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Chapter-V.pptx
- 1.
- 2. Introduction Centerof Gravity Center of Lines, Areas, and Volume Center of composite Bodies Determining the centroid by integration
- 3. Centroid: Centroid is definedas the point of the geometric center of an object, where the density is wholly distributed over the body. Fig.1. Centroid of triangle The point where the cutout is balanced perfectly is the center of the object.
- 4. The center ofgravity is defined as the exact place in a body around which the instants due to gravity are regarded as zero. It is the point at which the entire body is perfectly balanced in relation to gravity. It is the point where the gravitational force or weight of the body acts in any orientation of the body.
- 5. The center ofgravity is abbreviated as C.G or simply G. The gravitational field always affects the center of gravity because when the gravitational field’s value varies, the center of gravity’s value changes. Fig. 2. Point of center of gravity
- 6. A body iscomposed of an infinite number of particles of differential size. When the body is located within a gravitational field each of these particles have a weight 𝑑𝑊. These weights form a parallel force system. The resultant of this system is the total weight of the body, which passes through a single point called the center of gravity, 𝐺 Fig. 3. classification of force
- 7. To determine thelocation of the center of gravity of any body, the principle of moments is applied to the parallel system of gravitational forces. The moment of the resultant gravitational force 𝑊 about any axis equals the sum of the moments about the same axis of the gravitational forces 𝑑𝑊 acting on all particles of infinitesimal elements of the body. If the principle of moment is applied about the y-axis: 𝑥𝑊 = 𝑥𝑤 , 𝑡ℎ𝑒𝑛 𝑥 = 𝑥𝑤 𝑊 , 𝑓𝑜𝑟 𝑥 𝑎𝑛𝑑 𝑧 𝑎𝑥𝑒𝑠 𝑎𝑙𝑠𝑜, 𝑦 = 𝑦𝑤 𝑊 , 𝑧 = 𝑧𝑤 𝑊
- 8. The center ofmass is defined as the position at which the entire body is directed. The mass distribution is considered uniform around the center of the mass. Since the center of mass is independent of the gravitational field the body remains unaffected by change in the gravitational field’s force. In a simple rigid bodies with uniform density, the center of mass is located at the center or centroid. In order to study the dynamic response or accelerated motion of a body knowing the location of center of mass of the body is important.
- 9. With the substitutionof 𝑊 = 𝑚𝑔 and 𝑑𝑊 = 𝑔𝑑 the expressions for the coordinates of the center of gravity become: 𝑥 = 𝑥𝑑𝑚 𝑚 𝑦 = 𝑦𝑑𝑚 𝑚 𝑧 = 𝑧𝑑𝑚 𝑚 their respective position vectors: 𝑟 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 𝒓 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 𝒓 = 𝑟𝑚 𝑚
- 10. The major distinctionbetween center of gravity and the center of mass is that the center of gravity is the position at which the entire body weight is balanced. While the center of mass is the position at which the entire mass of the body is directed.
- 11. Difference between centerof mass and center of gravity. Center of Mass Center of Gravity Is the point where mass distribution is uniform in all directions. Is the point where weight is evenly distributed in all direction. Is based on the mass of the body Is based on the weight of the body It is the center where the entire bodily mass is concentrated It is the point at which the body’s entire weight is suspended Mass of the body is distributed uniformly throughout the body Weight of the body is distributed uniformly throughout the body. When the body moves from left to right mass operating to the left side is equals to mass acting to on the right side. When the body travels on an axis from left to right weight on the left side is equals with weight on the right side. Change in gravitational field has no effect on it Change in gravitational field has effect on it. When spinning around a point it produces angular momentum When spinning around an axis the net torque is zero due to gravitational force
- 12. For a slenderrod or wire of length L, cross-sectional area A, and density 𝜌, the body is approximated as a line segment, and 𝑑𝑚 = 𝜌𝐴𝑑𝐿. If 𝜌 and 𝐴 are constant over the length of the rod, the coordinates of the center of mass becomes the coordinates of the centroid 𝐶 of the line segment. Thus, 𝒙 = 𝒙𝒅𝑳 𝑳 𝒚 = 𝒚𝒅𝑳 𝑳 𝒛 = 𝒛𝒅𝑳 𝑳
- 13. If a linesegment (or rod) lies within the x–y plane and described by a thin curve 𝑦 = 𝑓 (𝑥), as shown in the figure below, then its centroid is determined from: 𝒙 = 𝒙𝒅𝑳 𝑳 𝒚 = 𝒚𝒅𝑳 𝑳
- 14. The length ofthe differential element is given by the Pythagorean theorem: 𝒅𝑳 = 𝒅𝒙 𝟐 + 𝒅𝒚 𝟐 This can be written as: 𝒅𝑳 = 𝒅𝒙 𝒅𝒙 𝟐 𝒅𝒙 𝟐 + 𝒅𝒚 𝒅𝒙 𝟐 𝒅𝒙 𝟐 = 𝟏 + 𝒅𝒚 𝒅𝒙 𝟐 𝒅𝒙 In other way: 𝒅𝑳 = 𝒅𝒙 𝒅𝒚 𝟐 𝒅𝒚 𝟐 + 𝒅𝒚 𝒅𝒚 𝟐 𝒅𝒚 𝟐 = 𝒅𝒙 𝒅𝒚 𝟐 + 𝟏 𝒅𝐲
- 15. When a bodyof density𝜌 has a small but constant thickness 𝑡, the body can model as a surface area 𝐴. The mass of an element becomes 𝑑𝑚 = 𝜌𝑡𝑑𝐴. If 𝜌 and 𝑡 are constant over the entire area, the coordinates of the center of mass of the body becomes the coordinates of the centroid C of the surface area. Thus, 𝑥 = 𝑥𝑑𝐴 𝐴 𝑦 = 𝑦𝑑𝐴 𝐴 𝑧 = 𝑧𝑑𝐴 𝐴 The numerators in the equations are the first moments of area.
- 16. For a generalbody of volume 𝑉 and density 𝜌, the element has a mass 𝑑𝑚 = 𝜌𝑑𝑉. The density 𝜌 cancels if it is constant over the entire volume. The coordinates of the center of mass also become the coordinates of the centroid C of the body. Thus, 𝒙 = 𝒙𝒅𝑽 𝑽 𝒚 = 𝒚𝒅𝑽 𝑽 𝒛 = 𝒛𝒅𝑽 𝑽
- 17. Principles used tochoose the differential element and setting up the integrals to simplify difficulties in the determination of centroid: Order of Element: a first-order differential element should be selected rather a higher-order element which cover the entire figure by a single integration. Example: A 1st order horizontal strip of area 𝑑𝐴 = 𝑙𝑑𝑦 requires only one integration with respect to y to cover the entire figure. The second-order element 𝑑𝑥𝑑𝑥 requires two integrations, with respect to x & y, to cover the entire figure.
- 18. A 1st orderelement in the form of a circular slice of volume 𝑑𝑉 = 𝜋𝑟2 𝑑𝑦 requires only one integration, & is preferable rather to choose a 3rd order element 𝑑𝑉 = 𝑑𝑥𝑑𝑦𝑑𝑧, which require three integrations..
- 19. Continuity: Choose an elementwhich can be integrated in one continuous operation to cover the entire figure. Example: To determine the centroid the horizontal strip is preferable to the vertical strip, which, requires two separate integrals because of the discontinuity in the expression for the height of the strip at 𝑥 = 𝑥1.
- 20. Discarding Higher-Order Terms: Higherorder terms should be dropped compared with lower-order terms. Example: In the figure above the vertical strip of area under the curve is given by the first-order term 𝑑𝐴 = 𝑦𝑑𝑥, and the second-order triangular area 1 2 𝑑𝑥𝑑𝑦 is discarded. In the limit, of course, there is no error.
- 21. Choice of Coordinates: Choosethe coordinate system which best matches the boundaries of the figure. Example: The boundaries of the area in the 1st figure is easily described in rectangular coordinates, whereas the boundaries of the circular sector in the 2nd figure is best suited to polar coordinates.
- 22. Centroidal Coordinate ofdifferentials Element: When a 1st or 2nd order differential element is chosen, it is essential to use the coordinate of the centroid of the element for the moment arm in expressing the moment of the differential element. Example: 𝑥 = 𝑥𝑐 𝑑𝐴 𝐴 𝑦 = 𝑦𝑐 𝑑𝐴 𝐴 𝑥 = 𝑥𝑐 𝑑𝑉 𝑉 𝑧 = 𝑧𝑐 𝑑𝑉 𝑉
- 23. #1. Locate thecentroid of a circular arc as shown in the figure. Solution: The x-axis is chosen as axis of symmetry. The co-ordinate system is polar coordinates. Take the differential element length of the arc is 𝐝𝐋 = 𝒓𝒅𝜽
- 24. The total lengthof the arc is: 𝐿 = 2𝛼𝑟 Thus, 𝐿𝑥 = 𝑥𝑐 𝑑𝐿 , ⇒ 2𝛼𝑟𝑥 = −𝛼 𝛼 𝑟𝑐𝑜𝑠𝜃 𝑟𝑑𝜃 2𝛼𝑟𝑥 = 𝑟2 −𝛼 𝛼 𝑐𝑜𝑠𝜃𝑑𝜃 𝑥2𝛼𝑟 = 𝑟2 𝑠𝑖𝑛𝜃 𝛼 −𝛼 = 𝑟2 𝑠𝑖𝑛𝛼 − 𝑠𝑖𝑛 −𝛼 ⇒ 𝑥 = 𝑟𝑠𝑖𝑛𝛼 𝛼 For 2𝛼 = 𝜋, 𝑥 = 2𝑟 𝜋
- 25. #2. Locate thecentroid of the rod bent into the shape of a parabolic arc as shown in figure. Solution: The co-ordinate system is cartesian coordinates. Take the differential element length of the arc is 𝐝𝐋 = 𝒅𝒙 𝒅𝒚 𝟐 + 𝟏𝒅𝒚
- 26. From the givenequation 𝑥 is evaluated as: 𝑥 = 𝑦2 ⇒ 𝑑𝑥 𝑑𝑦 = 2𝑦 ∴ 𝑑𝐿 = 2𝑦 2 + 1 𝑑𝑦 𝑥𝑑𝐿 = 𝑥𝑐 𝑑𝐿 , ⇒ 𝑥 2𝑦 2 + 1 𝑑𝑦 = 𝑥 2𝑦 2 + 1 𝑑𝑦 𝑥 2𝑦 2 + 1 𝑑𝑦 = 𝑦2 2𝑦 2 + 1 𝑑𝑦
- 27. Integration of theleft equation is: ⇒ 𝑥 4𝑦2 + 1 𝑑𝑦 = 𝑦2 4𝑦2 + 1 𝑑𝑦 4𝑦2 + 1 𝑑𝑦 , 𝑙𝑒𝑡 𝑢 = 2𝑦, 𝑢 2 = 𝑦, 𝑑𝑢 = 2𝑑𝑦, ⇒ 𝑑𝑢 2 = 𝑑𝑦 0 1 4𝑦2 + 1𝑑𝑦 = 1 2 0 2 𝑢2 + 1 𝑑𝑢 = 𝑢 2 𝑢2 + 1 + 12 2 𝑙𝑛 𝑢 + 𝑢2 + 1 2 0 Substitute values of u: 𝑥 1 2 𝑦 2 4𝑦2 + 1 + 1 2 𝑙𝑛 2𝑦 + 4𝑦2 + 1 1 0 = 1.479𝑥
- 28. Integration of theright equation is: 0 1 𝑦2 4𝑦2 + 1 𝑑𝑦 , 𝑙𝑒𝑡 𝑢 = 2𝑦, 𝑢 2 = 𝑦, 𝑑𝑢 = 2𝑑𝑦, ⇒ 𝑑𝑢 2 = 𝑑𝑦 0 1 𝑦2 4𝑦2 + 1 𝑑𝑦 = 0 2 𝑢 2 2 𝑢2 + 1 1 2 𝑑𝑢 = 1 8 0 2 𝑢2 𝑢2 + 1 𝑑𝑢 = 1 8 𝑢 8 2𝑢2 + 12 𝑢2 + 1 + 12 8 𝑙𝑛 𝑢 + 𝑢2 + 1 2 0 Substitute values of u: 𝑦 32 8𝑦2 + 1 4𝑦2 + 1 + 1 64 𝑙𝑛 2𝑦 + 4𝑦2 + 1 1 0 = 0.606 1.479𝑥 = 0.606 𝒙 = 𝟎. 𝟒𝟏𝒎 𝐀𝐧𝐬.
- 29. From the givenequation 𝑦 is evaluated as: 𝑥 = 𝑦2 ⇒ 𝑑𝑥 𝑑𝑦 = 2𝑦 ∴ 𝑑𝐿 = 2𝑦 2 + 1 𝑑𝑦 𝑦𝑑𝐿 = 𝑦𝑐 𝑑𝐿 , ⇒ 𝑦 2𝑦 2 + 1 𝑑𝑦 = 𝑦 2𝑦 2 + 1 𝑑𝑦 𝑦 2𝑦 2 + 1 𝑑𝑦 = 𝑦 2𝑦 2 + 1 𝑑𝑦
- 30. Integration of theleft equation is: ⇒ 𝑦 2𝑦 2 + 1 𝑑𝑦 = 𝑦 2𝑦 2 + 1 𝑑𝑦 𝑦 4𝑦2 + 1 𝑑𝑦 = 𝑦 4𝑦2 + 1 𝑑𝑦 4𝑦2 + 1 𝑑𝑦 , 𝑙𝑒𝑡 𝑢 = 2𝑦, 𝑢 2 = 𝑦, 𝑑𝑢 = 2𝑑𝑦, ⇒ 𝑑𝑢 2 = 𝑑𝑦 0 1 4𝑦2 + 1𝑑𝑦 = 1 2 0 2 𝑢2 + 1 𝑑𝑢 = 𝑢 2 𝑢2 + 1 + 12 2 𝑙𝑛 𝑢 + 𝑢2 + 1 2 0 Substitute values of u: 𝑦 1 2 𝑦 2 4𝑦2 + 1 + 1 2 𝑙𝑛 2𝑦 + 4𝑦2 + 1 1 0 = 1.479𝑦
- 31. Integration of theright equation is: 0 1 𝑦 4𝑦2 + 1 𝑑𝑦 , 𝑙𝑒𝑡 𝑢 = 4𝑦2 + 1, 𝑑𝑢 = 8𝑦𝑑𝑦, ⇒ 𝑑𝑢 8 = 𝑦𝑑𝑦 0 1 4𝑦2 + 1 𝑦𝑑𝑦 = 1 5 𝑢 1 8 𝑑𝑢 = 1 8 1 5 𝑢1 2 𝑑𝑢 = 1 12 𝑢3 2 5 1 Substitute values of u: 1 12 4𝑦2 + 1 3 2 1 0 = 0.8484 1.479𝑦 = 0.8484 𝒚 = 𝟎. 𝟓𝟕𝟒𝒎 𝐀𝐧𝐬.
- 32. #2. Determine thedistance from the base of a triangle of altitude ℎ to the centroid of its area. Solution: The x-axis is taken to coincide with the base 𝑏 of the triangle. The co-ordinate system is cartesian coordinates The differential stripe area is taken as 𝑑𝐴 = 𝑥𝑑𝑦
- 33. From similarity oftriangles: ℎ 𝑏 = (ℎ − 𝑦) 𝑥 ⇒ 𝑥 = 𝑏(ℎ − 𝑦) ℎ Thus, 𝐴𝑦 = 𝑦𝑐𝑑𝐴 , ⇒ 𝑏ℎ 2 𝑦 = 0 ℎ 𝑦𝑥𝑑𝑦 𝑏ℎ 2 𝑦 = 0 ℎ 𝑦 𝑏 ℎ (ℎ − 𝑦)𝑑𝑦
- 34. Integration with respectto 𝑦: ⇒ 𝒃𝒉 𝟐 𝒚 = 𝟎 𝒉 (𝒚𝒃 − 𝒃𝒚𝟐 𝒉 )𝒅𝒚 , ⇒ 𝒃𝒉 𝟐 𝒚 = 𝒃𝒚𝟐 𝟐 − 𝒃𝒚𝟑 𝟑𝒉 𝟎 𝒉 ⇒ 𝒃𝒉 𝟐 𝒚 = 𝒃𝒉𝟐 𝟐 − 𝒃𝒉𝟑 𝟑𝒉 ⇒ 𝒃𝒉 𝟐 𝒚 = 𝒃𝒉𝟐 𝟔 ⇒ 𝒚 = 𝒉 𝟑
- 35. #3. Locate thecentroid of the area of a circular sector with respect to its vertex. Solution: Choose the x-axis as the axis of symmetry. The co-ordinate system is polar coordinates Two methods of solution: I. Circular ring is taken as the differential area: Take the differential stripe area 𝑑𝐴 = 2𝑟𝑜𝛼𝑑𝑟𝑜
- 36. The centroid ofthe differential area is: 𝐱𝐜 = 𝐫𝐨𝐬𝐢𝐧𝛂 𝛂 Thus, 𝐀𝐱 = 𝐱𝐜𝐝𝐀 , ⇒ 𝟐𝛂 𝟐𝛑 𝛑𝐫𝟐 𝐱 = 𝟎 𝐫 𝐫𝐨𝐬𝐢𝐧𝛂 𝛂 𝟐𝐫𝐨𝛂𝐝𝐫𝐨 𝐫𝟐 𝛂𝐱 = 𝟐𝐬𝐢𝐧𝛂 𝟎 𝐫 𝐫𝐨 𝟐 𝐝𝐫𝐨 𝐫𝟐 𝛂𝐱 = 𝟐 𝟑 𝐫𝐨 𝟑 𝐬𝐢𝐧𝛂 𝒓 𝟎 𝐱 = 𝟐 𝟑 𝐫𝐬𝐢𝐧𝛂
- 37. II. Swinging triangleabout the vertex is taken as the differential area: Take the differential stripe area 𝑑𝐴 = 𝑟 2 𝑟𝑑𝜃 The centroid of the differential area is: 𝑥𝑐 = 2 3 𝑟𝑐𝑜𝑠𝜃 𝑨𝒙 = 𝒙𝒄𝒅𝑨 , ⇒ 𝟐𝜶 𝟐𝝅 𝝅𝒓𝟐 𝒙 = 𝟎 𝒓 2 3 𝑟𝑐𝑜𝑠𝜃 𝑟 2 𝑟𝑑𝜃 𝒓𝟐 𝜶𝒙 = 1 3 𝒓𝟑 −𝜶 𝜶 𝑐𝑜𝑠𝜃𝒅𝜽 𝒓𝟐 𝜶𝒙 = 𝟏 𝟑 𝒓𝟑 𝑠𝑖𝑛𝜃 𝜶 −𝜶 𝒙 = 2 3 𝑟 𝑠𝑖𝑛𝛼 𝜶 For a semicircular area, 𝟐𝜶 = 𝝅, 𝒙 = 4𝑟 3 𝝅
- 38. #4. Locate thecentroid of the area under the curve 𝑥 = 𝑘𝑦3 from 𝑥 = 0 to 𝑥 = 𝑎 Solution: Two methods of solution: I. A vertical differential element is taken as the differential area: Take the differential stripe area 𝑑𝐴 = 𝑦𝑑𝑥
- 39. The x-coordinate ofthe centroid is: 𝑨𝒙 = 𝒙𝒄𝒅𝑨 , ⇒ 𝒙 𝟎 𝒂 𝒚𝒅𝒙 = 𝟎 𝒂 𝑥𝑦𝑑𝑥 𝒚 = 𝒙 𝒌 𝟏 𝟑 , 𝒌 = 𝒂 𝒃𝟑 , ⇒ 𝒚 = 𝒃 𝒙 𝒂 𝟏 𝟑 𝒙 𝟎 𝒂 𝒃 𝒙 𝒂 𝟏 𝟑 𝒅𝒙 = 𝟎 𝒂 𝒃 𝒙 𝒂 𝟏 𝟑 𝒙𝒅𝒙 𝒙 𝟑 𝟒 𝒃 𝒂𝟏 𝟑 𝒙𝟒 𝟑 𝒂 𝟎 = 𝟑 𝟕 𝒃 𝒂𝟏 𝟑 𝒙𝟕 𝟑 𝒂 𝟎 𝒙 = 𝟒 𝟕 𝒂
- 40. The y-coordinate ofthe centroid is: 𝑨𝒚 = 𝒚𝒄𝒅𝑨 , ⇒ 𝒚 𝟎 𝒂 𝒚𝒅𝒙 = 𝟎 𝒂 𝒚 𝟐 𝑦𝑑𝑥 𝒚 = 𝒙 𝒌 𝟏 𝟑 , 𝒌 = 𝒂 𝒃𝟑 , ⇒ 𝒚 = 𝒃 𝒙 𝒂 𝟏 𝟑 𝒚 𝟎 𝒂 𝒃 𝒙 𝒂 𝟏 𝟑 𝒅𝒙 = 𝟏 𝟐 𝟎 𝒂 𝒃 𝒙 𝒂 𝟏 𝟑 𝟐 𝒅𝒙 𝒚 𝟑 𝟒 𝒃 𝒂𝟏 𝟑 𝒙𝟒 𝟑 𝒂 𝟎 = 𝟑 𝟏𝟎 𝒃𝟐 𝒂𝟐 𝟑 𝒙𝟓 𝟑 𝒂 𝟎 𝒚 = 𝟐 𝟓 𝒃
- 41. II. When ahorizontal differential element is taken as the differential area: Take the differential stripe area 𝑑𝐴 = 𝑎 − 𝑥 𝑑𝑦 : 𝐱𝐜 = 𝟏 𝟐 𝐱 + 𝐚 𝐱 = 𝐤𝐲𝟑 , 𝐤 = 𝐚 𝐛𝟑 , ⇒ 𝐱 = 𝐚 𝐲 𝐛 𝟑 𝐀𝐱 = 𝐱𝐜𝐝𝐀 ⇒ 𝐱 𝟎 𝐛 𝐚 − 𝐱 𝐝𝐲 = 𝟎 𝐛 𝟏 𝟐 𝐚 + 𝐱 𝐚 − 𝐱 d𝐲 𝐱 𝟎 𝐛 𝐚 − 𝐱 𝐝𝐲 = 𝟎 𝐛 𝟏 𝟐 𝐚𝟐 − 𝒙𝟐 d𝐲 𝒂𝒚 𝒃 𝟎 − 𝟏 𝟒 𝐚 𝐛𝟑 𝐲𝟒 𝐛 𝟎 𝐱 = 𝟏 𝟐 𝒂𝟐 𝒚 𝒃 𝟎 − 𝐚𝟐 𝟕𝐛𝟔 𝒚𝟕 𝐛 𝟎 𝟑 𝟒 𝐚𝐛 𝐱 = 𝟑𝐚𝟐 𝐛 𝟕 , 𝐱 = 4 7 a
- 42. To determine thecentroid of y-co-ordinate: Take the differential stripe area 𝑑𝐴 = 𝑎 − 𝑥 𝑑𝑦 : 𝒚𝐜 = 𝒚 𝐱 = 𝐤𝐲𝟑 , 𝐤 = 𝐚 𝐛𝟑 , ⇒ 𝐱 = 𝐚 𝐲 𝐛 𝟑 𝐀𝒚 = 𝒚𝐜𝐝𝐀 , ⇒ 𝒚 𝟎 𝐛 𝐚 − 𝐱 𝐝𝐲 = 𝟎 𝐛 𝒚 𝐚 − 𝐱 d𝐲 𝒚 𝟎 𝐛 𝐚 − 𝐱 𝐝𝐲 = 𝟎 𝐛 𝐚𝐲 − 𝒙𝒚 d𝐲 𝒂𝒚 𝒃 𝟎 − 𝟏 𝟒 𝐚 𝐛𝟑 𝐲𝟒 𝐛 𝟎 𝐱 = 𝟏 𝟐 𝒂𝒚𝟐 𝒃 𝟎 − 𝒂 𝟓𝐛𝟑 𝒚𝟓 𝐛 𝟎 𝟑 𝟒 𝐚𝐛 𝐱 = 𝟑𝒂𝒃𝟐 𝟏𝟎 , 𝐱 = 𝟐 𝟓 𝒃
- 43. #5. Locate thecentroid of the volume of a hemisphere of radius r with respect to its base. Solution: Two methods of solution: I. A circular slice of thickness 𝑑𝑦 parallel to the x-z plane is taken as the differential volume: Take the differential slice element volume 𝑑𝑉 = 𝜋𝑧2 𝑑𝑦, ⇒ 𝑑𝑉 = 𝜋 𝑟2 − 𝑦2 𝑑𝑦 The axis of symmetry lays on the y-axis.⇒ 𝒙 = 𝟎, 𝒛 = 𝟎
- 44. The centroid ofthe hemisphere can be determined as: 𝑽𝒚 = 𝒚𝒄𝒅𝑽 ⇒ 𝒚 𝟎 𝒓 𝝅 𝒓𝟐 − 𝒚𝟐 𝒅𝒚 = 𝟎 𝒓 𝒚𝝅 𝒓𝟐 − 𝒚𝟐 𝒅𝒚 𝝅𝒚 𝟎 𝒚 𝒓𝟐 𝒅𝒚 − 𝟎 𝒚 𝒚𝟐 𝒅𝒚 = 𝝅 𝟎 𝒓 𝒓𝟐 𝒚𝒅𝒚 − 𝟎 𝒓 𝒚𝟑 𝒅𝒚 𝒚𝝅 𝒓𝟐𝒚 − 𝟏 𝟑 𝒚𝟑 𝒓 𝟎 = 𝝅 𝟏 𝟐 𝒓𝟐𝒚𝟐 − 𝟏 𝟒 𝒚𝟒 𝒓 𝟎 𝟐 𝟑 𝒓𝟑𝝅𝒚 = 𝟏 𝟒 𝝅𝒓𝟒 𝒚 = 𝟑 𝟖 𝒓
- 45. II. A differentialelement of cylindrical shell of length y, radius z, & thickness 𝑑𝑧 is used 𝑑𝑉 = 2𝜋𝑧𝑑𝑧 𝑦, 𝑦 = 𝐫𝟐 − 𝐳𝟐 𝐕𝐲 = 𝐲𝐜𝐝𝐕 ⇒ 𝟐𝛑𝐲 𝟎 𝐫 𝐫𝟐 − 𝐳𝟐 𝐳𝐝𝐳 = 𝟐𝛑 𝟎 𝐫 𝐫𝟐 − 𝐳𝟐 𝟐 𝐫𝟐 − 𝐳𝟐 𝐳𝐝𝐳 𝟐𝛑𝐲 𝟎 𝐫 𝐫𝟐 − 𝐳𝟐 𝐳𝐝𝐳 = 𝛑 𝟎 𝐫 𝐫𝟐 𝐳𝐝𝐳 − 𝟎 𝐫 𝐳𝟑 𝐝𝐳
- 46. For the leftside integration equation using integration by substitution: Let ,𝒖 = 𝐫𝟐 − 𝐳𝟐 , 𝒅𝒖 = −𝟐𝒛𝒅𝒛, ⇒ 𝒛𝒅𝒛 = 𝒅𝒖 𝟐 𝛑𝐲 𝟎 𝐫 𝒖 𝟏 𝟐 𝐝𝒖 = 𝛑 𝟎 𝐫 𝐫𝟐 𝐳𝐝𝐳 − 𝟎 𝐫 𝐳𝟑 𝐝𝐳 𝟐 𝟑 𝛑𝐲 𝒖𝟑 𝟐 𝒓 𝟎 = 𝛑 𝟏 𝟐 𝐫𝟐 𝐳𝟐 − 𝟏 𝟒 𝐳𝟒 𝐫 𝟎 Substitute, 𝒖 = 𝐫𝟐 − 𝐳𝟐 𝟐 𝟑 𝛑 𝐲 𝐫𝟐 − 𝐳𝟐 𝟑 𝟐 𝐫 𝟎 = 𝛑 𝟏 𝟐 𝐫𝟐 𝐳𝟐 − 𝟏 𝟒 𝐳𝟒 𝐫 𝟎 𝟐 𝟑 𝐫𝟑 𝛑𝐲 = 𝟏 𝟒 𝛑𝐫𝟒 𝐲 = 𝟑 𝟖 𝐫
- 47. III. Using theangle 𝜃 as a variable with limits of 0 to 𝜋 2. For the a circular slice differential element: 𝑑𝑉 = 𝜋𝑧2 𝑑𝑦, 𝑑𝑦 = 𝑟𝑑𝜃 𝑠𝑖𝑛𝜃, 𝑧 = 𝑟𝑠𝑖𝑛𝜃 ⇒ 𝑑𝑉 = 𝜋𝑟3 𝑠𝑖𝑛3 𝜃𝑑𝜃 𝐕𝐲 = 𝐲𝐜𝐝𝐕 For the a circular slice differential element: 𝑑𝑉 = 2𝜋𝑧𝑑𝑧 𝑦, 𝑑𝑧 = 𝑟𝑑𝜃 𝑐𝑜𝑠𝜃, 𝑦 = 𝑟𝑐𝑜𝑠𝜃 ⇒ 𝑑𝑉 = 2𝜋𝑟3 𝑐𝑜𝑠2 𝜃𝑠𝑖𝑛𝜃𝑑𝜃 𝐕𝐲 = 𝐲𝐜𝐝𝐕
- 48. #6. A uniformsemicircular rod of radius r is supported in a bearing at its upper end and is free to swing in the vertical plane. Calculate the angle made by the diameter with the vertical for the equilibrium position. Solution: The centroid of the semicircular rod is located at 2𝑟 𝜋.
- 49. The angle 𝜃from the vertical can be calculated as: 𝑡𝑎𝑛𝜃 = 2𝑟 𝜋 𝑟 ⇒ 𝜃 = 𝑡𝑎𝑛−1 2 𝜋 = 32.48°
- 50. #7. Locate thecenter of gravity x of the homogeneous rod. If the rod has a weight per unit length of 100 N/m, determine the vertical reaction at A and the x and y components of reaction at the pin B.. Solution: The co-ordinate system is cartesian coordinates. Take the differential element length of the arc is 𝐝𝐋 = 𝟏 + 𝒅𝒚 𝒅𝒙 𝟐 𝒅𝒙.
- 51. From the givenequation 𝑥 is evaluated as: 𝑥 = 𝑦2 ⇒ 𝑑𝑦 𝑑𝑥 = 2𝑥 ∴ 𝑑𝐿 = 1 + 2𝑥 2 𝑑𝑥 𝑥𝑑𝐿 = 𝑥𝑐 𝑑𝐿 , ⇒ 𝑥 1 + 2𝑥 2 𝑑𝑥 = 𝑥 1 + 2𝑥 2 𝑑𝑥 𝑥 1 + 4𝑥2 𝑑𝑥 = 𝑥 1 + 4𝑥2 𝑑𝑥
- 52. Integration of theleft equation is: 𝑥 1 + 4𝑥2 𝑑𝑥 1 + 4𝑥2 𝑑𝑥 , 𝑙𝑒𝑡 𝑢 = 2𝑥, 𝑢 2 = 𝑥, 𝑑𝑢 = 2𝑑𝑥, ⇒ 𝑑𝑢 2 = 𝑑𝑥 0 1 1 + 4𝑥2𝑑𝑥 = 1 2 0 2 𝑢2 + 1 𝑑𝑢 = 𝑢 2 𝑢2 + 1 + 12 2 𝑙𝑛 𝑢 + 𝑢2 + 1 2 0 Substitute values of u: 𝑥 1 2 𝑥 2 1 + 4𝑥2 + 1 2 𝑙𝑛 2𝑥 + 1 + 4𝑥2 1 0 = 1.479𝑥
- 53. Integration of theright equation is: 0 1 𝑥 1 + 4𝑥2 𝑑𝑥 , 𝑙𝑒𝑡 𝑢 = 4𝑥2 + 1, 𝑑𝑢 = 8𝑥𝑑𝑥, ⇒ 𝑑𝑢 8 = 𝑥𝑑𝑥 0 1 1 + 4𝑥2 𝑥𝑑𝑥 = 1 5 𝑢 1 8 𝑑𝑢 = 1 8 1 5 𝑢1 2 𝑑𝑢 = 1 12 𝑢3 2 5 1 Substitute values of u: 1 12 4𝑥2 + 1 3 2 1 0 = 0.8484 1.479𝑥 = 0.8484 𝒙 = 𝟎. 𝟓𝟕𝟒𝒎 𝐀𝐧𝐬.
- 54. Total length ofthe rod is: 𝐝𝐋 = 𝟏 + 𝒅𝒚 𝒅𝒙 𝟐 𝒅𝒙 𝐋 = 1.479𝐦 Concentrated weight of the rod at the centroid is: 𝑊 = 100 𝑁 𝑚 ∗ 1.479𝑚 = 147.9𝑁 The reaction force at A can be calculated as: +⤹ ∑𝑀𝐵 = 0 0.574 ∗ 147.9 − 1 ∗ 𝐴𝑦 = 0 𝑨𝒚 = 𝟖𝟒. 𝟖𝟑𝟓𝑵 𝑨𝒏𝒔.
- 55. #8. Determine thelocation (x, y) of the centroid of the wire. Solution: The y-axis is chosen as axis of symmetry. ∴ 𝑥 = 0 The co-ordinate system is cartesian coordinates. Take the differential element length of the arc is 𝐝𝐋 = 𝟏 + 𝒅𝒚 𝒅𝒙 𝟐 𝒅𝒙.
- 56. From the givenequation 𝑥 is evaluated as: 𝑦 = 𝑥2 ⇒ 𝑑𝑦 𝑑𝑥 = 2𝑥 ∴ 𝑑𝐿 = 1 + 2𝑥 2 𝑑𝑥 𝑦𝑑𝐿 = 𝑦𝑐 𝑑𝐿 , ⇒ 𝑦 1 + 2𝑥 2 𝑑𝑥 = 𝑦 1 + 2𝑥 2 𝑑𝑥 𝑦 1 + 4𝑥2 𝑑𝑥 = 𝑥2 1 + 4𝑥2 𝑑𝑥
- 57. Integration of theleft equation is: 𝑥 1 + 4𝑥2 𝑑𝑥 1 + 4𝑥2 𝑑𝑥 , 𝑙𝑒𝑡 𝑢 = 2𝑥, 𝑢 2 = 𝑥, 𝑑𝑢 = 2𝑑𝑥, ⇒ 𝑑𝑢 2 = 𝑑𝑥 0 2 1 + 4𝑥2𝑑𝑥 = 1 2 0 4 𝑢2 + 1 𝑑𝑢 = 𝑢 2 𝑢2 + 1 + 12 2 𝑙𝑛 𝑢 + 𝑢2 + 1 4 0 Substitute values of u: 𝑦 1 2 𝑥 2 1 + 4𝑥2 + 1 2 𝑙𝑛 2𝑥 + 1 + 4𝑥2 2 0 = 2.585𝑦
- 58. Integration of theright equation is: 0 1 𝑦2 4𝑦2 + 1 𝑑𝑦 , 𝑙𝑒𝑡 𝑢 = 2𝑦, 𝑢 2 = 𝑦, 𝑑𝑢 = 2𝑑𝑦, ⇒ 𝑑𝑢 2 = 𝑑𝑦 0 1 𝑦2 4𝑦2 + 1 𝑑𝑦 = 0 2 𝑢 2 2 𝑢2 + 1 1 2 𝑑𝑢 = 1 8 0 2 𝑢2 𝑢2 + 1 𝑑𝑢 = 1 8 𝑢 8 2𝑢2 + 12 𝑢2 + 1 + 12 8 𝑙𝑛 𝑢 + 𝑢2 + 1 2 0 Substitute values of u: 𝑦 32 8𝑦2 + 1 4𝑦2 + 1 + 1 64 𝑙𝑛 2𝑦 + 4𝑦2 + 1 1 0 = 0.606 1.479𝑥 = 0.606 𝒙 = 𝟎. 𝟒𝟏𝒎 𝐀𝐧𝐬.
- 59. Integration of theright equation is: 0 2 𝑥2 1 + 4𝑥2 𝑑𝑥 , 𝑙𝑒𝑡 𝑢 = 2𝑥, 𝑢 2 = 𝑥, 𝑑𝑢 = 2𝑑𝑥, ⇒ 𝑑𝑢 2 = 𝑑𝑥 0 2 𝑥2 1 + 4𝑥2 𝑑𝑥 = 0 4 𝑢 2 2 𝑢2 + 1 1 2 𝑑𝑢 = 1 8 0 4 𝑢2 𝑢2 + 1 𝑑𝑢 = 1 8 𝑢 8 2𝑢2 + 12 𝑢2 + 1 + 12 8 𝑙𝑛 𝑢 + 𝑢2 + 1 4 0 Substitute values of u: 𝑥 32 8𝑥2 + 1 4𝑥2 + 1 + 1 64 𝑙𝑛 2𝑥 + 4𝑥2 + 1 2 0 = 0.606 2.585𝑦 = 9.052 𝒚 = 𝟑. 𝟓𝟎𝟐𝐟𝐭 … … 𝐀𝐧𝐬.
- 60. #9. Determine they-coordinate of the centroid of the shaded area. Check your result for the special case 𝑎 = 0. Solution: The y-axis is chosen as axis of symmetry. ∴ 𝑥 = 0 The co-ordinate system is cartesian coordinates. Take the differential element area of the triangle is 𝐝𝐀 = 𝟐𝒙𝒅𝒚.
- 61. From similarity oftriangle: 2𝑥 𝑦 = 2ℎ 𝑡𝑎𝑛30 ℎ , 2ℎ𝑥 = 2ℎ𝑦 𝑡𝑎𝑛30 2𝑥 = 2𝑦 𝑡𝑎𝑛30
- 62. From the givenequation 𝑦 is evaluated as: 2𝑥 = 2𝑦 𝑡𝑎𝑛30 ∴ 𝑑𝐴 = 2𝑦𝑡𝑎𝑛30 𝑑𝑦 𝑦𝑑𝐴 = 𝑦𝑐 𝑑𝐴 , ⇒ 𝑦 2𝑦𝑡𝑎𝑛30 𝑑𝑦 = 𝑦 2𝑦𝑡𝑎𝑛30 𝑑𝑦 = 𝑦2𝑡𝑎𝑛30 𝑎 ℎ 𝑦𝑑𝑦 = 2𝑡𝑎𝑛30 𝑎 ℎ 𝑦2 𝑑𝑦 𝑡𝑎𝑛30𝑦 𝑦2 ℎ 𝑎 = 2 3 𝑡𝑎𝑛30𝑦 𝑦3 ℎ 𝑎 ℎ2 − 𝑎2 𝑡𝑎𝑛30𝑦 = 2 3 ℎ3 − 𝑎3 𝑡𝑎𝑛30 𝑦 = 2 3 ℎ3 − 𝑎3 ℎ2 − 𝑎2 , 𝑤ℎ𝑒𝑛 𝑎 = 0, 𝑦 = 2 3 ℎ
- 63. #10. Determine thex- and y- coordinates of the centroid of the shaded area. Solution: The co-ordinate system is cartesian coordinates. Take the differential element area of the triangle is 𝐝𝐀 = 𝒚𝟐 − 𝒚𝟏 𝒅𝒙.
- 64. Centroid of theshaded area can be evaluated as: 𝐀𝒙 = 𝒙𝐜𝐝𝑨 , ⇒ 𝒙 𝑦2 − 𝑦1 𝑑𝑥 = 𝒙 𝑦2 − 𝑦1 𝑑𝑥 From the given equation: 𝑦1 = 𝑎𝑥 1 2 , 𝑦2 = 𝑥3 𝑎2
- 65. From the givenequation 𝑥 is evaluated as: 𝒙 0 𝑎 𝑎𝑥 1 2 − 𝑥3 𝑎2 𝑑𝑥 = 0 𝑎 𝑥 𝑎𝑥 1 2 − 𝑥3 𝑎2 𝑑𝑥 𝑥 2 3 𝑎1 2 𝑥3 2 − 1 4 𝑥4 𝑎2 𝑎 0 = 2 5 𝑎1 2 𝑥5 2 − 1 5 𝑥5 𝑎2 𝑎 0 5 12 𝑎2 𝑥 = 1 5 𝑎3 𝒙 = 𝟏𝟐 𝟐𝟓 𝒂 … … 𝑨𝒏𝒔.
- 66. From the givenequation 𝑦 is evaluated as: 𝐀𝒚 = 𝒚𝒄𝐝𝑨 , ⇒ 𝒚 𝑦2 − 𝑦1 𝑑𝑥 = 𝒚 𝑦2 − 𝑦1 𝑑𝑥, 𝒚𝒄 = 𝒚 = 𝟏 𝟐 𝒚𝟏 + 𝒚𝟐 𝒚 0 𝑎 𝑎𝑥 1 2 − 𝑥3 𝑎2 𝑑𝑥 = 1 2 0 𝑎 𝑎𝑥 1 2 + 𝑥3 𝑎2 𝑎𝑥 1 2 − 𝑥3 𝑎2 𝑑𝑥 = 𝒚 0 𝑎 𝑎𝑥 1 2 − 𝑥3 𝑎2 𝑑𝑥 = 1 2 0 𝑎 𝑎𝑥 − 𝑥6 𝑎4 𝑑𝑥 = 𝑦 2 3 𝑎1 2 𝑥3 2 − 1 4 𝑥4 𝑎2 𝑎 0 = 1 2 𝑎𝑥2 − 1 7 𝑥7 𝑎4 𝑎 0 5 12 𝑎2 𝑦 = 3 7 𝑎3 , ⇒ 𝒚 = 𝟑𝟔 𝟑𝟓 𝒂 … … 𝑨𝒏𝒔.
- 67. #11. The thicknessof the triangular plate varies linearly with y from a value 𝑡𝑜 along its base to 𝑦 = 0 to 2𝑡𝑜 at 𝑦 = ℎ. Determine the y coordinate of the center of mass of the plate. Solution: The co-ordinate system is cartesian coordinates. Take the differential element area of the triangle is 𝐝𝒎 = 𝝆𝒅𝑽.
- 68. Centroid of theshaded area can be evaluated as: 𝑑𝑚 = 𝜌𝑑𝑣, ⇒ 𝜌𝑑𝑣 = 𝑡𝜌𝑑𝐴, ⇒ 𝜌𝑑𝑣 = 𝜌𝑡 𝑏 − 𝑥 𝑑𝑦 V𝑦 = 𝑦cd𝑉 , ⇒ 𝑦 𝜌𝑡 𝑏 − 𝑥 𝑑𝑦 = 𝑦 𝜌𝑡 𝑏 − 𝑥 𝑑𝑦 From similarity of triangle: 𝑦 𝑥 = ℎ 𝑏 , 𝑥 = 𝑏𝑦 ℎ , 𝑡𝑜 ℎ = 𝑧 𝑦 , ⇒ 𝑧 = 𝑡𝑜𝑦 ℎ , ∴ 𝑡 = 𝑡𝑜 + 𝑡𝑜𝑦 ℎ
- 69. y coordinate ofthe center of mass of the plate: ⇒ 𝑦 𝑡𝑜 + 𝑡𝑜𝑦 ℎ 𝑏 − 𝑏𝑦 ℎ 𝑑𝑦 = 𝑦 𝑡𝑜 + 𝑡𝑜𝑦 ℎ 𝑏 − 𝑏𝑦 ℎ 𝑑𝑦 𝑦 0 ℎ 𝑡𝑜 + 𝑡𝑜𝑦 ℎ 𝑏 − 𝑏𝑦 ℎ 𝑑𝑦 = 0 ℎ 𝑦 𝑡𝑜 + 𝑡𝑜𝑦 ℎ 𝑏 − 𝑏𝑦 ℎ 𝑑𝑦 Integrating the left portion of the equation: 𝑦 0 ℎ 𝑡𝑜 + 𝑡𝑜𝑦 ℎ 𝑏 − 𝑏𝑦 ℎ 𝑑𝑦 = 𝑦 0 ℎ 𝑡𝑜𝑏𝑑𝑦 − 1 ℎ2 0 ℎ 𝑡𝑜𝑏𝑦2 𝑑𝑦 𝑦 𝑡𝑜𝑏𝑦 − 𝑡𝑜𝑏𝑦3 3ℎ2 ℎ 0 = 𝑦 2 3 𝑡𝑜𝑏ℎ
- 70. Integrating the rightportion of the equation: 𝑦 0 ℎ 𝑡𝑜 + 𝑡𝑜𝑦 ℎ 𝑏 − 𝑏𝑦 ℎ 𝑑𝑦 = 0 ℎ 𝑦 𝑡𝑜 + 𝑡𝑜𝑦 ℎ 𝑏 − 𝑏𝑦 ℎ 𝑑𝑦 0 ℎ 𝑡𝑜𝑏𝑦 − 1 ℎ2 0 ℎ 𝑡𝑜𝑏𝑦3 1 2 𝑡𝑜𝑏𝑦2 − 1 4 𝑡𝑜𝑏𝑦4 ℎ2 ℎ 0 = 1 4 𝑡𝑜𝑏ℎ2 Values of 𝑦: 𝑦 2 3 𝑡𝑜𝑏ℎ = 1 4 𝑡𝑜𝑏ℎ2 𝒚 = 𝟏 𝟒 𝐡 … … 𝐀𝐧𝐬.
- 71. #11. Locate themass center of the homogeneous solid body whose volume is determined by revolving the shaded area through 360𝑜 about the z-axis. Solution: The co-ordinate system is cartesian coordinates. Take the differential element area of the triangle is 𝐝𝑽 = 𝝅𝒓𝟐 𝒅𝒛.
- 72. Centroid of theshaded area can be evaluated as: 𝑑𝑉 = 𝜋𝑟2 𝑑𝑧, ⇒ 𝑑𝑉 = 𝑅2 𝑏6 𝑧6 𝜋𝑑𝑧, ⇒ 𝑟 = 𝑅 𝑏3 𝑧3 V𝑧 = 𝑧cd𝑉 , ⇒ 𝑧 𝑅2 𝑏6 𝑧6𝜋𝑑𝑧 = 𝑧 𝑅2 𝑏6 𝑧6𝜋𝑑𝑧
- 73. Centroid of theshaded area can be evaluated as: 𝑧 0 300 𝑅2 𝑏6 𝑧6 𝜋𝑑𝑧 = 0 300 𝑧 𝑅2 𝑏6 𝑧6 𝜋𝑑𝑧 𝑧 1 7 𝜋 𝑅2 𝑏3 𝑏7 300 0 = 1 8 𝜋 𝑅2 𝑏3 𝑏8 300 0 , 𝑤ℎ𝑒𝑟𝑒 𝑅 = 200 & 𝑏 = 300 1 7 3.24 ∗ 1012𝜋 𝑧 = 1 8 9.72 ∗ 1014𝜋 𝒛 = 𝟐𝟔𝟐. 𝟓𝒎𝒎 … … 𝑨𝒏𝒔.
- 74. #1. Determine thedistance 𝑥, 𝑦 to the center of gravity of the homogeneous rod.
- 75. #2. Locate thecentroid 𝑥 of the parabolic area.
- 76. #3. Determine thex-coordinate of the mass center of the portion of the spherical shell of uniform but small thickness.
- 77. #4. Locate thecenter of gravity 𝑧 of the solid.
- 78. A composite bodyconsists of a series of connected “simpler” shaped bodies, which may be rectangular, triangular, semicircular, etc. The body can be sectioned or divided into its composite parts. The weight and location of the center of gravity of each of these parts are known. This avoids the need for integration to determine the center of gravity for the entire body. Composite body 𝑥 = ∑𝑚𝑥𝑐 ∑𝑚 𝑦 = ∑𝑚𝑦𝑐 ∑𝑚 𝑧 = ∑𝑚𝑧𝑐 ∑𝑚 𝑚1 + 𝑚2 + 𝑚3 𝑥 = 𝑚1𝑥1 + 𝑚2𝑥2 + 𝑚3𝑥3
- 79. In practice theboundaries of an area or volume might not be expressible in terms of simple geometrical shapes as shapes which can be represented mathematically. For such cases method of approximation is used to determine centroid of a body. 𝑚1 + 𝑚2 + 𝑚3 𝑥 = 𝑚1𝑥1 + 𝑚2𝑥2 + 𝑚3𝑥3
- 80. Example: #1. CentroidC of the irregular area Example: #2. The centroid C of an irregular volume 𝑥 = ∑𝐴𝑥𝑐 ∑𝐴 𝑥 = ∑𝐴𝑥𝑐 ∑𝐴 𝑥 = ∑ 𝐴∆𝑥 𝑥𝑐 ∑ 𝐴∆𝑥 𝑥 = ∑𝑉𝑥𝑐 ∑𝑉
- 81. To determine locationof the center of gravity or the centroid of a composite geometrical object represented by a line, area, or volume: Using a sketch, divide the body or object into a finite number of composite parts of simpler shapes. Composite body with a hole is considered as body without the hole and the hole as an additional composite part having negative weight or size. Establish the coordinate axes on the sketch and determine the coordinates 𝑥𝑐, 𝑦𝑐 & 𝑧𝑐 of the center of gravity or centroid of each part. Determine 𝑥, 𝑦 & 𝑧 by applying the center of gravity equations. If an object is symmetrical about an axis, the centroid of the object lies on this axis.
- 82. #1. Locate thecentroid of the wire shown in Figure below. Solution: The wire is divided into three segments as shown
- 83. Tabulated values. Thus, Segment 𝐿𝑚𝑚 𝑥𝑐 𝑚𝑚 𝑦𝑐 𝑚𝑚 𝑧𝑐 𝑚𝑚 𝐿𝑥𝑐 𝑚𝑚2 𝐿𝑦𝑐 𝑚𝑚2 𝐿𝑧𝑐 𝑚𝑚2 1 𝜋𝑟 = 188.5 60 2𝑟 𝜋 = −38.2 0 11310 −7200.7 0 2 40 0 20 0 0 800 0 3 20 0 40 −10 0 800 −200 ∑ 𝟐𝟒𝟑. 𝟓 ∑ 𝟏𝟏𝟑𝟏𝟎 −𝟓𝟔𝟎𝟎. 𝟕 −𝟐𝟎𝟎 𝑥 = ∑𝐴𝑥𝑐 ∑𝐴 , ⇒ 𝑥 = 11310 243.5 , ∴ 𝑥 = 46.45𝑚𝑚 𝑦 = ∑𝐴𝑦𝑐 ∑𝐴 , ⇒ 𝑦 = −5600.7 243.5 , ∴ 𝑦 = −23𝑚𝑚 𝑧 = ∑𝐴𝑧𝑐 ∑𝐴 , ⇒ 𝑧 = −200 243.5 , ∴ 𝑧 = −0.82𝑚𝑚
- 84. #2. The weldedassembly is made of a uniform rod weighing 0.370 𝑙𝑏 𝑓𝑡 of length and the semicircular plate weighing 8 𝑙𝑏 𝑓𝑡2. Calculate the coordinates of the center of gravity of the assembly. Solution: The object is divided into four parts.
- 85. Solution: The object isdivided into four parts.
- 86. Tabulated values. Thus, Part 𝐿𝑓𝑡 𝐴 𝑓𝑡2 𝑊 𝑙𝑏 𝑥𝑐 𝑓𝑡 𝑧𝑐 𝑓𝑡 𝑊𝑥𝑐 𝑓𝑡2 𝑊𝑧𝑐 𝑓𝑡2 1 − 𝜋𝑟2 2 = 0.174 1.392 0 4𝑟 3𝜋 = 0.141 0 0.196 2 0.5 − 0.185 0.25 0 0.04625 0 3 𝜋𝑟 = 1.05 − 0.388 0.5 2𝑟 𝜋 = 0.212 0.194 0.082 4 0.5 − 0.185 0.25 0 0.04625 0 ∑ 2.15 ∑ 𝟎. 𝟐𝟖𝟔𝟓 𝟎. 𝟐𝟕𝟖 𝑥 = ∑𝑊𝑥𝑐 ∑𝑊 , ⇒ 𝑥 = 0.2865 2.15 , ∴ 𝑥 = 1.60𝑖𝑛 𝑧 = ∑𝑊𝑧𝑐 ∑𝑊 , ⇒ 𝑧 = 0.278 2.15 , ∴ 𝑧 = 1.55𝑖𝑛
- 87. #3. Locate thecentroid of the shaded area. Solution: The object is divided into four parts.
- 88. Tabulated values. Thus, Part 𝐴𝑖𝑛2 𝑥𝑐 𝑖𝑛 𝑦𝑐 𝑓𝑡 𝐴𝑥𝑐 𝑖𝑛3 𝐴𝑦𝑐 𝑖𝑛2 1 120 6 5 720 600 2 30 14 3.33 420 100 3 −14.14 6 1.27 −84.84 −18 4 −8 12 4 −96 −32 ∑ 127.86 2.15 ∑ 𝟗𝟓𝟔. 𝟏𝟔 𝟔𝟓𝟎 𝑥 = ∑𝐴𝑥𝑐 ∑𝐴 , ⇒ 𝑥 = 956.16 127.86 , ∴ 𝑥 = 7.478𝑖𝑛 𝑦 = ∑𝐴𝑧𝑐 ∑𝐴 , ⇒ 𝑦 = 650 127.86 , ∴ 𝑦 = 5.083𝑖𝑛
- 89. #4. Determine thecoordinates of the mass center of the body constructed of a uniform square plate, a uniform straight rod, a uniform quarter-circular rod, and a particle (negligible dimensions). The mass of each component is given in the figure.
- 90. The object isdivided into four parts.
- 91. Tabulated values. Thus, Part 𝑚𝑥𝑐 𝑦𝑐 𝑧𝑐 𝑚𝑥𝑐 𝑚𝑦𝑐 𝑚𝑧𝑐 1 0.3𝑚 0 0.5𝑏 0.5𝑏 0 0.15𝑏𝑚 0.15𝑏𝑚 2 0.15𝑚 0.637𝑏 0 0.637𝑏 0.096𝑏𝑚 0 0.096𝑏𝑚 3 0.25𝑚 𝑏 0 0 0.25𝑏𝑚 0 0 4 0.1𝑚 0.5𝑏 0 0 0.05𝑏𝑚 0 0 ∑ 𝟎. 𝟖𝒎 ∑ 𝟎. 𝟑𝟗𝟔𝒃𝒎 𝟎. 𝟏𝟓𝒃𝒎 𝟎. 𝟐𝟒𝟔𝒃𝒎 𝑥 = ∑𝑚𝑥𝑐 ∑𝑚 , ⇒ 𝑥 = 0.396𝑏𝑚 0.8𝑚 , ∴ 𝑥 = 0.495𝑏 𝑦 = ∑𝑚𝑦𝑐 ∑𝑚 , ⇒ 𝑦 = 0.15𝑏𝑚 0.8𝑚 , ∴ 𝑦 = 0.188𝑏 𝑧 = ∑𝑚𝑧𝑐 ∑𝑚 , ⇒ 𝑧 = 0.246𝑏𝑚 0.8𝑚 , ∴ 𝑧 = 0.308𝑏
- 92. #5. Locate thecenter of mass of the bracket-and-shaft combination. The vertical face is made from sheet metal which has a mass of 25 𝑘𝑔 𝑚2 . The material of the horizontal base has a mass of 40 𝑘𝑔 𝑚2 , and the steel shaft has a density of 7.83 𝑀𝑔 𝑚3 .
- 93. The object isdivided into five parts. The object is symmetrical with respect to x-axis ∴ 𝑥 = 0 Part s Area 𝒎𝟐 Volume 𝒎𝟑 M 𝒌𝒈 𝒎𝟐 𝝆 𝒌𝒈 𝒎𝟐 𝒎𝑨 𝒈𝒎 𝝆𝑽 𝒈𝒎 1 0.004 − 25 − 100 − 2 0.023 − 25 − 575 − 3 −0.0038 − 25 − −95 − 4 0.015 − 40 − 600 − 5 − 0.0002 − 7.83 1566
- 94. Tabulated values. Thus, Part 𝑚𝐴𝑔𝑚 𝑦𝑐 𝑚𝑚 𝑧𝑐 𝑚𝑚 𝑚𝑦𝑐 𝑔𝑚. 𝑚𝑚 𝑚𝑧𝑐 𝑔𝑚. 𝑚𝑚 1 100 0 21000 0 21000 2 575 0 −75 0 −43125 3 −95 0 −100 0 9500 4 600 0.05 −150 30000 −90000 5 1566 75 0 117450 0 ∑ 2746 ∑ 𝟏𝟒𝟕𝟒𝟓𝟎 −𝟏𝟐𝟏𝟔𝟓 𝑦 = ∑𝑚𝑦𝑐 ∑𝑚 , ⇒ 𝑦 = 147450 2746 , ∴ 𝑦 = 53.7𝑚𝑚 𝑧 = ∑𝑚𝑧𝑐 ∑𝑚 , ⇒ 𝑧 = −121650 2746 , ∴ 𝑧 = 44.3𝑚𝑚
- 95. #6. Determine thedistance 𝑧 to the centroid of the shape that consists of a cone with a hole of height ℎ = 50𝑚𝑚 bored into its base.
- 96. The object isdivided into two parts. The object is symmetrical with respect to x- and y-axis ∴ 𝑥 = 𝑦 = 0
- 97. Tabulated values. Thus, Therefore, thedistance measured from centroid of cylinder to the centroid of the shape is: 184.24𝑚𝑚 − 50𝑚𝑚 = 134.24𝑚𝑚 Part 𝑧𝑐 𝑚 𝑉 𝑚𝑚3 𝑉𝑧𝑐 𝑚4 1 0.175 0.03534 0.006185 2 0.025 −0.00393 −0.000398 ∑ 0.03141 𝟎. 𝟎𝟎𝟓𝟕𝟖𝟕 𝑧 = ∑𝑉𝑧𝑐 ∑𝑉 , ⇒ 𝑧 = 0.005787 0.03141 , ∴ 𝑧 = 184.24𝑚𝑚
- 98. #1. Determine thecoordinates of the mass center of the body which is constructed of three pieces of uniform thin plate welded together.
- 99. #2. The assemblyis made from a steel hemisphere,𝜌𝑠𝑡 = 7.83 𝑀𝑔 𝑚3, and an aluminum cylinder, 𝜌𝑎𝑙 = 2.7 𝑀𝑔 𝑚3. Determine the height h of the cylinder so that the mass center of the assembly is located at 𝑧 = 160𝑚𝑚
- 100. #3. Calculate the𝑥 −, 𝑦 −, and 𝑧 −coordinates of the mass center of the fixture formed from aluminum plate of uniform thickness.
- 101. The two theoremsof Pappus and Guldinus are used to find the surface area and volume of any body of revolution. It is a method for calculating the surface area generated by revolving a plane curve about a nonintersecting axis in the plane of the curve. Surface Area: The line segment of length 𝐿 in the 𝑥 − 𝑦 plane generates a surface when revolved about the x-axis.
- 102. An element ofthis surface is the ring generated by 𝑑𝐿. The area of this ring is its circumference times its slant height: 𝑑𝐴 = 2𝜋𝑦𝑑𝐿 ⇒ 𝐴 = 2𝜋 𝑦𝑑𝐿 Since, 𝑦𝐿 = 𝑦𝑑𝐿 , ⇒ 𝐴 = 2𝜋𝑦𝐿 If the area is only revolved through an angle u (radians), then 𝐴 = 𝜃𝑦𝐿 Where; 𝐴 is surface area of revolution 𝜃 is angle of revolution measured in radians, 0 ≤ 𝜃 ≤ 2𝜋 𝑦 is perpendicular distance from the axis of revolution to the centroid of the generating curve 𝐿 is length of the generating curve
- 103. Volume: A volume canbe generated by revolving a plane area about an axis that does not intersect the area. An element of the volume generated by revolving the area A about the x-axis is the elemental ring of cross section 𝑑𝐴 and radius 𝑦.
- 104. The volume ofthe element is: 𝑑𝑉 = 2𝜋𝑦𝑑𝐴, ⇒ 𝑉 = 2𝜋 𝑦𝑑𝐴 Since, 𝑦𝐴 = 𝑦𝑑𝐴 , ⇒ 𝑉 = 2𝜋𝑦𝐴 If the area is only revolved through an angle u (radians), then 𝑉 = 𝜃𝑦𝐴 Where; 𝑉 is volume of revolution 𝜃 is angle of revolution measured in radians, 0 ≤ 𝜃 ≤ 2𝜋 𝑦 is perpendicular distance from the axis of revolution to the centroid of the generating curve 𝐴 is generating area
- 105. #6. show thatthe surface area of a sphere is A = 4𝜋𝑅2 and its volume is V = 4 3 𝜋𝑅3. a b
- 106. Surface Area: Thesurface area of the sphere in Figure a is generated by revolving a semicircular arc about the x axis. The centroid of this arc is located at a distance 𝑦 = 2𝑅 𝜋 from the axis of revolution (x axis). Angle of revolution 𝜃 = 2𝜋 ∴ 𝐴 = 𝜃𝑦𝐿, ⇒ 𝐴 = 2𝜋 2𝑅 𝜋 𝜋𝑅 = 4𝜋𝑅2 Volume: The volume of the sphere is generated by revolving the semicircular area in Figure b about the x axis. The centroid of this area is located at a distance 𝑦 = 4𝑅 3𝜋 from the axis of revolution (x axis). Angle of revolution 𝜃 = 2𝜋 ∴ 𝑉 = 𝜃𝑦𝐴, ⇒ 𝐴 = 2𝜋 4𝑅 3𝜋 1 2 𝜋𝑅2 = 4 3 𝜋𝑅3
- 107. #2. Calculate thevolume 𝑉 of the solid generated by revolving the 60mm right-triangular area through 180o about the z-axis. If this body were constructed of steel, what would be its mass 𝑚?
- 108. Angle of revolution𝜃 = 180° 𝑉 = 𝜃𝑦𝐴 𝑉 = 𝜋 30 + 1 3 60 1 2 602 𝑉 = 282743.34𝑚𝑚3 The mass of the body is then, 𝑚 = 𝜌𝑉 𝜌𝑠𝑡𝑒𝑒𝑙 = 7830 𝑘𝑔 𝑚3 ∴ 𝑚 = 7830 𝑘𝑔 𝑚3 ∗ 282743.43𝑚𝑚3 1𝑚 1000𝑚𝑚 𝑚 = 2.21𝑘𝑔 … … 𝐴𝑛𝑠.
- 109. #3. Determine thesurface area and volume of the solid formed by revolving the shaded area 360o about the z axis.
- 110. Angle of revolution𝜃 = 360° 𝑉 = 𝜃𝑟𝐴 𝑟𝐴 = 𝑟1𝐴1 + 𝑟2𝐴2 𝑃𝑎𝑟𝑡 𝒓𝒄(𝒎) 𝐴𝑟𝑒𝑎 𝑨𝒓𝒄 1 0.636 1.767 1.123 2 0.75 3 2.25 ∑ 4.767 3.373 𝑟 = 0.707, 𝐴 = 4.767 𝑉 = 2𝜋 ∗ 4.767 𝑽 = 𝟐𝟗. 𝟗𝟓𝒎𝟑 … … 𝑨𝒏𝒔.
- 111. #1. Determine thevolume V generated by revolving the elliptical area through about the z-axis..
- 112. #2. Determine theheight h to which liquid should be poured into the cup so that it contacts half the surface area on the inside of the cup. Neglect the cup’s thickness for the calculation.