Buoyancy Momentum/Impulse Angular Momentum Torque
Newton s Third Law Problem: How can a horse pull a cart if the cart is pulling back on the horse with an equal but opposite force? Aren t these balanced forces resulting in no acceleration? NO!!!
Newton s Third Law Explanation: forces are equal and opposite but act on different objects they are not balanced forces the movement of the horse depends on the forces acting on the horse
Rocket Propulsion In a rocket engine, burning fuel produces hot gases. The rocket engine exerts a force on these gases and causes them to escape out the back of the rocket. By Newton s third law, the gases exert a force on the rocket and push it forward.
Momentum A moving object has a property called momentum that is related to how much force is needed to change its motion. The momentum of an object is the product of its mass and velocity Momentum is given the symbol p and can be calculated with the following equation:
Momentum Momentum quantity of motion Use the unit kg m/s p = mv m p v p: momentum (kg m/s) m: mass (kg) v: velocity (m/s)
Force and Changing Momentum By combining these two relationships, Newton s second law can be written in this way: In this equation mv f is the final momentum and mv i is the initial momentum
Newton s Third Law Action-Reaction Pairs Both objects accelerate. The amount of acceleration depends on the mass of the object. a F m Small mass more acceleration Large mass less acceleration
Conservation of Momentum Law of Conservation of Momentum The total momentum in a group of objects doesn t change unless outside forces act on the objects. p before = p after
Conservation of Momentum Elastic Collision KE is conserved Inelastic Collision KE is not conserved
Momentum Find the momentum of a bumper car if it has a total mass of 280 kg and a velocity of 3.2 m/s. GIVEN: p =? m = 280 kg v = 3.2 m/s p WORK: m v
Momentum The momentum of a second bumper car is 675 kg m/s. What is its velocity if its total mass is 300 kg? GIVEN: p = 675 kg m/s m = 300 kg v =? p WORK: m v
Conservation of Momentum A 5-kg cart traveling at 1.2 m/s strikes a stationary 2-kg cart and they connect. Find their speed after the collision. Cart 1: m = 5 kg v = 4.2 m/s Cart 2 : m = 2 kg v = 0 m/s BEFORE p = 21 kg m/s p = 0 p before = 21 kg m/s Cart 1 + 2: m = 7 kg v =? AFTER m p v v = p m v = (21 kg m/s) (7 kg) v = 3 m/s p after = 21 kg m/s
Impulse = Momentum F = ma
Angular Momentum L = mvr L = angular momentum, m = mass, v = velocity, and r = distance to center of motion L 1 = L 2 m 1 v 1 r 1 = m 2 v 2 r 2 3-15 Section 3.6
Angular Momentum Mass (m) is constant. As r changes so must v. When r decreases, v must increase so that m 1 v 1 r 1 = m 2 v 2 r 2 3-16 Section 3.6
Torque Torque is a twisting action that produces rotational motion or a change in rotational motion. Torque = rf 3-17 Section 3.6
Conservation of Angular Momentum Rotors on large helicopters rotate in the opposite direction Copyright Houghton Mifflin Company. All rights reserved. 3-18 Section 3.6
Conservation of Angular Momentum Figure Skater she/he starts the spin with arms out at one angular velocity. Simply by pulling the arms in the skater spins faster, since the average radial distance of the mass decreases. m 1 v 1 r 1 = m 2 v 2 r 2 m is constant; r decreases; Therefore v increases 3-19 Section 3.6
Buoyancy Upward force exerted by a fluid Archimedes principle: An object immersed wholly or partially in a fluid experiences a buoyant force equal in magnitude to the weight of the volume of fluid that is displaced. An object will float in a fluid if its average density is less than the density of the fluid An object will sink it is average density is greater than the density of the fluid. 3-20 Section 3.6
Discussion People can easily float in the Great Salt Lake in Utah. Why is that? 3-21 Section 3.6
Chapter 3 - Important Equations F = ma (2 nd Law) or w = mg (for weight) F 1 = -F 2 (3 rd Law) F = (Gm 1 m 2 )/r 2 (Law of Gravitation) G = 6.67 x 10-11 N-m 2 /kg 2 (gravitational constant) g = GM/r 2 (acc. of gravity, M=mass of sph. object) r = mv (linear momentum) P f = P i (conservation of linear momentum) L = mvr (angular momentum) L1= m 1 v 1 r 1 =L2 = m 2 v 2 r 2 (Cons. of ang. Mom.) 3-22 Review