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## Physics I 95.141 LECTURE 21 11/24/10

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**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. 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. • a) (10 pts) What is the moment of inertia of this system? 30cm 15cm 2kg**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. • 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? 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. • c) (5pts) What is the angular acceleration of the rotating cylinder/mass system? 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. • d) (10pts) Determine the Kinetic Energy, as a function of time, associated with i) the rotating system and ii) the falling mass. 30cm 15cm 2kg 10kg**Review Example**• What is the vector cross product of the two vectors:**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.**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 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 • Conservation of Angular Momentum • With no external torques acting on a system, the angular momentum of the system is conserved. • Vector Cross products**Review of Angular Motion**• We know equations of motion for angular motion • We know torques cause angular acceleration • Objects can have rotational kinetic energy • And angular momentum • So why the cross product?**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 FROM LECTURE 19**Torque and the Cross Product**• However, we since learned that • 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 FROM LECTURE 19**Angular Momentum of a Particle**• We have already defined angular momentum as • But this definition is for objects rotating with some angular velocity and moment of inertia around an axis of rotation. • More general, alternate, definition:**Equivalence of our two definitions**• Suppose we have a mass rotating around an axis • Use cross product y x m=2kg 2m**Equivalence of our two definitions**• Suppose we have a mass rotating around an axis • Use y x m=2kg 2m**So why use cross product?**• Cross products are messy…why would we ever use them, instead of the simpler • 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!**Example**• Calculate the angular momentum (about the origin) of the rock of mass m dropped from rest off the cliff. d (0,0)**Example**• What Torque is exerted (about the origin) on the rock? d (0,0)**Relationship of torque to angular momentum**• When we discussed linear momentum, we revised Newton’s 2nd Law to state • Similarly, we can write an expression for net torque in terms of angular momentum • Double check with our falling rock:**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.**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. L constant!**Does this make sense?**• What is happening? • What do we need to know? • System= Container + Skinner**Moments of Inertia**• Skinner=point mass • Shipping Container 2.5m 2.5m 12m Mass=3,500kg**σ Container**• Determine surface mass density σ • Divide into 6 slabs • Total Mass = 3500kgs 2.5m 2.5m 12m**I Container**• Divide into 6 slabs • Top & Bottom:**Ends**• Use parallel axis theorem h=6m**Sides**• Use parallel axis theorem**Conservation of Momentum**• Skinner seems to be making one rotation every 2 seconds.**Finally**• In the clip, it takes about 20 seconds to turn the container 90 degrees.