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

1.Your car rolls slowly in a parking lot and bangs into the metal base of a light pole. In terms of safety, is it better for your collision with the light pole to be elastic, inelastic, or is the safety risk the same for either case? Explain and labled the problem.

A 16-lb bowling ball is at the origin, while a 12-lb ball is at (5,0)m and a 14-lb ball is at (7,4)m. Determine the coordinates of the center of mass of the three bowling balls. A) Enter the x-coordinate of the center of mass. B) Enter the y-coordinate of the center of mass.

To keep the calculations fairly simple, but still reasonable, we shall model a human leg that is 92.0 cm long (measured from the hip joint) by assuming that the upper leg and the lower leg (which includes the foot) have equal lengths and that each of them is uniform. For a 70.0 kg person, the mass of the upper leg is 8.60 kg, while that of the lower leg (including the foot) is 5.25 kg. Find the location of the center of mass of this leg, relative to the hip joint, if it is (a) fully extended, and (b) bent at the knee to form a right angle with the upper leg.

Find the center of mass coordinates XCM y YCM of the arrangement of masses shown. Assume that the connecting bars are massless. (m= 3.0 kg; M = 1.5 kg; L = 2.0 m)

Determine the coordinates of the center of mass of the three-particle system below.  The masses are as follows, m1=3.0 kg, m2 = 4.0 kg and m3 = 8 kg.  (Use xcm= (x1m1+ x2m2+ ...xnmn)/(m1+m2+...mn) and a similar formula for y.)  Assume ys= 2 m and xs=2 m.