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Physics 1408-001 Principles of Physics Lecture notes are on the web http://highenergy.phys.ttu.edu/~slee/lectures/ Lecture 17 (Serway & Jewett Vol.1) – Ch. 9 February 21, 2007 HW Assignment #5 has been placed on WebAssign, and is due by 11:30pm on Tuesday, 2/27 Sung-Won Lee [email protected] The solutions for Exam 1 will be posted on the wall; Next to my office (Sci 117) The grades for Exam 1 is in progress!! Chapter 9 Linear Momentum and Collisions Energy 1. 2. 3. 4. 5. 6. 7. Announcement I Linear Momentum and its Conservation Impulse and Momentum Collisions in 1-dimension 2-dimensional Collisions The Center of Mass Motion of a System of Particles Rocket Propulsion 9.5 The Center of Mass • There is a special point in a system or object, called the center of mass, that moves as if all mass of the system is concentrated at that point • The coordinates of the center of mass are – where M is the total mass of the system • The center of mass can be located by its position vector, rCM • ri is the position of the i th particle, defined by Center of Mass, Example • Both masses are on the x-axis • Center of mass is on the x-axis • Center of mass is closer to the particle with the larger mass Center of Mass, Extended Object, Position • The position of the center of mass can also be found by: • The center of mass of any symmetrical object lies on an axis of symmetry and on any plane of symmetry Center of Mass, Example • An extended object can be considered a distribution of small mass elements, ∆m • The center of mass is located at position rCM 1 1 xcm = ∑ xi ∆m = M ∫ x dm M i 1 1 ycm = ∑ yi ∆m = M ∫ y dm M i As Fig shows, we can divide an extended object into many small cells or boxes, each with the very small mass ∆m The Center of Mass of a Rod Find the center of mass of a thin uniform rod of length L and mass M. Find the tangential acceleration of one tip of a 1.6 m rod that rotates about its center of mass with an angular speed of 6.0 rad/s2. dm dx M = so dm = dx M L L 1 1 M xcm = x dm = x dx ∫ ∫ M M L L L 1 1 = ∫ xdx = ⎡⎢⎣ 12 x 2 = 12 L 0 L0 L at = rα = 12 Lα = 12 (1.6 m)(6.0 rad/s 2 ) = 4.8 m/s 2 9.6 Motion of a System of Particles • Assume the total mass, M, of the system remains constant. • We can describe the motion of the system in terms of the v and a of the center of mass of the system. • We can also describe the momentum of the system and Newton’s 2nd law for the system. • The velocity of the center of mass of a system of particles is • The momentum can be expressed as Acceleration of the Center of Mass • The acceleration of the center of mass can be found by differentiating the velocity with respect to time Forces in a System of Particles • The acceleration can be related to a force • If we sum over all the internal forces, they cancel in pairs and the net force on the system is caused only by the external forces Momentum of System of Particles • The total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass Motion of the Center of Mass, Example • A projectile is fired into the air and suddenly explodes • The total linear momentum of a system of particles is conserved if no net external force is acting on the system • MvCM = ptot = constant when ΣFext = 0 9.7 Rocket Propulsion • The operation of a rocket depends upon the law of conserva tion of linear momentum as applied to a system of particles, where the system is the rocket plus its ejected fuel • With no explosion, the projectile would follow the dotted line • The initial mass of the rocket plus all its fuel is M + ∆m at time ti and velocity v • The initial momentum of the system is pi = (M + ∆m) v • After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion • At some time t + ∆t, the rocket’s mass has been reduced to M and an amount of fuel, m has been ejected • The rocket’s speed has increased by ∆v Rocket Propulsion, 2 • Because the gases are given some momentum when they are ejected out of the engine, the rocket receives a compensating momentum in the opposite direction • Therefore, the rocket is accelerated as a result of the “push” from the exhaust gases • The basic equation for rocket propulsion is • The increase in rocket speed is proportional to the speed of the escape gases (ve) – So, the exhaust speed should be very high The increase in rocket speed is also proportional to the natural log of th e ratio Mi/Mf – So, the ratio should be as high as possible, meaning the mass of t he rocket should be as small as possible and it should carry as mu ch fuel as possible • Chapter 10 Rotation of Rigid Object about a Fixed Axis 1. 2. 3. 4. 5. 6. 7. 8. 9. Angular Position, Velocity, and Acceleration Rotational Kinematics Angular and Linear Quantities Rotational Kinetic Energy Calculation of Moments of Inertia Torque Relationship between Toque and Angular Acceleration Work, Power, and Energy in Rotational Motion Rolling Motion of a Rigid Object Thrust • The thrust on the rocket is the force exerted on it by the ejected exhaust gases Thrust = • The thrust increases as the exhaust speed increases • The thrust increases as the rate of change of mass increa ses – The rate of change of the mass is called the burn rate Rigid Body Rotation A rigid body is an extended object whose size, shape, and distribution of mass do not change as the object moves and rotates. 10.1 Angular Velocity & Angular Acceleration Rotation + Translation Fig illustrates the three basic types of motion of a rigid body Fig. shows a wheel rotating on an axle. Notice that 2 points on the wheel turn through the same angle as the wheel rotates. That’s ∆θ1 = ∆θ2 during same ∆t. So, the 2 points have equal angular velocities. Tangential acceleration : at = dvt d (rω ) dω = =r dt dt dt Let’s define the angular acceleration: Table shows the resulting kinematic equations for constant angular acceleration. These equations apply to a particle in circular motion or to any rigid-body rotation !! d ω at = dt r The derivative in above eq. is the rate of change of velocity We see that the tangential and angular accelerations are related by 10.2 Rotational Kinematic Equations α≡ at = rα 10.3 Angular/Linear Quantities 10.4 Rotational Kinetic Energy • An object rotating about some axis with an angular speed , ω, has rotational kinetic energy even though it may not have any translational kinetic energy • Each particle has a kinetic energy of Ki = ½ mivi2 • Since the tangential velocity depends on the distance, r, from the axis of rotation, we can substitute vi = ωI r • The total rotational kinetic energy of the rigid object is the sum of the energies of all its particles 10.5 Moment of Inertia • The definition of moment of inertia is • The dimensions of moment of inertia: [kg.m2] • We can calculate the I of an object more easily by assuming it is divided into many small volume elements, each of mass ∆mi • We can rewrite the expression for I in terms of ∆m • With the small volume segment assumption, • If ρ is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known where I is called the moment of inertia Notes on Various Densities • Volumetric Mass Density: mass per unit volume: ρ = m/V • Face Mass Density: mass per unit thickness of a sheet of uniform thickness, t: σ = ρt • Linear Mass Density: mass per unit length of a rod of uniform cross-sectional area: λ = M/L = ρΑ Moment of Inertia of a Uniform Thin Hoop • Since this is a thin hoop, all mass elements are the same distance from the center Moment of Inertia of a Uniform Rigid Rod • The shaded area has a mass – dm = λdx = M/L dx • Then the moment of inertia is Moments of Inertia of Various Rigid Objects Moment of Inertia of a Uniform Solid Cylinder • Divide the cylinder into concentric shells with radius r, thickness dr and length L • Then for I The Parallel-Axis Theorem The I depends on the rotation axis. Suppose you need to know the I for rotation about the off-center axis in Fig. I = I cm + Md 2 The parallel-axis theorem If axis of interest is distance d from a parallel axis through the center of mass, the I is Moment of Inertia for a Rod Rotating Around One End • The moment of inertia of the ro d about its center is • D is 1/2 L • Therefore, End of Lecture 17 HW Assignment #4 will be due on 2/20. Before the next lecture, read the text book, chapter Chapter 10. Next lecture: Fridayday, 2/23