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Physics I 95.141 LECTURE 21 11/24/10 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Exam Prep Question • The system to the right consists of a cylinder (R=15cm, M=50kg) and 4 2kg (point) masses attached to 30cm massless rods. The system is free to rotate around an axis through its center of mass. • • • • a) (10 pts) What is the moment of inertia of this system? b) (10 pts) Assume a 10kg mass is attached to a massless cord, wrapped around the cylinder, and dropped from rest. What is the acceleration of the mass? c) (5pts) What is the angular acceleration of the cylinder/mass system? d) (10pts) Determine the Kinetic Energy, as a function of time, associated with i) the rotating system and ii) the falling mass. 95.141, F2010, Lecture 21 Department of Physics and Applied Physics 30cm 15cm 2kg 10kg Exam Prep Question • The system below consists of a cylinder (R=15cm, M=50kg) and 4 2kg (point) masses attached to 30cm massless rods. The system is free to rotate around an axis through its center of mass. • 30cm 15cm a) (10 pts) What is the moment of inertia of this system? 2kg 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Exam Prep Question • The system below consists of a cylinder (R=15cm, M=50kg) and 4 2kg (point) masses attached to 30cm massless rods. The system is free to rotate around an axis through its center of mass. • 30cm 15cm b) (10 pts) Assume a 10kg mass is attached to a massless cord, wrapped around the cylinder, and dropped from rest. What is the acceleration of the mass? 2kg 10kg 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Exam Prep Question • The system below consists of a cylinder (R=15cm, M=50kg) and 4 2kg (point) masses attached to 30cm massless rods. The system is free to rotate around an axis through its center of mass. • 30cm 15cm c) (5pts) What is the angular acceleration of the rotating cylinder/mass system? 2kg 10kg 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Exam Prep Question • The system below consists of a cylinder (R=15cm, M=50kg) and 4 2kg (point) masses attached to 30cm massless rods. The system is free to rotate around an axis through its center of mass. • 30cm 15cm d) (10pts) Determine the Kinetic Energy, as a function of time, associated with i) the rotating system and ii) the falling mass. 2kg 10kg 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Review Example • What is the vector cross product of the two vectors: A 1iˆ 2 ˆj 4kˆ B 2iˆ 3 ˆj 1kˆ 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Administrative Notes • Exam III – Wednesday 12/1 – In Class, 9am-9:50am – Chapters 9-11 • Practice Exams posted • Practice problems posted by end of day • Exam Review Scheduled for 11/29….subject to change. Will probably have to be shifted to Tuesday. 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Outline • • Vector Cross Products Conservation of Angular Momentum • What do we know? – Units – Kinematic equations – Freely falling objects – Vectors – Kinematics + Vectors = Vector Kinematics – Relative motion – Projectile motion – Uniform circular motion – Newton’s Laws – Force of Gravity/Normal Force – Free Body Diagrams – Problem solving – Uniform Circular Motion – Newton’s Law of Universal Gravitation 95.141, F2010, Lecture 21 Department of Physics and Applied Physics – – – – – – – – – – – – – – – – – – – – – – – – – – – Weightlessness Kepler’s Laws Work by Constant Force Scalar Product of Vectors Work done by varying Force Work-Energy Theorem Conservative, non-conservative Forces Potential Energy Mechanical Energy Conservation of Energy Dissipative Forces Gravitational Potential Revisited Power Momentum and Force Conservation of Momentum Collisions Impulse Conservation of Momentum and Energy Elastic and Inelastic Collisions2D, 3D Collisions Center of Mass and translational motion Angular quantities Vector nature of angular quantities Constant angular acceleration Torque Rotational Inertia Moments of Inertia Angular Momentum Review of Lecture 20 • Introduced concept of Angular Momentum L I • Conservation of Angular Momentum – With no external torques acting on a system, the angular momentum of the system is conserved. • Vector Cross products C A B 95.141, F2010, Lecture 21 Department of Physics and Applied Physics A Ax iˆ Ay ˆj Az kˆ B B x iˆ B y ˆj Bz kˆ iˆ A B Ax Bx ˆj Ay By kˆ Az Bz Review of Angular Motion • We know equations of motion for angular motion 2 o2 2 o o f 2 1 2 o o t t 2 o t • We know torques cause angular acceleration RF I • Objects can have rotational kinetic energy KErot I 1 2 2 • And angular momentum L I • So why the cross product? 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Torque and the Cross Product • When we first introduced torque as the product of the radius and the perpendicular component of the Force, we were only interested in the magnitude of the torque! • Magnitude given by RFsinθ…same as cross product R, F FROM LECTURE 19 R RF RF sin 95.141, F2010, Lecture 21 Department of Physics and Applied Physics F Torque and the Cross Product • However, we since learned that I • And we know that angular acceleration points in direction of axis of rotation… so Torque must as well! • Torque is cross product of R,F R F FROM LECTURE 19 R RF RF sin 95.141, F2010, Lecture 21 Department of Physics and Applied Physics F Angular Momentum of a Particle • We have already defined angular momentum as L I • But this definition is for objects rotating with some angular velocity and moment of inertia around an axis of rotation. • More general, alternate, definition: Lrp 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Equivalence of our two definitions • Suppose we have a mass rotating around an axis 10 rad s m=2kg 2m • Use cross product Lrp 95.141, F2010, Lecture 21 Department of Physics and Applied Physics y x Equivalence of our two definitions • Suppose we have a mass rotating around an axis 10 rad s m=2kg 2m • Use L I 95.141, F2010, Lecture 21 Department of Physics and Applied Physics y x So why use cross product? • Cross products are messy…why would we ever use them, instead of the simpler L I RF • Because the cross product allows us to determine the angular momentum of, or torque on, objects which are not necessarily moving with constant, or even circular motion! 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Example • Calculate the angular momentum (about the origin) of the rock of mass m dropped from rest off the cliff. d (0,0) 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Example • What Torque is exerted (about the origin) on the rock? d (0,0) 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Relationship of torque to angular momentum • When we discussed linear momentum, we nd revised Newton’s 2 Law to state dp F dt • Similarly, we can write an expression for net dL torque in terms of angular momentum dt • Double check with our falling rock: L dmgt kˆ 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Newton’s 2nd Law: Angular Form • The vector sum of all of the torques acting on a particle, object, or system, is equal to the time rate of change of the angular momentum of the particle, object or system. dL dt 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Conservation of Angular Momentum • We used the revised expression for conservation of linear momentum to argue that if there is no net external force on a system or object, then the momentum of the system or object is conserved. • Similarly: – If the net external torque acting on a system of object is zero, then the angular momentum of that object will remain constant. dL L constant! 0 dt 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Conservation of Momentum 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Does this make sense? • What is happening? • What do we need to know? – System= Container + Skinner L final Linitial 0 LSkinner Lcontainer Linitial 0 L I 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Moments of Inertia • Skinner=point mass • Shipping Container 2 I Skinner M Skinner RSkinner R 1.25m M 70kg 2.5m 2.5m 12m Mass=3,500kg 95.141, F2010, Lecture 21 Department of Physics and Applied Physics σ Container • Determine surface mass density σ • Divide into 6 slabs 2.5m • Total Mass = 3500kgs A 2 Abottom 2 Asides 2 Aends 2.5m A 2(2.5 12) 2(2.5 12) 2(2.5 2.5) A 132.5m 2 M kg 26.4 2 A m 95.141, F2010, Lecture 21 Department of Physics and Applied Physics 12m I Container • Divide into 6 slabs Top & Bottom: M topbot Atopbot M topbot 1584kg 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Ends • Use parallel axis theorem 2.5m h=6m 95.141, F2010, Lecture 21 Department of Physics and Applied Physics I sides 121 M 2 Mh 2 M ends 330kg Sides • Use parallel axis theorem I sides 121 M 2 Mh 2 M sides 1584kg h 1.25m 12m 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Conservation of Momentum • Skinner seems to be making one rotation every 2 seconds. 95.141, F2010, Lecture 21 Department of Physics and Applied Physics Finally • In the clip, it takes about 20 seconds to turn the container 90 degrees. 95.141, F2010, Lecture 21 Department of Physics and Applied Physics