Physics: Principles with Applications, 6th edition Lecture PowerPoints Chapters 7, 8 Physics: Principles with Applications, 6th edition Giancoli © 2005 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

Momentum Center of Mass Up to this point, we have assumed that all object move as if they were particles The center of mass of an object is that point within the object that acts like a particle when the object is moving

Momentum Center of Mass “One point in a moving object which would act like a particle subjected to the same net force” To determine center of mass (CM) Place two objects of masses mA and mB along x-axis Defined at xCM = (mAxA + mBxB) / M , where M is total mass Motion of CM of a system of particles is directly related to net force acting on system as a whole Linear momentum of a system of particles is product of mass M and velocity of CM

Momentum Center of Mass

Momentum Center of Mass For center of mass in two dimensions Calculate xCM using x values Calculate yCM using y values Total mass is located at coordinates xCM, yCM

Momentum Momentum and Relation to Force An object’s momentum is product of an object’s mass and its velocity P = mv If you only see work “momentum”, it means linear momentum Velocity is a vector, so momentum is a vector Direction is the direction of the velocity vector Units for momentum are kg m/s (no special name) The rate of change of momentum of an object is equal to the net force applied to it ΣF = Δp / Δt

Momentum Momentum and Relation to Force Example 7-1: Force of a tennis serve (see notes) If a tennis ball leaves a racket at 55m/s, and the 0.060kg ball is in contact with the racket for 4 x 10-3s, what is the average force on the ball?

Momentum Momentum and Relation to Force

Momentum Conservation of Momentum Take two billiard balls, headed in opposite directions, which collide momentum before = momentum after mA vA + mB vB = mA v’A + mB v’B

Momentum Conservation of Momentum The Law of Conservation of Momentum: “The total momentum of an isolated system of objects remains constant” System is a set of objects we choose An isolated system is one without external forces If net external forces not zero, then momentum won’t be conserved In this case, redefine the system to include the objects exerting these forces Then momentum will be conserved

Momentum Conservation of Momentum Example 7-4: Rifle recoil (see notes) Calculate the recoil velocity of a 5.0kg rifle that shoots a 0.020kg bullet at 620m/s

Momentum Conservation of Momentum

Momentum Collisions and Impulse Since F = Δp / Δt, F Δt = Δp F Δt has a name: Impulse Impulse is often represented by the letter J Since Δp = m Δv, can also use F Δt = m Δv Units are kg m/s or N s Total change in momentum is equal to the impulse Useful for forces that act over a very short period of time, even though force is not constant

Momentum Collisions and Impulse

Momentum Collisions and Impulse Example 7-6: Bend your knees (see notes) a) Calculate the impulse experienced when a 70 kg person lands on firm ground after jumping from a height of 3.0 m. b) Estimate average force landing with stiff legs (happens over 1 cm) and c) with bent legs (happens over 50 cm)

Momentum Collisions and Impulse a) v = 7.7 m/s just as he hits the ground F Δt = m Δv ; 70 kg x (0 m/s – 7.7 m/s) = -540 N s Force acts upward (opposite downward momentum) b) Person decelerates from 7.7 m/s to 0 m/s over 0.01 m Average velocity is (7.7 m/s + 0 m/s) / 2 = 3.9 m/s Since v = d / Δt, Δt = d / v = 10-2 m / 3.9 m/s = 2.6x10-3 s Average F = -540 N s / 2.6x10-3 s = 2.1 x 105 N c) Same as b except distance for deceleration is 0.50 m Δt = d / v = 5 x 10-1 m / 3.9 m/s = 0.13 s Average F = -540 N s / 0.13 s = 4.2 x 103 N

Momentum Collisions and Impulse Rock climbers use nylon ropes on steep cliffs If a rock climber loses grip, will begin to fall Momentum will ultimately be halted by means of the rope The ropes are made of nylon or similar stretchy material If the rope stretches when pulled taut by the falling climber's mass, force applied on climber over longer time. Longer time for breaking momentum means less force exerted on the falling climber. Source: www.physicsclassroom.com Momentum Real-World Applications

Momentum Conservation of Energy and Momentum in Collisions If two (hard) objects collide, their KE is the same before and after the collision; called an elastic collision Few perfect elastic collisions, since energy ‘lost’ as sound and heat But collision of two billiard balls very close to being perfectly elastic, so treated as such

Momentum Elastic Collisions in One Dimension Head-on collision two small objects along one line KE and momentum conserved, so collision elastic After algebraic manipulation (not shown) Relative speed of objects same magnitude but opposite direction after collision

Momentum Elastic Collisions in One Dimension Example 7-7: Pool or billiards (see notes) Billiard ball A of mass mA moving with speed vA collides head-on with billiard ball B of equal mass (mB = mA) at rest (vB=0). What are speeds of balls after collision? During collision, ball A comes to a stop; ball B moves in same direction at ball A’s initial velocity

Momentum Collisions in Two or Three Dimensions Need to find x, y components of each path Momentum is conserved in x direction and in y direction

Momentum Conservation of Energy and Momentum in Collisions If two objects collide, and KE is not conserved, called an inelastic collision Even though KE not conserved, total energy is conserved

Momentum Inelastic Collisions In inelastic collisions, KE is not conserved If objects stick after collision, is said to be completely inelastic Example 7-9: Railroad cars (see notes) Railroad car A mA=10,000 kg) traveling at vA = 24.0 m/s strikes an identical car B (mB = mA) at rest (vB = 0 m/s). Calculate how much of initial KE is converted to other forms.

Momentum Inelastic Collisions Before collision, only car A moving, so KE = ½ m(vA)2 = ½ 10000kg x (24 m/s)2 = 2.88 x 106 J After collision, both cars move together at 12 m/s,* KE= ½ (20000kg) (12 m/s)2 = 1.44 x 106 J So energy converted to other forms = 2.88 x 106 J – 1.44 x 106 J = 1.44 x 106 J * By conservation of momentum

Momentum Multimedia

Torque and Angular Momentum Objects, such as wheels rotate around an axis of rotation It is easier to measure rotation in radians (rad) instead of degrees 1 radian = angle subtended by arc with length = radius Θ = l / r If l = r, Θ = 1 rad Note: counterclockwise rotation is + clockwise rotation is -

Torque and Angular Momentum Rotational motion Average angular velocity Analogous to linear velocity ω = Δθ/Δt Units are rad/s Angular velocity is a vector quantity

Torque and Angular Momentum Rotational motion Average angular velocity Direction of angular velocity according to right-hand rule

Torque and Angular Momentum Angular acceleration Analogous to linear acceleration α = Δω/Δt Units are rad/s2

Torque and Angular Momentum Applying the same force at different points, rA and rB If rA = 3•rB, FB must be 3•FA Perpendicular distance from axis of rotation to line along which force acts is called the lever arm or moment arm Force times lever arm = moment of force about the axis = torque (τ) τ = r┴F = rF┴ (┴ means perpendicular) τ = Frsinθ Longer lever arm means less force must be applied means more torque

Torque and Angular Momentum Both force and radius are vectors Torque is the vector product of multiplying force x radius: τ = F d sin ϴ Direction of torque determined by right-hand rule

Torque and Angular Momentum Moment of Inertia For a mass in a circle of radius r around an axis of rotation, F = mrα, where m is mass, r is radius and α is angular acceleration Multiply both sides by r to get τ = mr2α mr2 represents rotational inertia and is called moment of inertia (I) Consider moments of inertia for all particles making up a rigid rotating object: Στ = Iα Analogous to F = ma

Torque and Angular Momentum Moments of inertia for some various objects of uniform composition

Torque and Angular Momentum Angular momentum (L): L = Iω, where I is moment of inertia and ω is angular velocity about axis of rotation Under some circumstances, angular momentum is conserved “The total angular momentum of a rotating object remains constant if the net torque acting on it is zero” In this case, Iω is constant

Torque and Angular Momentum Angular momentum (L) Is a vector quantity Direction of angular momentum determined by right-hand rule

Torque and Angular Momentum