Physics 218 Exam 3 Fall 2010, Sections

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1 Physics 28 Exa 3 Fall 200, Sections Do not fill out the inforation below until instructed to do so! Nae Signature Student ID E-ail Section # : SOUTIONS ules of the exa:. You have the full class period to coplete the exa. 2. When calculating nuerical values, be sure to keep track of units. 3. You ay use this exa or coe up front for scratch paper. 4. Be sure to put a box around your final answers and clearly indicate your work to your grader. 5. Clearly erase any unwanted arks. No credit will be given if we can t figure out which answer you are choosing, or which answer you want us to consider. 6. Partial credit can be given only if your work is clearly explained and labeled. 7. All work ust be shown to get credit for the answer arked. If the answer arked does not obviously follow fro the shown work, even if the answer is correct, you will not get credit for the answer. Put your initials here after reading the above instructions:

2 Part Part (25) Part 2 (25) Part 3 (25) Part 4 (25) Bonus (5) Exa Total Table to be filled by the graders Score 2

3 Part : (25p) Energy Proble.: Two cylinders, each with ass and radius, are nailed to the wall and allowed to rotate freely but initially at rest. A slab of hoogenous aterial with ass M and length is initially at rest and placed between the two cylinders as shown in the picture below. The slab then falls between the cylinders aking the rotate without slipping. M g M Question..: (4p) Identify the external forces to the cylinders+slab syste. The gravity on the slab and the force that each nail exerts on the cylinders to keep the in place. Question..2: (4p) Is the Initially Soetie after echanical energy of the cylinders+slab syste conserved? Explain your reasoning. The internal friction force does not produce work as the disks are rotating without slipping. Yes, the echanical energy is conserved as the work of the non-conservative forces is zero. Y Question..3: (5p) At the oent in which the slab losses contact with the cylinders, find the agnitude of the angular velocity of the cylinders. M g Because echanical energy is conserved, E i E f Mg 2 Mg 2 + 2( 2 ω f 2 ) + 2 Mv 2 Mg ω f Mv 2 X Since the slab went without slipping we know v ω f Mg ω 2 f + 2 Mω 2 f 2 ω 2 f ( + 2 M2 ) ω f Mg + 2 M2 ω f Mg M2 2Mg 2 ( + M) Question..4: (2p) Is the direction of the angular velocity of each cylinder opposite or the sae to each other? Obviously the cylinders rotate in opposite ways. 3

4 Part 2: (25p) Energy and Kineatics Question 2..: A disk of ass and radius is spinning with angular velocity ω. The disk is spinning and slipping with respect to the surface of the ground. Initially the disk is just spinning but not oving in the horizontal direction. There is friction between the disk and the table with kinetic coefficient µ K. g Ground Question 2..2: (5p) Draw a coordinate syste, the orientation in which the disk rotates, and a free body diagra of the disk. Question 2..3: (0p) Find the angular acceleration of the disk. Is it constant?, is it parallel or antiparallel to the angular velocity? If you write a torque specify the point fro where the torque is coputed. Z Y w N f X Fro the center of the disk the torque is τ +f α α f µ N K µ g K 2µ g K 2µ g K 2 which is clearly constant as the all the coponents are constant. In the coordinate syste depicted above the angular velocity is negative, while the angular acceleration found is positive, so the angular acceleration is antiparallel to the angular velocity. As expected as the object is deaccelerating. Question 2..4: (0p) After an N nuber of revolutions the disk is found rotating without slipping. Using the work- energy theore find the angular velocity at which the disk is now rotating. Assue the distance travelled in the horizontal direction is zero. The work - energy theore states that : K i + i +W non cons K f + f In our case we have K i 0, i 2 Iω 2 and f 2 Iω 2 f,k f 2 v 2 f ω f After N revolutions the work of the torque is W non cons τ2πn µ K g2πn where the inus sign coes fro the fact that the torque was going against the angle displaceent. 2 Iω 2 µ K g2πn 2 Iω 2 f ω 2 f Iω 2 2µ K g2πn ω 2 f (I + 2 ) ω f Iω 2 4µ K gπn (I + 2 ) 2 ω 2 2 4µ K gπn ( ) ω 2 8µ K gπn 3 4

5 Part 3: (25p) Angular Moentu Proble 3.: A tande rotor helicopter has two sets of blades powered by the sae otor that rotate in opposite direction with angular velocity of agnitude ω b. The oent of inertia of each set of helicopter blades with respect to each set s center is I b and the oent of inertia of the helicopter around its center of ass is,. The helicopter is initially hovering at high altitude without oving in the horizontal or vertical position and without rotation of the body of the helicopter. Ignore any air resistance. Question 3..: (5p) Find the total angular oentu of the helicopter, and show how you reached that nuber. I b CM I b The total angular oentu is I b ω b I b ω b 0 as the body is not rotating and the blades are rotating in opposite direction. Question 3..2: (0p) If a alfunction suddenly stops one set of blades to a screeching halt, find the angular velocity the body of the helicopter develops around its center of ass. The angular oentu has to be conserved as there are no external forces. i 0 f I b ω b + ω h ω h I bω b Question 3..3: (0p) To stop the spinning, the helicopter posses a jet engine that can provide a constant torque τ jet on the body of the helicopter and with respect to the center of ass in the direction of the angular velocity of the reaining set. Find the tie it takes to bring the broken helicopter to the status of no- body rotation. Basically find the tie the torque τ jet reduces the angular velocity fro ω h to zero. The torque τ jet produces a angular acceleration of α τ jet ω f 0 ω h + αt t ω h α I bω b τ jet I bω b τ jet 5

6 Part 4: (25p) Collisions Proble 4.: A square paddle is set through axis A to a table and allowed to rotate free. The oent of inertia of the paddle around axis A is I A. Initially the paddle is at rest. A point- like particle of ass is oving with velocity v 0 towards the paddle at a distance with respect to the axis of the paddle as shown in the figure. The particle collisions with the paddle and sticks to the paddle s sticky surface. Question 4..: (5p) Is any A y V 0 x coponent of the angular oentu of the ball+paddle syste as coputed fro the point where the axis intercept the table, the sae before and after the collision?, why? Yes. Since no external force can exert torque on the z coponent the angular oentu on the z coponent ust be conserved. In addition the axis can provide force in the x and y direction, but the torque of these forces as coputed fro the axis is also zero, so the angular oentu is conserved in x,y and z. Question..: (0p) Find the angular velocity of the collision. paddle+particle syste after the Fro conservation of angular oentu: v 0 (I A + 2 )ω f ω f v 0 (I A + 2 ) Question..2: (0p) What is the ratio of the energy before the collision to the energy after the collision? Is this an elastic or an inelastic collision? E i 2 v 2 0, E f 2 (I A + 2 )ω 2 f 2 (I 2 v 2 A + 2 ) 0 2 (I A + 2 ) 2 v (I A + 2 ) v ( I A + 2 ) E f E i ( I A +) 2 Because the final kinetic energy is less than the initial kinetic energy the collision was inelastic. 6