Lecture 13 REVIEW Physics 106 Spring 2006 http://web.njit.edu/~sirenko/ What should we know? Vectors addition, subtraction, scalar and vector multiplication Trigonometric functions sinθ, cos θ, tan θ, θ = tan -1 (a/b), etc. Integration and Derivatives (basic concepts) 2x = (x 2 ) SI Units Newton s Laws F = ma F 12 = -F 21 Energy Conservation Kinetic Energy, Potential Energy, and Work Circular motion and Centripetal Force a c = v 2 /R 04/19/2006 Andrei Sirenko, NJIT 1 04/19/2006 Andrei Sirenko, NJIT 2 What should we know? Angular variables angular velocity, angular acceleration, etc. Rotational Inertia Kinetic Energy of Rotation Angular Momentum and Torque Newton s law of Gravitation and planetary motion Newton s Laws I. If no net force acts on a body, then the body s velocity cannot change. II. The net force on a body is equal to the product of the body s mass and acceleration. III. When two bodies interact, the force on the bodies from each other are always equal in magnitude and opposite in direction (F 12 = -F 21 ) Force is a vector Force has direction and magnitude Mass connects Force and acceleration; Satellite orbits, Potential and Kinetic Energy of a satellite Oscillations and Pendulums 04/19/2006 Andrei Sirenko, NJIT 3 04/19/2006 Andrei Sirenko, NJIT 4
Components of Vectors: Vector Multiplication Dot product Length (Magnitude) 04/19/2006 Andrei Sirenko, NJIT 5 04/19/2006 Andrei Sirenko, NJIT 6 Vector Cross Product The value of cross product: c = a b sinφ φ=0 c = 0 φ=π/2 c = a b (max) Order is important: 04/19/2006 Andrei Sirenko, NJIT 7 04/19/2006 Andrei Sirenko, NJIT 8
What does zero mean? t = 0 beginning of the process x = 0 is arbitrary; can set where you want it x 0 = x(t=0); position at t=0; do not mix with the origin v (t) = 0 x does not change x(t) - x 0 = 0 v 0 = 0 v(t) = at; x(t) - x 0 = at 2 /2 a = 0 v(t) = v 0 ; x(t) - x 0 = v 0 t Uniform Circular Motion Centripetal acceleration Period a 0 v(t) = v 0 +at; x(t) - x 0 = v 0 t +at 2 /2 help: t = (v - v 0 )/a x - x 0 = ½(v 2 -v 02 )/a a = (v - v 0 )/t x - x 0 = ½ (v + v 0 )t Acceleration and velocity are positive in the same direction as displacement is positive 04/19/2006 Andrei Sirenko, NJIT 9 04/19/2006 Andrei Sirenko, NJIT 10 ma c = mv 2 /R = ΣF (all forces along the direction towards the center) Gravitational Force: mg down to the ground Kinetic Energy: Potential Energy: Gravitation: Tension Force: along the string T Elastic(due to spring force): Normal Force: N perpendicular to the support Static Friction Force maximum value F fr max = µ st N ma = N mg ma = mv 2 /R 04/19/2006 Andrei Sirenko, NJIT 11 1 U K y mg 2 Conservation of Mechanical Energy 04/19/2006 Andrei Sirenko, NJIT 12
Kinetic Energy: Potential Energy: Gravitation: Elastic(due to spring force): Examples for Energy Conservation + + Kinetic Energy changes Gravitational Potential Energy Elastic Potential Energy Total Mechanical Energy = Const. K f -K i = W = mgy - W friction 1 2 U = 0 K = 0 U K Conservation of Mechanical Energy W friction E f -E i = - W friction =f k d cos180 =-mg µ d cos18 04/19/2006 Andrei Sirenko, NJIT 13 04/19/2006 Andrei Sirenko, NJIT 14 Particle: Linear Momentum Completely Inelastic Collision Collisions in 1D Conservation of Linear Momentum works! System of Particles: Extended objects: 04/19/2006 Andrei Sirenko, NJIT 15 Half the original Kinetic Energy 04/19/2006 Andrei Sirenko, NJIT 16
Rotational Kinematics: Linear Displacement Angular Displacement Linear Velocity Angular Velocity Linear Acceleration Angular Acceleration Radian 1 Radian = 180 / π 57.3 The arc length is equal to the radius Circle: 360 = 2π radians 6.283 radians ½Circle: 180 = π radians 3.1415 radians 04/19/2006 Andrei Sirenko, NJIT 17 Radians = degrees (π /180) 1 degree = π /180=0.0174532925 radians. 180 = 3.14156 radians 90 = 1.5708 radians 45 = 0.7854 radians 04/19/2006 Andrei Sirenko, NJIT 18 Rotation: Angular Displacement Angular Velocity Angular Acceleration Kinetic Energy of Rotation K = ½ mv 2 Point mass (no rotation); v of the COM v = ωr System of particles or an object 04/19/2006 Andrei Sirenko, NJIT 19 04/19/2006 Andrei Sirenko, NJIT 20
Rotational Inertia Rotational Inertia I (a) I (b) I rotational equivalent of mass m Main difference between m and I: Rotational Inertia depends on the direction of rotation! 04/19/2006 Andrei Sirenko, NJIT 21 04/19/2006 Andrei Sirenko, NJIT 22 Parallel-Axis Theorem for Rotational Inertia Calculate I com for the axis going through the COM Use Parallel-Axis Theorem to calculate I Torque: τ The value of torque: τ = r F sinφ φ=0 τ=0 φ=π/2 τ= r F (max) In vector notation form: τ = [r F] 04/19/2006 Andrei Sirenko, NJIT 23 04/19/2006 Andrei Sirenko, NJIT 24
Newton s Second Law for Rotation Torque causes the change in ω τ net = I α Rotational equivalent of F = ma Angular Displacement Angular Velocity Angular Acceleration Force F = ma τ net = I α 04/19/2006 Andrei Sirenko, NJIT 25 04/19/2006 Andrei Sirenko, NJIT 26 04/19/2006 Andrei Sirenko, NJIT 27 04/19/2006 Andrei Sirenko, NJIT 28
Work and Rotational Kinetic Energy Work kinetic kinetic energy theorem Work, rotation about fixed axis Angular Momentum: Definition: Angular Momentum for rotation System of particles: l = r m v sinφ l = I ω [kg m 2 /s] Work, constant torque Power, rotation about fixed axis Torque: 04/19/2006 Andrei Sirenko, NJIT 29 04/19/2006 Andrei Sirenko, NJIT 30 Sample Problem XII 5 Newton s 2 nd Law A penguin of mass m falls from rest at point A,, a horizontal distance D from the origin O of an xyz coordinate system. a) What is the angular momentum of l of the penguin about O? b) About the origin O,, what is the torque τ on the penguin due to the gravitational force F g? 04/19/2006 Andrei Sirenko, NJIT 31 04/19/2006 Andrei Sirenko, NJIT 32
Linear Momentum Angular Momentum [kg m 2 /s] Linear Momentum Conservation: Both, elastic and Inelastic collisions [kg m/s] Both are vectors Angular Momentum Conservation: 1. Define a reference frame 2. Calculate P before the collision 3. Compare with P after the collision m I v ω For rotating body: If the external torque is equal to zero, L is conserved 1. Define a rotational axis and the origin 2. Calculate L before interaction or any changes in I 3. Compare with L after the interaction or any change in I 04/19/2006 Andrei Sirenko, NJIT 33 04/19/2006 Andrei Sirenko, NJIT 34 Conservation of Angular Momentum Angular momentum of a solid body about a fixed axis Rolling Smooth rolling motion Law of conservation of angular momentum (Valid from microscopic to macroscopic scales!) 04/19/2006 Andrei Sirenko, NJIT 35 If the net external torque τ net acting on Rotation and Translation Reference frame a system is zero, the angular momentum L of the system remains constant, no matter what changes take place within the system 04/19/2006 Andrei Sirenko, NJIT 36
Kinetic Energy Sample Problem X12 1: 1: A uniform solid cylindrical disk (M( = 1.4 kg, r = 8.5 cm) roll smoothly across a horizontal table with a speed of 15 cm/s. What is its kinetic energy K? Stationary observer Parallel axis axis theorem A rolling object has two types of kinetic energy: a rotational kinetic energy due to its rotation about its center of mass and a translational kinetic energy due to translation of its center of mass. Forces The acceleration tends to make the wheel slide. A static frictional force f s acts on the wheel to oppose that tendency. 04/19/2006 Andrei Sirenko, NJIT 37 04/19/2006 Andrei Sirenko, NJIT 38 Angular Momentum Conservation: If the external torque is equal to zero, L is conserved M r M 1. L i = L bullet = m v r sin(π/2) =??? 2. L f = I ω = (Mr 2 + Mr 2 + mr 2 ) ω f = =2 kg m 2 /s Free falling / sliding without friction: 04/19/2006 Andrei Sirenko, NJIT 39 1. M = 1 kg 2. m = 10 g = 0.01 kg 3. r = 1 m 4. ω i = 0 ω f = 1 rad/s 5. v bullet =? K f /K i =? 1. Define a rotational axis and the origin 2. Calculate L before interaction or any changes in I 3. Compare with L after the interaction or any change in I m 3. L i = L f (angular momentum conserv.) 4. v bullet = ω f (2Mr 2 + mr 2 )/mr = 200 m/s 5. K i = ½ m v 2 bullet = 200 J 6. K f = ½ I ω 2 = 1 J 7. K f /K i = 1/200 04/19/2006 Andrei Sirenko, NJIT 40
Balance of Forces: Equilibrium Balance of Torques: Newton s Law of Gravitation (known since 1665) 1. The vector sum of all the external forces that act on the body must be zero. 2. The vector sum of all the external torques that act on the body, measured about any possible point, must be zero. 3. The linear momentum P of the body must be zero. 4. The gravitational force F g on a body effectively acts on a single point, called the center of gravity (cog) of the body. If g is the same for all elements of the body, then the body s s cog is coincident with the body s s center of mass. 04/19/2006 Andrei Sirenko, NJIT 41 04/19/2006 Andrei Sirenko, NJIT 42 Only six planets, including the Earth, were known until the 18th Century Planets and Satellites: Kepler's Laws Copernicus 1510 04/19/2006 Andrei Sirenko, NJIT 43 04/19/2006 Andrei Sirenko, NJIT 44
Potential Energy : U U between r 1 and r 2 is the work done by the Gravitation Force during the move from r 1 to r 2 : Potential Energy : Is it U = mgh or, anyway a? It is the same thing, just different zero levels. is more universal (always correct) Reference point r = 0 r 1 1 r 2 1 2 2 r h U = mgh works for h << r, zero at the Earth surface always works, zero at U = 0; at infinity! (far away) 04/19/2006 Andrei Sirenko, NJIT 45 U = GMm/r GMm/(r+h) = GMm(r+h-r)/(r (r+h)) = mh [GM/(r (r+h))] mgh 04/19/2006 Andrei Sirenko, NJIT 46 R Escape Speed: From energy conservation: E 1 = mv 2 /2 GmM/R E 2 = 0 (velocity is small) v 2 = 2GM/R= 2gR v = (2GM/R) ½ 11,200 m/s R First Satellite Speed: Newton s s cannon in 1687 in Principia Mathematica v satellite (gr) ½ v satellite 8,000 m/s g 8.70 m/s 2 An object in orbit is weightless not because 'it is beyond the earth's e gravity' but because it is in 'free-fall' fall' - just like a skydiver. 04/19/2006 Andrei Sirenko, NJIT 47 04/19/2006 Andrei Sirenko, NJIT 48
Potential and Kinetic Energy Potential Energy Potential Energy Satellites: Orbits and Energy Kinetic Energy for the orbital motion Kinetic Energy for the orbital motion Total Energy Total Energy: 04/19/2006 Andrei Sirenko, NJIT 49 04/19/2006 Andrei Sirenko, NJIT 50 Simple Harmonic Motion Simple Harmonic Motion (SHM) 1. Amplitude is different 2. Period (or frequency) is different. 3. Phase is different. 04/19/2006 Andrei Sirenko, NJIT 51 04/19/2006 Andrei Sirenko, NJIT 52
Displacement, Velocity, and Acceleration of SHM Displacement, Velocity, and Acceleration of Simple Harmonic Motion 04/19/2006 Andrei Sirenko, NJIT 53 04/19/2006 Andrei Sirenko, NJIT 54 Energy of SHM + mgx I α = τ 04/19/2006 Andrei Sirenko, NJIT 55 04/19/2006 Andrei Sirenko, NJIT 56
Any rigid body behaves like SHO close to stable equilibrium Compare to: for SHO 04/19/2006 Andrei Sirenko, NJIT 57 04/19/2006 Andrei Sirenko, NJIT 58