QUESTIONS PHYSICS FOR ENGINEERS. 2006-2007. MID-TERM EXAM A2E AGR FOR P1. A homogeneous sphere is thrown on a rough horizontal floor. Its initial center of mass velocity is v, meanwhile its initial angular velocity is zero. What happens is: SURNAMES a) The sphere rolls from the beginning on the horizontal surface. b) The velocity of the center of mass remains constant, the angular velocity increases. c) The velocity of the center of mass drops, the angular velocity remains null. d) The sphere slides, but cannot roll because the friction is not zero. FIRST NAME e) None of the above. P2. A skater on ice has her arms extended while she spins at 2 rps. When she cross her arms on her body, her moment of inertia halves. As a result, we’ll see that: a) Her angular velocity halves. b) Her angular velocity duplicates. c) Her angular velocity reduces to a fourth. d) Her angular velocity increases four times. e) Her angular velocity remains constant, however her angular momentum duplicates. P3. The figure depicts a ring of radious R and mass M, where the mass is homogenously shared out along its rim. The momenta of inertia about the different axis depicted in the figure are: P4. The variation of the angular momentum with time quantifies: a) The sum of external forces a) X Y Z b) The sum of the momenta of inertia b) c) The sum of the torques of external forces c) d) The variation of the angular velocity d) e) None of the above e) None of the above Warning: to answer questions P1 to P5 you have to use the answer form, only the answers including there will be considered. For the choosen right answer, use the symbol . To invalidate a previously marked cell, use the symbol X X For questions P1 to P5, each correct answer adds +1; each wrong answer takes off -0.25

P5. A torque of 10 Nm is acting on a spinning disc, which turns around a normal fixed axis passing through its simmetry center. If the rotational work is 5.24 J, then we can say that the angle gone over is (within ±1º): ANSWER FORM   a b c d e P1 P2 P3 P4 P5  a) 90º  b) 30º  c) 60º  d) 45º  e) 180º PROBLEM A thin rigid rod of lenght 2L = 80 cm is composed by two sections, each of lenght L, made on different materials whose linear densities are A = 2 kg/m and B = 5 kg/m. The rod is hung horizontally from both ends on two threads A and B, as shown in the figure. Find: a) The tension supported by each thread (1.5 p) b) The moment of inertia about end A (1.5 p) c) Assume the thread B is cut. Find the angular acceleration and the angular velocity of the rod when it reaches the vertical position (2 p). A B L

Center of mass of the whole rod: PROBLEM SOLUTION a) Both sections are homogeneous and have the same lenght, so the position of both center of masses will be A B Y Mass of each section: L Center of mass of the whole rod: Total mass:

b) Moment of inertia about A PROBLEM SOLUTION b) Moment of inertia about A L A B c) Angular acceleration and angular velocity A

When the rod is vertical PROBLEM SOLUTION Chain rule When the rod is vertical (Another consideration: when the rod passes through the vertical, its weight doesn’t exert any torque and there aren’t any other forces exerting some torque, so its angular acceleration should be zero) Numerical solution