Physics 8 Exam 3 Spring 00, Sections 5-55 Do not fill out the information below until instructed to do so! Name Signature Student ID E-mail Section # Rules of the exam:. You have the full class period to complete the exam.. When calculating numerical values, be sure to keep track of units. 3. You may use this exam or come 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 marks. 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 must be shown to get credit for the answer marked. If the answer marked does not obviously follow from 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:
Part Part (30) Part (0) Part 3 (5) Part 4 (5) Bonus (7) Exam Total Table to be filled by the graders Score
Part : (30p) Collisions Problem.: A catcher system is composed of a mass m8kg attached to a massless rod of length dm, which in turn is attached to a massless cup (see figure). The catcher system is moving to the right with speed vm/s, at an angle φ30 but without rotating. At time t0 a particle with mass mkg is traveling to the left with speed v3m/s in direct collision course to the cup of the catcher system. A numerical answer with proper units is expected for each question. Y v 3m/s m Kg m 8Kg φ30 v m/s Question..: (5p) With the given coordinate system find the position of the center of mass in the y component. X r,y m r,y + m r,y m + m 0 + Kg m sin(30) 0Kg 0.m Question..: (5p) Find the velocity of the center of mass in the x and y components. v m v, + m v 6Kgm /s 6Kg m/s x ˆ m /s x ˆ + 0 y ˆ m + m 0Kg Question..3: (0p) Eventually the cup catches the mass m and they stick together. Find the angular velocity of the catcher+m system around its center of mass. (Hint: you may use L r P + I ω with I 6.4 Kg m ) There are not external forces, so the torques of the external forces is zero and therefore the angular momentum is conserved. r m v + r m v 0 + 6Kgm /s z ˆ 6Kgm /s z ˆ r P + I ω Kg m /s z ˆ + I ω L f 6Kgm /s z ˆ Kg m /s z ˆ + I ω 8Kgm /s z ˆ I ω ω 8Kgm /s z ˆ.5s z ˆ I Question..4: (0p) Find the energy the system lost in the collision. Where did that energy go? m v + m v 6J + 9J 5J E f m ( + m )v + I E f 5J 0J 5J ω 5J + 5J 0J The energy went into the work of nonconservative forces when the cup catched the mass m 3
Part : (0p) Momentum Problem.: A satellite orbiting around the earth is composed of a platform supporting a camera and a disk. The moment of inertia of the satellite around its center of mass axis is Is. Initially the disk is not rotating and the camera is pointing towards the starts. The moment of inertia of the disk with respect to its axis is Id. The gravitational force of the earth can be considered constant within the satellite. I d I s Camera Earth C.m. Stars Question..: (0p) The department of defense asks you to forget about science and point the camera towards the earth. Find out the angle you have to rotate the disk to make the camera point towards the earth (you can imagine a constant angular speed if that helps you) The only external force acting upon the satellite is from the earth gravity. Because the gravitational force can be considered constant within the satellite then the center of gravity equals the center of mass. As seen from the center of mass the gravity does not provide any torque, therefore the angular momentum has to be conserved. In addition, the satellite has to rotate an angle of dφ sf π to have the camera pointing towards the earth. I d ω di ω Si 0 + 0 0 I d ω df ω Sf d dφ df dφ sf + I S 0 I d dφ df dφ sf 0 I d dφ df dφ sf dφ df I S dφ I sf I S π d I d Question..: (7p) If the satellite orbits the earth with a period of T minutes, what should the velocity of the disk ωd be such that the camera is always pointing towards the earth? I d ω df ω Sf 0 ω df I S ω I Sf d I S π I d T Question..3: (3p) A software error makes the disk rotate with an angular velocity ωerror. Find the angular velocity of the satellite around its center of mass axis. I d ω error ω S 0 ω S I d ω I error S 4
Part 3: (5p) Equilibrium Problem 3.: A disk of radius R and mass m is standing in an inclined plane with angle α. There is friction force between the sphere and the plane. On the side of the cylinder a nail holds a string attached to a mass m. Question 3..: (4p) Specify your coordinate R m friction µs g system and draw the free- body diagram of the disk. Y N m α f T mg X Question 3..: (9p) Find the value of the mass m such that the disk does not roll, in terms of the parameters of the problem. In the process obtain the friction force as well. ext τ 0 F ma ball 0 on ball τ : RT Rf 0 T f y : N mgcos(α) T cos(α) 0 x : mg + T f 0 mg + T( ) 0 T mg m g m m Question 3..3: (6p) Find the normal force exerted by the plane on the disk in terms of the parameters of the problem. Simplify as much as possible the final expression. from y : N mgcos(α) T cos(α) 0 N mgcos(α) + mg cos(α) N mgcos(α) + mgcos(α) Question 3..4: (6p) Find the minimum value of µ s needed for the disk to be in equilibrium, in terms of the parameters of the problem. Simplify as much as possible the final expression. f must be equal to T, so the minimum µ K must satisfy T mg < µ mgcos(α) KN µ K < µ cos(α) K µ K > tan(α) 5
Part 4: (5p) Energy Considerations Problem 4.: A solid sphere of mass m and radius R is initially at rest at the top of a hill with height h when it begins to fall. There is friction between the hill and the sphere, and the sphere rolls down without slipping until reaching the base of the hill. (Isphere m R /5) h g R m g Question 4..: (5p) What is the total work of the friction force and why? The total work of the friction is zero, basically because the sphere rolls without slipping. base Question 4..: (5p) Is the energy of the sphere going to be conserved, and why? Yes. The energy of the sphere (potential + kinematic) will be conserved because the work of the non- conservative forces is zero. Question 4..3: (5p) Is the momentum of the sphere going to conserved, and why? No. The forces external to the sphere, like gravity, normal to the hill, and friction forces, do not add up to zero, and so the momentum does not have to be conserved. Question 4..4: (5p) Find the angular velocity of the sphere at the base of the hill. Because the energy has to be conserved, the initial energy has to be equal to the final energy. mgh E f mv cm + Iω mr ω + 7 E f mgh mω R 0 ω 0gh 7R 7 5 mr ω mω R 0 Question 4..5: (5p) At the base of the hill, find the ratio of the translational kinetic energy to the rotational kinetic energy. Ratio K trans K rot mv cm Iω mr ω 5 5 mr ω 6