document 2006 from gicumbi to ttc nyamata handball play
Parallel axis theorem and their use on Moment Of Inertia
- 1. –h 2 h X X y –h 2 b h X X' X X' y Statement and Derivation of parallel axis theorem Parallel axis theorem states that the moment of inertia of a plane area about any axis parallel to the centroidal axis of that area is equal to the sum of moment of inertia about a parallel centroidal axis of that plane area and the product of the area and square of the distance between the two axes. [Ix'x' = Ixx +Ah 2 ] Proof: Consider a lamina having area 'A', X – X be the centroidal axis of this lamina and x'- x' be the another axis which is parallel to the centroidal x – x axis and at a distance of h from centroidal axis . Consider an elemental area dA at a distance 'y' from the centroidal axis x – x. Ixx = M.O.I of the lamina about x - x axis (centroidal axis) we know, IXX = y 2 dA Here, IX'X' = moment of inertia of the lamina about x'-x' axis so, IX'X' = (h+y) 2 dA [(h+y) is the distance of elemental area from x' – x' axis.] =(h 2 +2hy +y 2 ) dA. =h 2 dA +2h.ydA+ y 2 dA = Ah 2 +2h× 0 + IXX [Ix'x' = Ixx +Ah 2 ] Hence proved the theorem Q. Determine moment of inertia about centroidal xx and yy axes of the plane figure shown in figure below. Axis दिएको छैन भने जता मान्िा पनन हुन्छ ! तर सके सम्म object first quadrant मा पने गरर मान्नु ! Solution: first divide the whole figure into known geometrical figure and taking left bottom corner as a origin. For figure (i) A1 = 0.14 × 1.5 = 0.21 m 2 x1 = 0.14 2 = 0.07m y1 = 1.5 2 = 0.75 m 20cm 1.5m 14cm 1.2m 1.17m1.5m 0.91m 0.13m 1.06m0.14m 20cm 1 2 3 14cm (y.dA = 1 st moment of area = centroid = 0 i.e ydA= -h/2 h/2 y.xdy = 0 )
- 2. y1 y h x = 0.39m 1.17m1.5m 0.91m 0.13m 1.06m0.14m 20cm 1 2 3 14cm For figure (2) A2 = 1.06 × 0.13 = 0.1378m 2 x2 = 0.14 + 1.06 2 = 0.67 m y2 = 0.13 2 = 0.065m For figure (3) A3 = 1 2 × 0.91 × 1.17 = 0.532m 2 x3 = 0.14 + 0.91 3 = 0.443m y3 = 0.13 + 1.17 3 = 0.52m - X = A1x1 + A2 x2 + A3x3 (A1 + A2 + A3) = 0.21 × 0.07 + 0.1378 × 0.67 + 0.532 × 0.443 (0.21 + 0.1378 + 0.532) = 0.39m - Y = A1y1 + A2 y2 + A3y3 (A1 + A2 + A3) = 0.21 × 0.75 + 0.1387 × 0.065 + 0.532 × 0.52 (0.21 + 0.1378 + 0.532) = 0.503m M.O.I about centriodal x - x axis Ixx = [ ]Ixx rectangle1 + [ ]Ixx rectangle2 + [ ]Ixx rectangle3 M.O.I about centriodal axis is calculating by using parallel axis theorem. (Note: i.e. M.O.I about centroidal axis of individual figure is transferred in to the centroidal axis of whole figure.) Ixx = b1h1 3 12 + A1 (y1 – –y) 2 + b2h2 3 12 + A2 (y2 – –y) 2 + b3h3 3 36 + A3 (y3 – –y) 2 = 0.14 × 1.5 3 12 + 0.21 (0.75 – 0.503) 2 + 1.06 × 0.13 3 12 + 0.1378 (0.06575 – 0.503) 2 + 0.91 × 1.17 3 36 + 0.532 (0.53 – 0.503) 2 = 0.0521 + 0.0266 + 0.0408 [Ixx= 0.11957 m 4 ] Iyy = [ ]Iyy rectangle1 + [ ]Iyy rectangle2 + [ ]Iyy rectangle3 = h1b1 3 12 + A1 (x1 – –x) 2 + h2b2 3 12 + A2 (x2 – –x) 2 + h3b3 3 36 + A3 (x3 – –x) 2 = 1.5 × 0.14 3 12 + 0.21 (0.07 – 0.39) 2 + 0.13 × 1.06 3 12 + 0.1378 (0.67 – 0.39) 2 + 1.17 × 0.91 3 36 + 0.532 (0.443 – 0.39) 2 = 0.02184 + 0.0237 + 0.0276 = 0.07317 m 4