Review for Test 3 q Responsible for Chapter 8 8 5 9 except 9 8 10 11 12 notes only and 13 except 13 7 13 8 also chapters 1 8 12 Notes from class Problems worked in class Homework assignments q Test format 18 problems 5 points each multiple choice some T F 2 bonus probs Time 75 minutes q Test materials Pencil eraser and calculator No formulae sheet or paper Closed textbook notes Student ID Rules for the Test q One and only one empty seat between each student q Test version number on scantron must match test version number on roll sheet Check carefully q No talking during test except to proctors or instructor q Put name and 810 811 number on scantron bubble them and check carefully q All devices except simple calculator in OFF mode Material Covered Chapter 9 Impulse and Momentum Impulse linear momentum Impulse momentum theorem Conservation of Linear momentum 1 and 2D collisions elastic inelastic Center of mass Velocity of the center of mass q Chapts 10 11 Rotation Kinematics and Dynamics arc length angular displacement velocity and acceleration rotational kinematic equations tangential velocity tangential acceleration rolling motion Torque Newton s 2nd Law I Moment of inertia Rotational work and kinetic energy Material Covered cont d Chapter 11 cont d Conservation of Energy with rotation Angular momentum Conservation of angular momentum applications Static Equilibrium applications of 0 F 0 center of gravity q Chapter 13 Oscillations Simple harmonic motion SHM Spring mass system as SHM Simple pendulum physical pendulum Energy for SHM Potential energy curves equilibrium points potential slope 8 5 Example Problem A 0 200 m bar with a mass of 0 750 kg is released from rest in the vertical position A spring is attached initially unstrained and has a spring constant of 25 0 N m Find the tangential speed with which the free end strikes the horizontal surface drawing to be provided Solution Bar rotating with axis at one end rotational KE no translational KE Bar falls from some height gravitional PE Ug A spring is attached to bar spring PE Us Bar rigid body need moment of inertia Use Conservation of Energy E KR U g Us i 0 xi 0 yi h since this would h mean all mass of rod Ei mg Ei mgh is at y h but mass is i 2 distributed So take y f 0 U g f 0 mass to be located at 2 2 1 1 center of gravity E f 2 I f 2 kx f 2 1 I rod 3 mL v f v t r f L f 2 vt 1 2 Ef mL 2 kx f L 2 2 1 1 E f 6 mv t 2 kx f 1 1 2 3 2 xf From geometry of problem 2 2 x f 0 2 0 1 0 1 0 1236 m Return to Conservation of Energy and solve for vt Ef 1 6 mv vt vt vt 2 t h L mv kx Ei mg mg 2 2 L 1 2 mg 2 kx f 2 k 2 3 gL 3 x f m 25 0 3 9 80 0 20 3 0 1236 2 0 750 5 88 1 528 2 09 m s 1 6 2 t 1 2 2 f
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