Equation of Motion. Start with the definition of force (rate of change of momentum) and use what we know about momentum.

Size: px

Start display at page:

Download "Equation of Motion. Start with the definition of force (rate of change of momentum) and use what we know about momentum."

Catherine York
6 years ago
Views:

1 Notes Syllabus update: If you are >10 min late to a discussion section, you will not be counted as participating in the discussion. Results in 50% loss in the next group quiz score. Quiz 2: Next Friday, Oct 20. Go to Tate B50 if your last name begins with A-Lim. Go to Fraser 101 if your last name begins with Liu-Z. Can use non-programmable calculator and 1 sheet (2 pages) of notes. A crib sheet will also be provided. Covers up to and including Force.

2 Force Force exerted on an object is the time rate of change of object s momentum. Example: if you are running into a concrete wall, you experience more force than if running into a mattress. You have the same momentum in both cases, but in the first case you come to a stop much more suddenly. There are many examples of force: Contact forces: pushing, pulling, springs, tension on the rope, friction Field forces: gravity, electromagnetic force There could be multiple forces on an object, they add vectorially. Example: multiple people pulling a rope.

3 Equation of Motion Start with the definition of force (rate of change of momentum) and use what we know about momentum. F F p t dp dt = t 0 = dv = m dt = ma dp dt Force = mass x acceleration (Newton s Second Law). Also known as the Equation of Motion. We can solve it for acceleration, which then tells us how the object moves. Unit of force: [mass x acceleration] = kg m/s 2 = N (Newton). Note: Law of Inertia is also known as Newton s First Law. F =

4 Action and Reaction Recall for an isolated system, the total change in momentum must be zero. Suppose two objects in an isolated system interact with each other (like two carts on a track). As they interact (collide), some momentum is exchanged between them. Object 1 exerts force on object 2, and object 2 exerts force on object 1. p1 + p2 = 0 p1 p2 + = 0 t t dp1 dp2 + = 0 dt dt F + F = Note that F 1 is the force exerted by object 2 on object 1. We can write F 1 = F 21. Similarly, F 2 = F 12. Pulling this together: F 21 F 12 = Action causes equal and opposite reaction, Newton s Third Law. Demo: 1H10.20 Newton's Sailboat

5 Addition of Forces Force is a rate of change of momentum, so it is a vector. Multiple forces add vectorially. Free body diagrams: sketches of various forces acting on an object. Very useful in analyzing problems. Example: Mass hanging on a rope T, tension in the rope Example: Book on a table. N, normal force (resistance from the table) F g, gravitational force F g, gravitational force

6 Translational Equilibrium Forces add vectorially: F total = Fi dp = dt = ma If the net sum of all forces on an object is zero, the object will have zero acceleration. It will stay at rest or move at a constant velocity. The object is then in a translational equilibrium. Or, going backwards: If an object experiences zero acceleration, the sum of forces acting on it is zero.

7 Forces in Systems Now consider a system of objects of masses m i. We know that the momenta add: p = p Rate of change of total momentum is the sum of all external forces on component masses and of all internal forces among the internal masses. If there are no external forces, we know the momentum is conserved, dp tot /dt = 0, so the sum of all internal forces must be zero. 0 = i Fi, int tot dp dt F tot tot = = i i i i i dp dt F = i i F i, ext + i F i,int Not surprising: for every action there is an equal and opposite reaction!

8 Forces in Systems p On the other hand, we saw: tot dr i = pi = mivi = mi = i i i dt dt i d m r i i = d dt ( Mr cm ) = Mv cm Differentiating both sides: So, the center of mass motion is completely described by the sum of all external forces acting on the components of the system. For a single object: dptot = dt F = tot Ftot Fi, ext i dp dt dv = M dt = Ma cm = F = Fi i cm = Ma = Ma cm

9 Impulse For a non-isolated system, the change in momentum is equal to an impulse received by the system from the environment. p = J p J Ftot = = t t J = F t tot This is the impulse equation. It states that the impulse received by a system is equal to the product of the net force acting on the system and the time interval over which the force was applied. Note that this is true only instantaneously, or if the force is constant in time. Otherwise, if the force changes over time, we would have to integrate F tot (t).

10 Hooke s Law Consider a spring, fixed on one end. The other end has a natural position x 0, corresponding to the natural length of the spring. Compressing the spring, means the end of the spring moves to some new position x. The spring will try to push back it will exert force on whatever is compressing it. Empirically, the force is proportional to how much the spring is compressed. Hooke s law: F spring = k ( x x0 ) Note that the elastic force of the spring always tries to oppose the displacement of the spring end. Linear Restoring Force. A common restoring behavior in various situations.

Gravity Many levels of understanding gravity: Works well on Earth, sufficient to describe common life situations. Newton s law of gravitation, works on Earth and explains celestial mechanics.

11 Gravity Many levels of understanding gravity: Works well on Earth, sufficient to describe common life situations. Newton s law of gravitation, works on Earth and explains celestial mechanics. Gauss law form of Newton s law: flux of the gravitational field through an enclosed surface is proportional to the mass enclosed by the surface. Einstein equations: gravity is curvature in space-time due to presence of matter (or energy)! But, still not unified with all other known forces: electromagnetism, weak and strong nuclear forces. Maybe we still didn t get it right! F = mgj ˆ GMm F = rˆ 2 r Φ = g da = 4πGM 1 2 S + 8πG Λ = T c Rµν gµν R gµν 4 µν

12 Force of Gravity We have stated that all objects on Earth feel the same downward acceleration, implying that they would feel the gravitational force: F g = ma = mg = mgyˆ where g = 9.8 m/s 2 at the Earth s surface. Often called the weight. Weight units are N or lb. Often incorrectly people talk about pounds as mass (eg relating kg and pound). The amplitude of the gravitational force is not exactly the same everywhere on Earth. Earth is not exactly round, slightly flattened at the poles. Gravity is 0.5% stronger at the poles. Also, we have seen that the centripetal acceleration at the equator is ~0.3% of the gravitational acceleration. This further weakens the gravitational pull at the equator.

13 Newton s Law of Gravity Newton proposed a law of gravity in order to explain the planetary motion. Gravitational force between two objects is proportional to the mass of each of the two objects, and inversely proportional to the square of the distance between the objects. And it is always attractive! F GMm = 2 r The constant G is known as Newton s constant, and it is one of the fundamental constants of nature. Measured (first by Cavendish in 1771): G = x m 3 kg -1 s -2 Assumes point masses, but also works for spherical objects. rˆ

14 Newton s Law of Gravity Hence, on Earth g = GM 2 R where M is the mass of the Earth and R is the radius of the Earth. So, measuring G and knowing the radius of the Earth allows us to estimate the mass of the Earth: M = gr G 2 = m s ( m) 11 3 m kg s 2 2 = kg Cavendish was the first to weigh the Earth.

15 Equivalence Principle Recall Newton s Second Law and Law of Gravitation: F = mgj ˆ m is the property of the object that relates the force applied on it and its acceleration. It is a measure of inertia. Equivalence principle: the two m s are identical! Inertial and Gravitational mass are the same. This is not obvious! Didn t have to be this way! Equivalence Principle is the basis of General Relativity. Careful experiments are conducted to check this principle so far see no evidence that the principle is violated in nature. Pendula and torsion balances with different masses. F GMm = 2 r rˆ m is the property of the object that defines the amplitude of the gravitational force the object feels

16 Electrostatic Force Also known as the Coulomb s Law: F Coul = k Qq 2 r rˆ Here, Q and q are the electrical charges of the two objects, r is their separation, and k is a coefficient. The same functional form as Newton s Law of Gravity. But could be attractive (if Q and q have opposite signs) and repulsive (if Q and q have the same signs). Note that the force depends on a new property of the object (i.e. its charge), and has nothing to do with the mass of the object. In SI units: Charge is measured in Coulombs (C). Measured: k = 8.99 x 10 9 N m 2 / C 2.

17 So, What Now? Use Newton s laws to study problems involving forces and accelerations. Careful thinking helps, following a few steps: Identify masses that interact with each other Draw a force diagram for each mass Define a coordinate system Write down equations of motion (F=ma) for each mass Possibly add constraints, eg due motion in one dimension. Solve equations.

18 Example: Skater and Box A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N in the x- direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10 m? How long does it take to cover this distance?

20 Example: Freight Train Three freight cars each of mass M are pulled with force F by the locomotive. Find force on each car, neglect friction.

22 Example: Inclined Plane A box is placed at rest on the top of a ramp of height h that makes an angle θ with the horizontal. What is the speed of the box when it hits the bottom?

24 Example: Inclined Plane (2) Now allow the ramp to move along the frictionless horizontal surface. How are the accelerations of the block and of the ramp related? Compute these accelerations.

27 Clicker Question You are on a sail-boat, on a perfectly still day. You get bored waiting for the wind, so you decide to provide one. You plug in your fan to the boat s battery and now you have air flow! You aim the fan at the sail. Which of the following is true? a) The boat happily sails forward (in the direction of the fan s airflow) b) The boat sails backward (opposite the airflow direction) c) The boat does not move d) The boat immediately sinks due to embarrassment

28 Clicker Question Following up on the previous question, now you turn your fan away from the sail. Which of the following is true? a) The boat sails in the direction of the fan s airflow b) The boat sails in the direction opposite the airflow c) The boat does not move d) The marine wildlife gathers to watch the show

30 Atwood Machine A If the pulley is accelerating upwards at rate A, what is the acceleration of each mass and what is the tension in the string. Demo: 1G10.40 Atwood's Machine

32 Atwood + Inclined Plane What are the acceleration of the two masses and the tension in the string? Which way does the system move?

34 Hanging a Stoplight A 35.0 kg stoplight is hanging as shown from a pole. What are the tensions in the two wires?

36 Broken Ankles? Suppose a person of mass M jumps to the ground from height h, and her center of mass moves downward a distance s during the collision time (with the ground). Will her ankles break?

38 Momentum Flow When holding a water hose, need to apply a force to counteract the water. How large is the force? One way to approach this is to assume the water to be made up of droplets, each of mass m and speed v, separated by distance l. Then, I I F droplet hand avg = Fdt = mv = F T = mv = avg mv T = mv l 2 In other words, a succession of collisions results in an average force that our hand has to apply to the water hose. Common situation when dealing with fluids or gases, or even photons and other beams of particles in physics.

39 Momentum Flux Now, instead of droplets, suppose we have a continuous flow of matter, through a cross sectional area A. If the matter has volume density ρ, then: Mass per unit length is: ρa Momentum per unit length is: ρav (equivalent of mv/l in the droplet example) Then the force can be written as: Or, in vector notation: F = dp dt Often it is useful to think in terms of momentum flow per unit crosssectional area, or momentum flux density: 2 F = = ρav 2 dp dt vˆ = ρav J 2 = ρv vˆ

40 We can then write: A Momentum Flux dp ( F = = J A)vˆ dt where the vector has amplitude equal to the area surface and direction that s perpendicular to the surface. Some ambiguity: the direction could be into or out of the system. Convention to take it to be into the system. J A So is positive if the momentum flows into the system.

41 Pressure of a Gas Consider a gas of n particles per unit volume, with each particle of mass m, inside of a container. What is the pressure (force per unit area) on the surface of the container due to this gas?

43 Suspended Garbage Can An inverted garbage can of weight W is suspended in air by water from a geyser. The water shoots up from the ground at speed v 0 at a constant rate K kg/s. What is the height at which the garbage can rides?

45 Continuously Varying Mass Start with an object of mass M, velocity v. Suppose it s mass is changing continuously: Either shedding its mass (like a rocket). Or mass is falling on it (travelling through some medium). Model it as ΔM falling on it (or leaving it) at velocity u. F F F F F net net net net net t = p2 p1 = ( M + M )( v + v) ( Mv + Mu) t = M v + M ( v u) + M v v M M = M + ( v u) + v t t t dv dm dm = M + ( v u) + (0) dt dt dt dv dm = M v rel v u v dt dt rel = Newton s Second Law For Continuously Variable Mass

46 Growing Raindrop A raindrop of initial mass M 0 starts to fall from rest under gravity. It gains mass as it falls through the cloud at a rate proportional to its instantaneous mass and instantaneous velocity dm = dt kmv Show that the speed of the drop eventually becomes constant and give an expression for the terminal speed.

48 Rocket Assuming that the rocket of initial mass M 0 is continuously burning fuel at a constant rate R, ejecting the gas at velocity u with respect to the rocket (downward), find the velocity of the rocket as a function of time during the launch from Earth.

49 Note the second term. Because of it, it is better to burn the fuel as quickly as possible.

Practice. Newton s 3 Laws of Motion. Recall. Forces a push or pull acting on an object; a vector quantity measured in Newtons (kg m/s²)

Practice. Newton s 3 Laws of Motion. Recall. Forces a push or pull acting on an object; a vector quantity measured in Newtons (kg m/s²) Practice A car starts from rest and travels upwards along a straight road inclined at an angle of 5 from the horizontal. The length of the road is 450 m and the mass of the car is 800 kg. The speed of