A rod of length 23.00 cm has linear density (mass per length) given by 1 = 50.0 + 17.5x where x is the distance from one end, and 2 is measured in grams/meter. (a) What is its mass? (b) How far from the x = 0 end is its center of mass? m

Problem 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.

Problem Two. A rod of length L = 2.0 m is placed on the x-axis from x = 0 to x = L. The rod has a mass density given by 2 = 2] 1+- x- where 2, = 2.0k. L' Find the location of the center of mass of the rod in meters. 4.) (A) 1.5 (B) 1.3 (С) 1.1 (D) 1.7 (E) 0.45 Consider if the rod is subjected to an applied force given by F = Aî +Bx²j , where A=10 N and B = 25 N2 . If the end of the rod at x = 0 is attached to a hinge, find the angular acceleration of the rod in rad/s². m 5.) (A) 0.68 (B) 0.92 (C) 0.26 (D) 0.34 (E) 0.45 Find the direction of the net torque vector acting on the rod. 6.) (A) +z (В) —х (C) +x (D) –z (E) -у

A 1.95-kg particle has a velocity (1.95 î – 3.07 ĵ) m/s, and a 2.91-kg particle has a velocity (1.06 î + 5.99 ĵ) m/s. (a) Find the velocity of the center of mass. i) m/s (b) Find the total momentum of the system. |î) kg · m/s

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.

(a) Show that the CM of a uniform thin rod of length / and mass M is at its center. (b) Determine the CM of the rod assuming its linear mass density A (its mass per unit length) varies linearly from A = at the left end to double that value, A = 2A0, at the right y end. dm λdx X 0 х dx

A uniform rod of length 0.8 m and mass 1.8 kg, has two point masses at each end. The point mass on the left end has a mass 1.4 kg, and the one on the right end has a mass 3 kg. Calculate the location of the center of mass of this system in terms of the distance from the left end.

The 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.

A uniform rod of length 0.8 m and mass 2.2 kg, has two point masses at each end. The point mass on the left end has a mass 1.5 kg, and the one on the right end has a mass 2.5 kg. Calculate the location of the center of mass of this system in terms of the distance from the left end.

A water molecule consists of an oxygen atomwith two hydrogen atoms bound to it. Theangle between the two bonds is 106◦.  If each bond is 0.093 nm long, how far fromthe oxygen atom is the center of mass of themolecule? Take the mass of an oxygen atomto be 16 times the mass of a hydrogen atom.Answer in units of nm