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PHY183-Lecture23

PHY183-Lecture23 - Physics for Scientists& Engineers 1...

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Unformatted text preview: Physics for Scientists & Engineers 1 Spring Semester 2011 Lecture 23 Coordinate Systems and Calculation of Center of Mass March 3, 2011 Physics for Scientists&Engineers 1 1 1 Calculating the Center of Mass How do we calculate the location of the center of mass of an arbitrarily shaped object? We represent the object by small identical cubes Each cube has its own center of mass and we can calculate the center of mass of the system of cubes March 3, 2011 Physics for Scientists&Engineers 1 2 2 Density Distributions and Center of Mass Clearly not all the cubes in the hammer have the same mass Density for each small cube is defined as (ri ) = mi / Vi If we assume that the density of each cube is different but each cube has the same size then we can write 1 R= M 1 ri mi = M i =1 n n i =1 ri (ri ) V Reducing the size of the cubes gives us R= 1 r ( r )dV MV March 3, 2011 Physics for Scientists&Engineers 1 3 3 Position of the Center of Mass With r = ( x, y, z ) We have X= 1 x (r )dV ; MV Y= 1 M and R = ( X , Y , Z ) y (r )dV ; V Z= 1 z (r )dV MV March 3, 2011 Physics for Scientists&Engineers 1 4 4 Position of the Center of Mass If the density is constant for the whole object = M / V (for constant ) we can write R= Or X= 1 xdV ; VV Y= 1 ydV ; VV Z= 1 zdV VV M rdV = V 1 rdV (for constant ) VV March 3, 2011 Physics for Scientists&Engineers 1 5 5 Volume Integrals in Cartesian Coordinates It is by far easiest to express the volume element dV in Cartesian coordinates Then dV is simply the product of the three individual coordinate elements dV = dxdydz The three-dimensional volume integral written in Cartesian coordinates becomes V March 3, 2011 ymax xmax f ( r ) dV = f ( r ) dx dy dz zmin ymin xmin zmax Physics for Scientists&Engineers 1 6 6 ThreeThree-Dimensional Non-Cartesian NonCoordinate Systems In chapter 1, we introduced a three-dimensional orthogonal coordinate system The Cartesian coordinate system with coordinates x, y, and z For some applications it is useful to represent the position vector in different coordinate systems Two other commonly used three-dimensional orthogonal coordinate systems Spherical coordinates Cylindrical coordinates Details in the book Example in the book for center of mass, but these will not be used for homework or exams for center of mass. March 3, 2011 Physics for Scientists&Engineers 1 7 7 Center of Mass of Variable Density Rod Consider a long thin rod The linear mass density of the rod is given by (x) in kg/m This problem is inherently one dimensional Start with general form and go to one dimension 1 1 R= r (r )dV X = M x ( x ) dx M = ( x ) dx MV x x So the x-coordinate is X= x ( x ) dx x ( x ) dx x Physics for Scientists&Engineers 1 8 March 3, 2011 8 Clicker Quiz Consider a long thin rod. One end of the rod is located at x1 = 0 and the other end of the rod is located at x2 = 5.00 m. The linear mass density of the rod is given by (x) = bx where b = 2.00 kg/m2. What is the mass of the rod? A) 5.00 kg B) 10.0 kg C) 15.0 kg D) 20.0 kg E) 25.0 kg March 3, 2011 Physics for Scientists&Engineers 1 9 9 Clicker Quiz Solution Consider a long thin rod. One end of the rod is located at x1 = 0 and the other end of the rod is located at x2 = 5.00 m. The linear mass density of the rod is given by (x) = bx where b = 2.00 kg/m2. What is the mass of the rod? A) 5.00 kg B) 10.0 kg C) 15.0 kg D) 20.0 kg E) 25.0 kg x2 m = ( x ) dx = ( bx ) dx = b 2 x1 x1 x1 x2 x2 2 x2 ( 5.00 m ) 2 m = ( 2.00 kg/m ) = 25.0 kg 2 March 3, 2011 Physics for Scientists&Engineers 1 10 10 Clicker Quiz 2 Consider a long thin rod. One end of the rod is located at x1 = 0 and the other end of the rod is located at x2 = 5.00 m. The linear mass density of the rod is given by (x) = bx where b = 2.00 kg/m2. What is the x-coordinate of the center of mass of the rod? A) 0.500 m B) 1.33 m C) 2.50 m D) 3.33 m E) 4.00 m March 3, 2011 Physics for Scientists&Engineers 1 11 11 Clicker Quiz 2 Solution Consider a long thin rod. One end of the rod is located at x1 = 0 and the other end of the rod is located at x2 = 5.00 m. The linear mass density of the rod is given by (x) = bx where b = 2.00 kg/m2. What is the x-coordinate of the center of mass of the rod? A) 0.500 m B) 1.33 m C) 2.50 m D) 3.33 m E) 4.00 m x2 ( x ) xdx x1 x2 x2 ( bx ) xdx m X= ( x ) dx x1 = x1 b x3 = m 3 x x2 1 3 2.00 kg/m ( 5.00 m ) X= = 3.33 m 25.0 kg 3 March 3, 2011 Physics for Scientists&Engineers 1 12 12 Center of Mass Example Now let's tackle a more complicated object A disk (constant ) with a rectangular hole in it h = 11.0 cm R = 11.5 cm d = 8.0 cm w = 7.0 cm Right side of hole coincides with central axis Where is the center of mass of this object? March 3, 2011 Physics for Scientists&Engineers 1 13 13 Center of Mass Example First we determine the symmetry planes Center of mass must Clearly the center of mass be on this line! will lie along the x-axis We could carry out an integral like we did for the half-sphere There is an easier way! Treat the object as a solid disk with a piece missing March 3, 2011 Physics for Scientists&Engineers 1 14 14 Center of Mass Example We can write the X coordinate as X= xdVd - xhVh xdVd - xhVh = Vd - Vh Vd - Vh where we have canceled out the density Elegant procedure: treat missing piece as object with negative density Putting in the numbers: (0 cm)(4570 cm 3 ) - (-3.5 cm)(616 cm 3 ) X= = 0.545 cm (4570 cm3 ) - (616 cm 3 ) March 3, 2011 Physics for Scientists&Engineers 1 15 15 Center of Mass of Parabolic Plate A uniform plate of height h is cut in the form of a parabolic section The lower boundary of the plate is defined by y = ax2 Find the location of the center of mass of the plate SOLUTION We can see that the center of mass will be located at x = 0 The y-coordinate is given by Y= ydA A dA A Physics for Scientists&Engineers 1 16 March 3, 2011 16 Center of Mass of Parabolic Plate We can express the differential area dA = xdy We are given y = ax 2 so x = y a The differential area is represented by the red rectangle and has area y dA = 2 a dy The y-coordinate is then y2 Y = 0h 2 0 h March 3, 2011 h 1 5/2 dy y 3/2 dy h 3 =0 = 5/ 2 = h h 1 3/2 5 y h y1/2 dy dy 3/ 2 a 0 y a Physics for Scientists&Engineers 1 17 17 General Considerations Note that the center of mass does not have to be inside the object The red dot represents the center of mass March 3, 2011 Physics for Scientists&Engineers 1 18 18 ...
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