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1206 - Concepts of Physics Wednesday, December 02nd 2009 Notes • Assignment #8 due today !!! • Assignment #9 due on Monday - for those who would like to get points • Please pick up your assignments ... • This afternoon is a good time to do that Notes • I will be in meetings - starting tomorrow (Thursday, December 03) until Tuesday, December 8th • Therefore it will be harder to just come by my office. • You can as always sent emails - it might be evening (late) until I can answer them keep that in mind Office hours Note! Since meeting locations changed these changed too • Thursday, December 03 -- 13:00 - 14:30 • Friday, December 04 -- not available • Monday, December 07 -- 11:30 - 13:30 • Tuesday, December 08 -- not available ( I will be at Laurentian, so “appointments” possible) • Wednesday, December 09 -- 9:00 - 10:15 Notes • Today - random exercises and repeats in preparation for final exam • Friday - similar to today • Monday - the “Physics of the 2012 movie” • Wednesday - your chance to ask last minute questions and discussion of solutions for assignment #9 • I might switch Monday and Wednesday ... Conservation Laws • Conservation laws are very important • They are also useful for solving Problems • We have talked about several of them: Energy conservation Linear momentum conservation Angular momentum conservation The principle of conservation of energy states that energy cannot be created or destroyed, although it can be changed from one form to another. Thus in any isolated or closed system, the sum of all forms of energy remains constant. The energy of the system may be interconverted among many different forms— mechanical, electrical, magnetic, thermal, chemical, nuclear, and so on—and as time progresses, it tends to become less and less available; but within the limits of small experimental uncertainty, no change in total amount of energy has been observed in any situation in which it has been possible to ensure that energy has not entered or left the system in the form of work or heat. For a system that is both gaining and losing energy in the form of work and heat, as is true of any machine in operation, the energy principle asserts that the net gain of energy is equal to the total change of the system's internal energy. J. P. Joule and others demonstrated the equivalence of heat and work by showing experimentally that for every definite amount of work done against friction there always appears a definite quantity of heat. The experiments usually were so arranged that the heat generated was absorbed by a given quantity of water, and it was observed that a given expenditure of mechanical energy always produced the same rise of temperature in the water. The resulting numerical relation between quantities of mechanical energy and heat is called the Joule equivalent, or is also known as mechanical equivalent of heat. In view of the principle of equivalence of mass and energy in the restricted theory of relativity, the classical principle of conservation of energy must be regarded as a special case of the principle of conservation of mass-energy. However, this more general principle need be invoked only when dealing with certain nuclear phenomena or when speeds comparable with the speed of light (3 × 108 m/s) are involved. Example - your turn Mass of the ball m = 10 kg, the height h is 0.20 m. Calculate potential and kinetic energy for the blue, dark purple and light purple position of the ball. Explain how you made use of the principle of energy conservation. Example - your turn Mass of the ball m = 10 kg, the height h is 0.20 m. Calculate potential and kinetic energy for the blue, dark purple and light purple position of the ball. Explain how you made use of the principle of energy conservation. There is no rotation, therefore we only need to use the linear terms. There is also no elastic potential energy. Gravitational potential energy is given as PE = m g h and kinetic energy is defined as 1/2 m v2. Energy conservation tells us that the sum of potential and kinetic energy at every point must be the same. At the blue position, the ball is not moving, therefore all energy is in the potential energy, which we can calculate with he given values: PE = m g h = 10 kg * 9.8 m/s2 * 0.20 m = 19.6 J The dark purple ball has not potential energy - everything has been converted into kinetic energy and back to potential energy for the light purple location. Pool balls are a good example inelastic means shape changes!! Objects stick together Astronomers observe the planet Mars as the Martians fight a nuclear war. The Martian bombs are so powerful that they rip the planet into three separate pieces of liquified rock, all having the same mass. If one fragment flies off with velocity components(all in the center of mass frame) what is the magnitude of the third one's velocity? v1x = 0 and v2x = 10000 km/h v1y = 10000 km/h and v2y = 0 km/h In the center of mass frame, the planet initially had zero momentum. After the explosion, the vector sum of the momenta must still be zero. Vector addition can be done by adding components, so m v1x + m v2x + m v3x = 0 m v1y + m v2y + m v3y = 0 where we have used the same symbol m for all the terms, because the fragments all have the same mass. The masses can be eliminated by dividing each equation by m, and we find v3x = v3y = - 10000 km/h which gives a magnitude of |v3| = sqrt(v3x2 + v3y2) = 14000 km/h A 120 kg lineman moving west at 2 m/s tackles an 80 kg football fullback moving east at 8 m/s. After the collision, both players move east at 2 m/s. Draw a vector diagram in which the before- and aftercollision momenta of each player is represented by a momentum vector. Label the magnitude of each momentum vector. YOUR TURN .. . A 120 kg lineman moving west at 2 m/s tackles an 80 kg football fullback moving east at 8 m/s. After the collision, both players move east at 2 m/s. Draw a vector diagram in which the before- and aftercollision momenta of each player is represented by a momentum vector. Label the magnitude of each momentum vector. Angular momentum is the tendency of a rotating object to keep rotating at the same speed about the same axis of rotation. Angular momentum is measured with respect to a certain axis of rotation. Objects executing motion around a point possess a quantity called angular momentum. This is an important physical quantity because all experimental evidence indicates that angular momentum is rigorously conserved in our Universe: it can be transferred, but it cannot be created or destroyed. For the simple case of a small mass executing uniform circular motion around a much larger mass (so that we can neglect the effect of the center of mass) the amount of angular momentum takes a simple form. As the adjacent figure illustrates the magnitude of the angular momentum in this case is L = mvr, where L is the angular momentum, m is the mass of the small object, v is the magnitude of its velocity, and r is the separation between the objects. Formation of a neutron star A star rotates with a period of 30 days about an axis through its center. After the star undergoes a supernova explosion, the stellar core, which had a radius of 1.0 x 104 km, collapses into a neutron star of radius 3.0 km. Determine the period of rotation of the neutron star. (Assume that during the collapse no external torque acts on it and it remains spherical with the same relative mass distribution and its mass remains constant. ) Formation of a neutron star A star rotates with a period of 30 days about an axis through its center. After the star undergoes a supernova explosion, the stellar core, which had a radius of 1.0 x 104 km, collapses into a neutron star of radius 3.0 km. Determine the period of rotation of the neutron star. (Assume that during the collapse no external torque acts on it and it remains spherical with the same relative mass distribution and its mass remains constant. ) The distribution of mass is symmetric, so the moment of inertia can be expressed as i = k m r2, where k is a numerical constant. For a solid sphere k = 2/5. The star undergoes a change in period, we will call the initial period Ti and the final one Tf. The period is the time required for the star’s equator to make one complete revolution around the axis of rotations. The star’s angular speed is given by ω = 2π/T. We can use the conservation of angular momentum (L = iω). Formation of a neutron star A star rotates with a period of 30 days about an axis through its center. After the star undergoes a supernova explosion, the stellar core, which had a radius of 1.0 x 104 km, collapses into a neutron star of radius 3.0 km. Determine the period of rotation of the neutron star. (Assume that during the collapse no external torque acts on it and it remains spherical with the same relative mass distribution and its mass remains constant. ) The distribution of mass is symmetric, so the moment of inertia can be expressed as i = k m r2, where k is a numerical constant. For a solid sphere k = 2/5. The star undergoes a change in period, we will call the initial period Ti and the final one Tf. The period is the time required for the star’s equator to make one complete revolution around the axis of rotations. The star’s angular speed is given by ω = 2π/T. We can use the conservation of angular momentum (L = iω). Conservation of angular momentum: Li = Lf --> iiωi = ifωf use expression for ω: ii (2π/Ti) = if (2π/Tf) use expression for i: k m ri2 (2π/Ti) = k m rf2 (2π/Tf) Simplify and solve for Tf: Tf = (rf/ri)2 Ti All values on the right side are given: Tf = (3.0 km/10000 km) (30 days) = 2.7 x 10-6 days = 0.23 s Doppler Submarines A submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1400 Hz. The speed of sound in the water is 1533 m/s. A second submarine (sub B) is located such that both submarines are traveling directly toward each other. The second submarine is moving at 9.00 m/s. (a) What frequency is detected by an observer riding on sub B as the subs approach each other? (b) The subs barely miss each other and pass. What frequency is detected by an observer riding on sub B as the subs recede from each other? (c) While the subs are approaching each other, some of the sound from sub A reflects from sub B and returns to sub A. If this sound were to be detected by an observer on sub A, what is its frequency? Doppler Submarines A submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1400 Hz. The speed of sound in the water is 1533 m/s. A second submarine (sub B) is located such that both submarines are traveling directly toward each other. The second submarine is moving at 9.00 m/s. (a) What frequency is detected by an observer riding on sub B as th subs approach each other? (b) The subs barely miss each other and pass. What frequency is detected by an observer riding on sub B as the subs recede from each other? (c) While the subs are approaching each other, some of the sound from sub A reflects from sub B and returns to sub A. If this sound were to be detected by an observer on sub A, what is its frequency? As the title already say’s we have to use what we learned about the Doppler effect. Both subs are moving, therefore we start with the general case formula and we need to remember the sign conventions depending on direction of movement. f’ = {(v + v(O))/(v - v(S))} f Doppler Submarines A submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar wave at a frequency of 1400 Hz. The speed of sound in the water is 1533 m/s. A second submarine (sub B) is located such that both submarines are traveling directly toward each other. The second submarine is moving at 9.00 m/s. (a) What frequency is detected by an observer riding on sub B as th subs approach each other? (b) The subs barely miss each other and pass. What frequency is detected by an observer riding on sub B as the subs recede from each other? (c) While the subs are approaching each other, some of the sound from sub A reflects from sub B and returns to sub A. If this sound were to be detected by an observer on sub A, what is its frequency? As the title already say’s we have to use what we learned about the Doppler effect. Both subs are moving, therefore we start with the general case formula and we need to remember the sign conventions depending on direction of movement. f’ = {(v + v(O))/(v - v(S))} f (a) f’ = {(1533 m/s + 9.00 m/s)/(1533 m/s) - (8.00 m/s)} 1400 Hz = 1416 Hz (b) same formula, but now the sign of the two speeds v(O) and v(S) is changed, so f’ = {(1533 m/s + (- 9.00 m/s))/(1533 m/s) - (- 8.00 m/s)} 1400 Hz = 1385 Hz (c) The sound of apparent frequency 1416 Hz for (a) is reflected from a moving source (sub B) and then detected by a moving observer (sub A), so f’ = {(1533 m/s + (+ 8.00 m/s))/(1533 m/s) - (+ 9.00 m/s)} 1400 Hz = 1432 Hz The mistuned piano strings Two identical piano strings of length 0.750 m are each tunes exactly to 440 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings. The mistuned piano strings Two identical piano strings of length 0.750 m are each tunes exactly to 440 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings. A small change in tension on one string will cause a small change in frequency compared to the original string. Therefore we expect to “hear” a beats effect. We need to set up the ration of the fundamental frequencies for the two strings using f = (v/2L) for each of them. The mistuned piano strings Two identical piano strings of length 0.750 m are each tunes exactly to 440 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings. A small change in tension on one string will cause a small change in frequency compared to the original string. Therefore we expect to “hear” a beats effect. We need to set up the ration of the fundamental frequencies for the two strings using f = (v/2L) for each of them. Set up ratio: f2/f1 = (v2/2L)/(v1/2L) = v2/v1 substitute the wave speeds with sqrt(T/M), where M mass per unit length, and T tension we get: f2/f1 = sqrt{(T2/M)/(T1/M)} = sqrt(T2/T1) T2 = 1.010 T1 (1% increase) Therefore f2/f1 = 1.005 So f1 = 1.005 f1 = 1.005 * 440 Hz = 442 Hz The difference is 2 Hz One block pushes another Two blocks of masses m1 and m2 with m1 > m2 are placed in contact with each other on a frictionless, horizontal surface. A constant horizontal force F is applied to m1, as shown. F m1 m2 (a) Find the magnitude of the acceleration of the system (b) Determine the magnitude of the contact force between the two blocks (c) When the force is applied toward the left on m2, is the magnitude of the contact force the same as it was when the force was applied toward the right on m1? One block pushes another Two blocks of masses m1 and m2 with m1 > m2 are placed in contact with each other on a frictionless, horizontal surface. A constant horizontal force F is applied to m1, as shown. F m1 m2 (a) Find the magnitude of the acceleration of the system (b) Determine the magnitude of the contact force between the two blocks (c) When the force is applied toward the left on m2, is the magnitude of the contact force the same as it was when the force was applied toward the right on m1? (a) “particle” under a net force. The two blocks are connected - so they can be treated as one particle with mass m = m1 + m2. We have to apply Newton’s second law to find the acceleration. The particle will move in +x direction. ΣFx = F = (m1 + m2) ax therefore ax = F/(m1+m2) One block pushes another Two blocks of masses m1 and m2 with m1 > m2 are placed in contact with each other on a frictionless, horizontal surface. A constant horizontal force F is applied to m1, as shown. F m1 m2 (a) Find the magnitude of the acceleration of the system (b) Determine the magnitude of the contact force between the two blocks (c) When the force is applied toward the left on m2, is the magnitude of the contact force the same as it was when the force was applied toward the right on m1? (b) To find the contact force P we have to consider the blocks individually. First we draw a free-body diagram for each block, where P12 is the contact force from m1 on m2. FN1 F m1 m1 g FN2 P21 P12 m2 m2 g Apply Newton’s second law to m2: ΣFx = P12 = m2 ax Substitute ax from (a) P12 = {m2/(m1 + m2)} F Apply Newton’s second law to m1: ΣFx = F - P21 = F - P12 = m1 ax P12 = {m2/(m1 + m2)} F One block pushes another Two blocks of masses m1 and m2 with m1 > m2 are placed in contact with each other on a frictionless, horizontal surface. A constant horizontal force F is applied to m1, as shown. F m1 m2 (a) Find the magnitude of the acceleration of the system (b) Determine the magnitude of the contact force between the two blocks (c) When the force is applied toward the left on m2, is the magnitude of the contact force the same as it was when the force was applied toward the right on m1? Both blocks are connected - therefore they must experience the same acceleration. We have a net force acting on the “system” or “particle, so we will use Newtons second law: F = m a for each component (horizontal and vertical)