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Transcript
Natural Sciences I 1 lecture 4: Newton's Laws of Motion Last week you learned how to work with the concepts of distance, time, motion, forces (vectors) and acceleration. An important general conclusion that emerged from the lectures and reading is that an object tends to continue in its state of motion (or lack thereof) unless acted upon by an unbalanced (net) force. This conclusion describes Galileo's concept of inertia. We'll see shortly that Isaac Newton adopted this conclusion as his First Law of Motion -- and added a couple more. THE LAWS OF MOTION Newton gave credit to Galileo for his First Law of Motion, which is already familiar to you (Galileo thought in terms of frictionless planes (see p. 41 of your text); Newton generalized to real or imagined bodies anywhere on Earth or in space). Every object retains its state of rest or its state of uniform, straight-line motion unless acted upon by an unbalanced force (In his own statement of the first law, Newton referred to the unbalanced force as an "external agency"). This law amounts to a statement that objects tend to resist any change in motion – whether its getting them moving in the first place or changing their velocity vectors in any way. This tendency to resist changes in motion is defined by the inertia of a body, and it is the mass that is a measure of inertia. You can draw upon numerous everyday experiences to test or build your intuition about inertia (the text calls attention to the experiences of a passenger in a bus – a subway might be even better!). As noted in your text, Sir Isaac Newton was born in 1642, the same year that Galileo died. As a student, Newton was shy (almost reclusive) and very absorbed in thinking about science and mathematics – everything from optics and the nature of light to the motions of the planets. In today's society he would probably be something of an outcast among his fellow students. Indeed, he did not receive widespread recognition and respect as a young man. He invented calculus when he was in his early to mid-twenties, but the significance of this contribution was not appreciated until he was much older. Only when Edmund Halley convinced him, in his midforties,to publish his ideas about gravity and the motions of the planets (his Principia) did Newton become widely recognized as one of the greatest scientific minds of all time. 2 The Second Law of Motion We've seen this one already, too. Newton actually used the second law in part to define what he meant by the "external agency" to which he referred in the first law. The formal statement of the second law is The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to the mass of the object In symbols... a F and a 1/m We can combine these two mathematical statements into a F/m or, multiplying both sides of the relationship by m F ma This is getting close to the formal mathematical statement of the second law as you may know it already. We need to do one more thing to convert the statement of proportionality above into an equation. The metric unit of force, called a newton, is defined as the force required to produce an acceleration of 1.0 m/s2 when applied to an object having a mass of 1.0 kg. With this definition, when force is expressed in newtons (N), mass in kg and acceleration in m/s2, we can equate the two sides F=ma which is not only relatively easy to remember, but also a useful aid to keeping your units straight: 1 N = (1 kg) 1 m 2 s = 1 kg m 2 s 3 Here's an intuition builder for the second law that involves a bicycle. To move along at constant speed, the cyclist must exert just enough force on the pedals to exactly compensate the forces tending to slow her down – road and air resistance (Note that a mechanical device is used to convert downward force on the pedals into a horizontal force tending to move the bicycle forward). Fair Fair + Froad Fapplied If the rider is moving at constant speed, Fapplied and (Fair + Froad) must be exactly balancing each other. The second law enters in if the rider wishes to accelerate to a faster speed. If a given additional force to the pedals results in an acceleration of 2 mi/h/s, then twice that additional force would produce and acceleration of 4 mi/h/s. Similarly, if the mass of the rider were doubled (e.g., by a heavy backpack), she would need to exert twice the force to achieve a given acceleration. Solution... Here's an example problem with actual numbers... A bicycle and rider having a total mass of 60 kg accelerates at 0.5 m/s2. How much extra force – i.e., above that needed to maintain the pre-existing speed – was applied in order to achieve this acceleration? What's known and what's asked for... m = 60 kg a = 0.5 m/s2 F = ? F=ma = (60 kg) (0.5 m/s2) = (60.0)(0.5) (kg) kg m 2 s = 30 N = 30 m 2 s 4 Mass & Weight It's probably not a bad time to return to the question of the difference between mass and weight. Weight is the downward force produced by gravity acting on an object. Now you know that F = m a, so you can relate mass to weight and vice versa. weight = (mass) x (acceleration due to gravity) w=mg So, weight is not equal to mass, but it is proportional to mass through the gravitational constant. The Third Law of Motion Newton's third law of motion concerns how a force is produced. Remember that a force can result only from the interaction between two or more objects (this is true whether the force involves contact between objects or operates at a distance, as in the case of gravity or magnetism). Your text uses the example of a spacewalking astronaut interactF ing with a satellite. The two (astronaut on are orbiting the Earth tosatellite) F gether, and any unbalanced (satellite on forces act on both. There astronaut) are no sources of friction, (This is a etc. GPS satellite!) In keeping with Newton's second law, a force would have to be applied to change the motion of the satellite (F = m a). If the astronaut pushes on the satellite for a period of 1 second, the satellite would accelerate during that 1-s period, then move away at some constant speed. Your intuition will probably tell you that the astronaut will also move away from the original position, but in the opposite direction. (Question: How could you determine the relative accelerations of the two bodies – satellite and astronaut?) 5 Newton's thought experiments probably didn't involve astronauts and satellites, but he nevertheless came to the realization that forces occur in matched pairs. You might well arrive at the same conclusion by finding a partner and doing some experiments on ice...or on roller blades! Newton began his statement of the third law as follows: To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts... Your text paraphrases in modern language to avoid ambiguities (Newton's phraseology suggests that one force might be the cause of another)... Whenever two objects interact, the force exerted on one object is equal in size and opposite in direction to the force exerted on the other object. F(on A due to B) = F(on B due to A) Example... Changing the motion of spacecraft with a thruster (a directed jet of exhaust gases or compressed gas) is a good example of the third law in action. Given: 1 kg of gas in a nozzle expands under pressure and accelerates to a speed of 100 m/s in 0.01 sec. Question: If the spacecraft has a mass of 2000 kg, what is its acceleration? Procedure: You're asked to calculate the acceleration of the spacecraft, but you know from the third law that this must be related to the acceleration of the gas, for which you have data. So, Fon gas = mgas a agas = !vgas / !t 2 100 m/s = 0.01 s = (1 kg) (10,000 m/s ) 2 = 10,000 m/s = 10,000 (kg m)/s = 10,000 N 2 (next page) 6 Example (cont'd) From the third law of motion, we can write Fon spacecraft = Fon gas and so aspacecraft = Fon spacecraft / mspacecraft 2 aspacecraft = 10,000 (kg m)/s / 2000 kg 2 = 5 m/s Additional Question: If the acceleration vector is aligned with the velocity vector, could you calculate the magnitude of the change in speed caused by this use of the thruster? MOMENTUM is a concept closely related to the laws of motion. It's familiar from everyday usage (e.g., in sports), and is relatively easy to grasp intuitively. Momentum is defined as the product of the mass of an object and its speed. p=mv so momentum is directly proportional to mass (the higher the mass, the greater the momentum) and directly proportional to speed (the higher the speed, the greater the momentum). It is interesting to note that Newton originally stated the second law in terms of momentum: F !p/!t. The most important thing to remember about momentum is that it is conserved during interactions between objects. This is actually a combination of the second and third laws, and is stated in words: In the absence of external forces, the total momentum of a group of interacting objects remains the same This is called the Law of Conservation of Momentum. Momentum conservation can be observed in the behavior of billiard balls on a pool table. Actual measurements would be difficult to make, but you might be able to convince yourself that if a cue ball moving at a particular speed directly strikes another stationary ball, the sum of the velocities of the two balls will be equal to the original velocity of the cue ball (You would have to make a number of simplifying assumptions to show this – what would they be?). What can you say about the momenta of the astronaut and the satellite (once that have separated from one another) in the cartoon at the bottom of page 4? Conservation of Momentum on a grand scale: the Coriolis Effect N A missile fired straight south appears to veer to the right because of the Earth's rotation under it. The same thing happens with airflow in the atmosphere. gh re su es pr hi This same effect is responsible for the counterclockwise spiral pattern of airflow in hurricanes (in the northern hemisphere) low pressure View looking down on a storm system Flow of air in the absence of a Coriolis effect 7 8 FORCES and CIRCULAR MOTION The First Law of Motion states that objects will move in a straight line unless acted upon by an unbalanced force. We would therefore conclude that an object in circular motion (e.g., the Earth around the Sun) is continually acted upon by an unbalanced force. That force is referred to as centripetal (or "center seeking") force. [Note: the apparent outward force exerted by an object in circular motion – such as a ball being swung in a circular path on a rope – is referred to as centrifugal...but this is not an accurate concept] The magnitude of the centripetal force required to keep an object in a circular path is related to the inertia of the object – i.e., to its mass and acceleration – as in the second law. The relevant equation is ce n 2 aC = v /r (see appendix for derivation) Note that the smaller the radius, the greater the acceleration needed to maintain uniform circular motion. rad trip e ius tal (r) velocity vector for ce Substituting a = F/m into the above equation gives the magnitude of the centripetal force required to keep and object of mass m and velocity v in circular motion at a distance of radius r : mv F = r 2 9 The Universal Law of Gravitation You are now quite familiar with problems and calculations involving the acceleration due to gravity. However, our experience here on the surface of the Earth is a specific case of the much more general phenomenon of universal gravitation. Newton figured this one out, too, and his efforts along these lines are what led to the famous "falling apple" anecdote. He was actually trying to understand why the Moon stayed in a circular path around the Earth rather than taking off on a straight line into space. Today his result is known as the Universal Law of Gravitation: Every object in the universe is attracted to every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them In symbols... masses of 2 bodies m1 m2 F d distance between them 2 We make this an equation by adding a constant of proportionality, G m2 m1 F F d Newton was able to show that the appropriate distance is from the center of one object to the center of the other. The force doesn't originate at the center, but the effect is the same as if it did. F = G m1 m2 d 2 The Universal Law of Gravitation is an example of an "inversesquare" law. Note how dramatically the force drops off with increasing distance between two bodies 10 attractive force 8 10 The gravitational attraction of a massive body can be viewed as a gravitational "well" with progressively steepening sides moving inward 6 4 2 0 2 4 6 8 10 12 14 16 18 relative distance Relationship between G and g... Remember that the weight of an object at the surface of the Earth is a force w = F = mg substitute into the equation for universal gravitation cancel the m's and you have g = G me d 2 mg = G m me d mass of the Earth 2 Question: What is d in this specific case? 11 See if you can use the Universal Law of Gravitation to calculate the acceleration due to Earth's gravity at various distances above the Earth... 6 6600 9.2 2 0 40,000 0.25 380,000 0.0028 0 core/mantle boundary 0 2000 4000 depth (km) 6000 0, 00 0 40 0, 00 10 0, 00 Earth's center 2 0 -3 0 4 0 -2 30 6 -1 0, 00 8 log (g) 10 g (m/s2 ) Earth's surface 1 change in g inside the Earth... 20 at Moon 0 33,600 km 0, 00 2.45 0, 00 12,800 10 6400 km Moon 0 200 km geostationary satellite 0, 00 9.8 0 6400 0 4 at surface 40 me = 5.9742E24 kg 8 0, 00 G = 6.67E-11 Nm /kg g (m/s ) 2 20 r (km) 2 30 2 g (m/s 2 ) distance above Earth Earth's surface 10 distance from Earth's center (kilometers) 12 Appendix: Derivation of the acceleration of an object moving in uniform circular motion With reference to the diagram above, we can calculate the force required to maintain mass m moving at velocity v in a circular path of radius r. Note, first, that the triangles a-b-c and a'-b'-c' are similar triangles because they have a common angle " between sides of equal length. This allows us to write r / s = v / !v where v represents the common magnitude of the vectors v1 and v2. For a very small time interval !t, the line s becomes indistinguishable from the arc u. Since s is the distance traveled in time !t at velocity v, at very small !t (i.e., in the limit !t 0) s = u = v!t Substituting s = v!t into the equation above yields r / v!t = v / !v and rearranging !v / !t = v 2/ r but remember that a = !v / !t, so a = v2 / r and F = m V 2 / r Additional topics (time permitting): 1) more on circular motion (ball on a tether) – page 13 2) Roche limit – page 13 13 More on circular motion... v If the demonstrator lets the ball go, it will fly in the direction of v, not directly away from him. 2 F = m v /r mass m a = v2/ r Gravity at close range: the Roche limit What would happen if the Moon got to close to the Earth? TOP VIEW Question: Suppose the demonstrator in the picture above were in outer space rather than standing on the Earth. What can you conclude about the behavior of the system (man + swinging ball) on the basis of your knowledge of Newton's third law? Does your conclusion have any implications for the Earth-Moon system? Answer: The difference in the pull of Earth's gravity across the Moon would cause it to break apart.