Find the mass and center of mass of the lamina that occupies the region D and has the given density function p: Dis the triangular region with vertices (0,0), (2, 1), (0,3); p(x, y) = x + y
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- Find the center of mass for the particle system with 6 identical masses (x, y in meter)Find the centre of mass of the 2D shape bounded by the lines y = 10.9 between z = 0 to 2.9. Assume the density is uniform with the value: 2.9kg. m 2 Also find the centre of mass of the 3D volume created by rotating the same lines about the z-axis. The density is uniform with the value: 2.3kg. m-³ (Give all your answers rounded to 3 significant figures.) a) Enter the mass (kg) of the 2D plate: Enter the Moment (kg.m) of the 2D plate about the y-axis: Enter the z-coordinate (m) of the centre of mass of the 2D plate: b) Enter the mass (kg) of the 3D body: Enter the Moment (kg.m) of the 3D body about the y-axis: Enter the x-coordinate (m) of the centre of mass of the 3D body:Let's consider a ring with a diameter of 0.1 m and a total load of -2 mC. The ring lies in the x-y plane, with its center at the origin. What is the force on a 1 mC load located at z = 3 m? In what direction would the force be on this load if it were in the x-y plane, but outside the ring?
- Set up the integrals to find the ycoordinate of the center of mass for a region of density (x;y)inside the circle x^2 + y^2 = 4 and below y = sqrt (2)Find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density. 7 y = 1 + x2 y = 0 X = -1 X = 1 p = k m = (x, y) = Need Help? Watch It Read ItConsider a uniform solid hemisphere of radius r and mass m which is kept in contact with a smooth wall as shown in the figure. Find the normal reaction exerted on the sphere by the wall
- Find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density or densities. (Hint: Some of the integrals are simpler in polar coordinates.) m = y = x² y = 0 X = 4 P = (x, y) = ( kxyPlease asapProblem 3: (a) Use spherical coordinates to find the center of mass (CM) of a uniform solid hemisphere of radius R, whose flat face lies in the ry plane with its center on the origin. [Note: dV = ² sin 0 dr do do.] (b) Use your result from part (a) to calculate the CM of a hemispherical "bowl" with outer radius R and inner radius kR, k < 1. (Depending on your work in part (a), you may not even need to do another integral.) (c) Use your result from the previous part to find the CM for an infinitely thin hemispherical shell of radius R.
- Explorers in the jungle find an ancient monument in the shape of a large isosceles triangle as shown. The monument is made from tens of thou- sands of small stone blocks of density = 800 kg/m3 . The monument is 15.7 m high and 64.8 m wide at its base and is everywhere 3.60 m thick from front to back. Before the monument was built many years ago, all the stone blocks lay on the ground. (a) Choosing a coordinate system with x = 0 at the center of the pyramid, y = 0 at the base of the pyramid, and z = 0 at the face of the pyramid shown in the diagram, determine the coordinates of the center of mass of the pyramid in the x, y, and z directions. (b) How much work did laborers do on the blocks to put them in position while building the entire monument? Note: The gravitational potential energy of an object–Earth system is given by Ug = MgyCM, where M is the total mass of the object and yCM is the elevation of its center of mass above the chosen reference level.Hamiltonian of a system is given by: Sum of momentum and speed Sum of KE and PE Difference between KE and PE The square root of momentum + speedThe height varies from h to zero according to this function: y(x) = h ( – 1)´ . The constants h and e replace 1.00 m and 3.00 m. There is also a thickness t and a density p. You need two integrals, the total mass and the center of mass. Possibly surprisingly, you don't actually need the numbers t, h, and p. Ax y(x) X The column at x has a mass Am = (density * volume) = y(x) p t Ax. You add all the Am values to get %3D the total mass M. The sum becomes an integral: М — pt y(x) dx For the center of mass, you add each column's x Am, and divide by M: pt Xc х у(x) dx Calculate xc. The only quantity you'll need is e = 5 m.