Physics 231 Lecture 8 Shelf-life of ideas at the Bookstore

Physics 231 Lecture 8 Shelf-life of ideas at the Bookstore
Main points of todays lecture: Frictional forces: kinetic friction: static friction Work: Kinetic energy: Work-energy theorem: x mass for lifting mass for sliding mass for holding ball for twirling inclinded plane with cart: potential energypil various toys pile driver Checking Understanding
A rod is suspended by a string as shown. The lower end of the rod slides on a frictionless surface. Which figure correctly shows the equilibrium position of the rod? Answer: B Slide 5-17 Some issues about pulleys
TL TR The pulley holding up mass M has four forces on it, M2g, MCg, TL and TR. Note TL= TR=T. Both tensions pull upward on the pulley and both weights pull down. You can pull up with twice the effective force this way. If youhad three loops, you could increase your effective force by a factor of 6. Friction Friction impedes the motion of one object along the surfaces of another. It occurs because the surfaces of the two objects temporarily stick together via microwelds.The frictional force can be larger if the two surfaces are at rest with respect to each other. Experimentally we have two cases: kinetic friction: static friction The coefficient of static friction generally exceeds that for kinetic friction: Frictional forces always oppose the motion of one surface with respect to the other. compare sliding on table with newton scale static vs kinetic friction. Example with static friction
Consider the figure below, with M1=105 kg and M2=44.1 kg. What is the minimum static coefficient of friction necessary to keep the block from slipping. If M1 doesnt move do demo with track and pulley and then turn off pump Putting it together If this isnt true, M1 will slip Example kinetic friction with ramp
The block shown below starts sliding down the ramp. Assuming the coefficient of kinetic friction k = 0.3, how long does it take for the block to travel 2m to the bottom of the ramp ? Draw the forces. Choose an appropriate coordinate system. Calculate the components. Use Newtons 2nd to get t y s=2m x 400 400 kinetic friction k = 0.3 vo=0 =40o / Work Physicists have a precise definition of work done by a force on an object. It is computed from the displacement of the object while being acted on by the force. x Note: work is only done if there is a displacement in the direction of the force. if = 0o, W = Fs if = 90o, W = 0. if = 180o, W = -Fs The units for work are Nm = J (Joules). These are the units we will use in this chapter. Other units are cal (calorie). 1 cal = 4.186J. Calorie (Food calories) is another unit. 1 Calorie = 1000 cal. Work is not the same thing as effort. What work is being done by me?
Indicate with your response unit whether: a) positive work is being done b) no work is being done c) negative work is being done What work is being done by gravity? Indicate with your response unit whether: a) positive work is being done b) no work is being done c) negative work is being done What work is being done by me?
Indicate with your response unit whether: a) positive work is being done b) no work is being done c) negative work is being done What work is being done by friction? Indicate with your response unit whether: a) positive work is being done b) no work is being done c) negative work is being done What work is being done by the string?
a) positive work b) negative work c) zero work d) cannot be determined Movie Example A workman lifts a 4 kg brick 1.5 meter vertically. a) What is the work done by the workman? b) What is the work done by gravity? c) If the workman lowers it back to the ground, what work does he do in lowering the brick? d) What would be his total work? Assume all motions are at constant velocity. m 4 kg y parts a&b 1.5 m -1.5 m Work-energy theorem, kinetic energy
We define, to be the kinetic energy. In terms of KE, Proof: Let us consider the case of a constant force F in the x direction. Then By Newtons second law, F=max. Then But There is no force in the y direction, ay=0, vy=vy,0 and 1/2mvy 2= 1/2mvy,0 2. Thus Checking Understanding
Each of the boxes, with masses noted, is pulled for 10 m across a level, frictionless floor by the noted force. Which box experiences the largest change in kinetic energy? Answer: D x Slide 10-31 Checking Understanding
Each of the boxes, with masses noted, is pulled for 10 m across a level, frictionless floor by the noted force. Which box experiences the smallest change in kinetic energy? Answer: C x Slide 10-33 Clicker question A 1000 kg car is moving at 30 m/s, when the driver applies the brakes. After the brakes are applied, the car rolls to a stop in 100 m. How much work is done by the brakes in bringing the car to a stop? a) N b) J c) -4.5x105 J d) 1.5x105 J Example A 1 kg block is thrown downward with an initial velocity of 10 m/s. What is the kinetic energy and speed of the block when it strikes the ground 10 m below. 10 m/s 10 m m 1 kg v0 10 m/s s 10 m KEf ? v Conceptual problem A cart on an air track is moving at 0.5 m/s when the air is suddenly turned off.The cart comes to rest after traveling 1 m.The experiment is repeated, but now the cart is moving at 1 m/s when the air is turned off. How far does the cart travel before coming to rest? a) 1 m b) 2 m c) 3 m d. 4 m e) impossible to determine v1,0 0.5m/s xtry1 1m v2,0 1 m/s xtry2 ? Physics 231 Lecture 9 Main points of todays lecture: Kinetic energy:
Work-energy theorem: Potential energy: Conservation of energy mass for lifting mass for sliding mass for holding ball for twirling inclinded plane with cart: potential energypil various toys pile driver Work-energy theorem, kinetic energy
We define, to be the kinetic energy. In terms of KE, Proof: Let us consider the case of a constant force F in the x direction. Then By Newtons second law, F=max. Then But There is no force in the y direction, ay=0, vy=vy,0 and 1/2mvy 2= 1/2mvy,0 2. Thus Checking Understanding
Each of the boxes, with masses noted, is pulled for 10 m across a level, frictionless floor by the noted force. Which box experiences the largest change in kinetic energy? Answer: D x Slide 10-31 Clicker question A 1000 kg car is moving at 30 m/s, when the driver applies the brakes. After the brakes are applied, the car rolls to a stop in 100 m. How much work is done by the brakes in bringing the car to a stop? (Hint: consider the change in kinetic energy.) a) N b) J c) -4.5x105 J d) 1.5x105 J Example A 1 kg block is thrown downward with an initial velocity of 10 m/s. What is the kinetic energy and speed of the block when it strikes the ground 10 m below. 10 m/s 10 m m 1 kg v0 10 m/s s 10 m KEf ? v Reading Quiz If you raise an object to a greater height, you are increasing kinetic energy. heat. potential energy. chemical energy. thermal energy. Answer: C Slide 10-10 ) ] ] Potential energy } } =work
For certain forces, the work done by the force in going from position (x0,y0) to position (x,y) depends only the displacement and not on the path taken. Such a force is called conservative. gravity is such a force. Consider the work on a mass m under a displacement s at an angle with respect to the vertical as shown below: (x0,y0) (x,y) } } ] ] ) =work From this we can see that being higher initially means that you can have a higher final kinetic energy. Thus, mg(y0-y) is the part of the stored potential energy, which was changed into kinetic energy as the object moves from its initial to its final position. The potential energy only depends on the difference in height between the initial and final positions, i.e. on the vertical component of the displacement. Example The potential energy of the car is 1.95x105 J larger when
What difference between the PE of a 2000 kg car raised 10 m in the air and that of the same car on the ground? How much work would it require to lift it to that height? The potential energy of the car is 1.95x105 J larger when the car is 10 m above the ground than it is on the ground. It would take 1.95x105 J of work to lift it the 10 m. Gravitational potential energy
The most important point of a potential energy is that is only a function of the position and not of the path taken to get there. If we break the path of the pail in to vertical sections the potential energy change is just mgy and horizontal sections where the potential energy remains constant, we can see that the potential energy depends on the total vertical displacement and is independent of the path over which it is achieved. Thus we can define PE=0 for some y0 and then PE=mg(y-y0) thereafter. PE is Wgrav. It is the work one would need to do to move the pail from A to B. (x0,y0) PE-PE0=mg(y-y0) (x,y) Conceptual problem Hint: the force is conservative.
At the bowling alley, the ball-feeder mechanism must exert a force to push the bowling balls up a 1.0-m long ramp.The ramp leads the balls to a chute 0.5 m above the base of the ramp. Approximately how much force must be exerted on a 5.0-kg bowling ball? a) 200 N b) 50 N c) 25 N d)5.0 N e)impossible to determine Hint: the force is conservative. It doesnt matter how you get up there. Conservation of energy
Up to an additive constant, we can define PE=mgy. It is equal in magnitude but opposition in sign to the work being done by gravity.. Then We can define the total energy E as the sum of kinetic and potential energy. Then we have Thus, when all work being done by forces in a problem can be expressed in terms of a potential energy, the total energy is conserved (i.e. remains constant). This is true for gravity and many other forces, but not for friction, for example. Checking Understanding
Three balls are thrown off a cliff with the same speed, but in different directions.Which ball has the greatest speed just before it hits the ground? Ball A Ball B Ball C All balls havethe same speed Answer: D Slide 10-26 Example A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. As the drawing shows, one person is observed to hit the water 5.00 m from the end of the slide in a time of s after leaving the slid. Ignoring friction and air resistance, find the height H in the drawing. x 5 m t 0.5 s h ? v0 vf H Conceptual problem Two marbles, one twice as heavy as the other, are dropped (not thrown) to the ground from the roof of a building. Just before hitting the ground, the heavier marble has a) as much kinetic energy as the lighter one. b) twice as much kinetic energy as the lighter one. c) half as much kinetic energy as the lighter one. d) four times as much kinetic energy as the lighter one. e) impossible to determine clicker question A C B a) 5.2 m/s b) 1.4 m/s c) 23 m/s d) 7.7 m/s
A kg bead slides on a curved wire, starting from rest at point A in the figure below. If the wire is frictionless, find the speed of the bead at C. a) 5.2 m/s b) 1.4 m/s c) 23 m/s d) 7.7 m/s A B C y0 5m yf 2m v0 vf ? Physics 231 Lecture 10 Main points of todays lecture:
Cons. of Energy with Gravity Potential energy of spring Work, energy and non-conservative and external forces Power Conceptual problem Two marbles, one twice as heavy as the other, are dropped (not thrown) to the ground from the roof of a building. Just before hitting the ground, the heavier marble has a) as much kinetic energy as the lighter one. b) twice as much kinetic energy as the lighter one. c) half as much kinetic energy as the lighter one. d) four times as much kinetic energy as the lighter one. e) impossible to determine clicker question A C B a) 5.2 m/s b) 1.4 m/s c) 23 m/s d) 7.7 m/s
A kg bead slides on a curved wire, starting from rest at point A in the figure below. If the wire is frictionless, find the speed of the bead at C. a) 5.2 m/s b) 1.4 m/s c) 23 m/s d) 7.7 m/s A B C y0 5m yf 2m v0 vf ? Work and PE for non-constant forces
When the force is not constant, the work can still be computed for small displacements x over which force is approximately constant: To get the total work, we add the contributions for the steps: W is just the area under the curve, if the force is conservative the change in potential energy is W. It is equal to the work one would need to do against the force to move the object over the chosen path. Work and potential energy for a spring
The force is a spring is not constant: Lets move the mass attached below by hand from x=0 to x with constant velocity This means Fhand+Fspring=0, Fhand= -Fspring=kx Fhand=kx x=0 x The PE is the area under this curve. It is a triangle: A=1/2baseheight Example x0 0.1m xf v0 k 2 N/m m 0.5 kg vf ? The spring above is stretched from it equilibrium length x=0 to a maximum length of xmax=0.1 m, then it is released. The spring constant is k=2 N/m. What is the speed of a 0.5kg mass when it returns to the equilibrium length? a) 2x10-2 m/s b) 2x10-1 m/s c) 1x10-3 m/s d) 1x10-2 m/s Conceptual quiz A spring-loaded toy dart gun is used to shoot a dart straight up in the air, and the dart reaches a maximum height of 24 m.The same dart is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the dart go this time, neglecting friction and assuming an ideal spring? a) 48 m b) 24 m c) 12 m d) 6 m e) impossible to determine Call x to be the initial displacement of the mass on the spring vo vf y0 h1,f 24 m h2,f ? Work-energy theorem with both conservative, non-conservative and external forces
In an isolated system, with conservative forces, mechanical energy is conserved! Proof: The work-energy theorem: For conservative forces, Wcons= - PE: Example A basketball of mass 0.6 kg is dropped from rest from a height of 1.22 m. It rebounds to a height of 0.69 m. How much mechanical energy was lost during the collisions with the floor? y0 1.22 m yf 0.69 m m 0.6 kg v0 vf Elost ? Example A skier starts from rest at the top of a hill that is inclined at 20 with the horizontal. The hillside is 200 m long, and the coefficient of friction between snow and skis is At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. What is the distance d that the skier glides along the horizontal portion of the snow before coming to rest? h 200 m / / Forms of Energy Mechanical Energy Thermal Energy Other forms include
Slide 10-12 Some Energy Transformations
Echem Ug K Eth Echem Ug Us K Ug Slide 10-15 Energy Transformations conserve energy
Kinetic energy K = energy of motion Potential energy U = energy of position Thermal energy Eth = energy associated with temperature System energy E = K + U + Eth + Echem + ... Energy is a conserved property of an isolated system. Energy can be transformed within the system without loss or gain of the energy of the isolated system. Slide 10-14 Power Power is the rate at which work is being done. The average power is If we chose the x-axis to lie along the force and displacement is characterized by an average velocity: The SI units for power are J/s = W (Watts). Another unit is hp (horsepower). 1 hp = 746 W. Bicycle challenge rules
Contestants must warm up by riding for 20 s at P > 200 W. Then each contestant is judged by the peak or maximum power generated anytime in the next 10 s. Can I have two volunteers? Should the instructor participate? a) yes b) no Carmichael(Armstrongs coach) told me that a decent pro cyclist would have put out at (an average power)least four hundred watts (during a 4 hour period), and that the stragglers at the end of the peloton (known as the gruppetto) (which dont feel as much drag force from air resistance) would clock in at perhaps three hundred and fifty. Armstrongin top Tour shapewould have come close to five hundred. (from New Yorker article) Example A 50.0-kg student climbs a 5.00-m-long rope at a constant speed and stops at the top. (a) What must her speed be in order to match the power output of a 200-W light bulb? (b) Quiz: How much work does she do? a) 2450 J b) 245 J c) 24.5 J d) 2.45 J clicker question A car accelerates uniformly from rest to 29 m/s in 12 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if the weight of the car is 1.2x104 N. a) 1.2x103 W b) 6.2x105 W c) 3.8x103 W d) 4.3x104 W Conceptual problem A cart on an air track is moving at 0.5 m/s when the air is suddenly turned off.The cart comes to rest after traveling 1 m.The experiment is repeated, but now the cart is moving at 1 m/s when the air is turned off. How far does the cart travel before coming to rest? a) 1 m b) 2 m c) 3 m d. 4 m e) impossible to determine v1,0 0.5m/s xtry1 1m v2,0 1 m/s xtry2 ? Physics 231 Lecture 11 Main points of todays lecture:
Examples of energy and power Impulses: forces that last only a short time Momentum Impulse-Momentum theorem Momentum conservation Momentum and external forces Bicycle challenge rules
Contestants must warm up by riding for 20 s at P > 200 W. Then each contestant is judged by the peak or maximum power generated anytime in the next 10 s. Can I have two volunteers? Should the instructor participate? a) yes b) no power bike demo Carmichael(Armstrongs coach) told me that a decent pro cyclist would have put out at (an average power)least four hundred watts (during a 4 hour period), and that the stragglers at the end of the peloton (known as the gruppetto) (which dont feel as much drag force from air resistance) would clock in at perhaps three hundred and fifty. Armstrongin top Tour shapewould have come close to five hundred. (from New Yorker article) Example A 50.0-kg student climbs a 5.00-m-long rope at a constant speed and stops at the top. (a) What must her speed be in order to match the power output of a 200-W light bulb? P 200 W m 50 kg vy ? (b) Quiz: How much work does she do? a) 2450 J b) 245 J c) 24.5 J d) 2.45 J clicker question Therefore:
A car accelerates uniformly from rest to 29 m/s in 12 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if the weight of the car is 1.2x104 N. a) 1.2x103 W b) 6.2x105 W c) 3.8x103 W d) 4.3x104 W v0` v 29m/s mg 1.2x104N P ? Therefore: Example A skier starts from rest at the top of a hill that is inclined at 20 with the horizontal. The hillside is 200 m long, and the coefficient of friction between snow and skis is At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. What is the distance d that the skier glides along the horizontal portion of the snow before coming to rest? h 200 m / / Reading Quiz Impulse is a force that is applied at a random time.
a force that is applied very suddenly. the area under the force curve in a force-versus-time graph. the interval of time that a force lasts. Answer: C Slide 9-5 Impulse: useful concept for forces that last a very short time
There are many processes in which forces last a very short time and are difficult to mathematically describe. Examples are: Kicking, striking batting, dribbling a ball. Various types of explosions, firearms, etc. The typical time dependence of such forces whose actions can be best described by the associated impulse is described below: dropping balls pile driver balistic pendulum 0.01s Impulse and Momentum The linear momentum of a particle of mass m is: The change in velocity is related to the change in momentum, i.e. impulse and to the average acceleration: It is related to the average impulsive force: Note: the shorter the force acts to achieve the impulse, the larger the impulsive force must be. Conceptual question Consider two carts, of masses m and 2m, at rest on an air track. If you push first one cart for 3 s and then the other for the same length of time, exerting equal force on each, the momentum of the light cart is a) four times b) twice c) equal to d) one-half e) one-quarter the momentum of the heavy cart. Partially inelastic collisions: earth and ball
A 0.4 kg ball is dropped from rest at a point 1.5 m above the floor. The ball rebounds straight upward to a height of 0.8m. How much energy is lost? What is the magnitude and direction of the impulse applied to the ball by the floor? If the ball is in contact with the floor for 0.01 seconds, what is the impulse force on the ball and on the earth? h0 1.5 m hf 0.8 m m 0.4 kg Choose the gravitation potential energy to vanish at y=0 dropping ball Bouncing balls Assuming each ball has the same mass, which ball experiences the larger impulse? a) the first ball b) the second ball Hint: Which ball has the largest change in velocity? An elastic collision has twice the impulse as a totally inelastic collisions. happy ball sad ball Quiz choose east to be positive
Jack swings at a 0.2 kg ball that is moving west with a velocity of 40 m/s and hits a line drive. The leaves his bat with a velocity of 40 m/sdue east. Assuming the ball is in contact with the bat for s, what is the average impulse force of the bat on the ball? a) 800N east b) 1600 N east c) 1600 N west d) 800 N west m 0.2 kg t 0.01 s v0 -40 m/s vf 40 m/s ? choose east to be positive Reading Quiz The total momentum of a system is conserved always.
if no external forces act on the system. if no internal forces act on the system. never; momentum is only approximately conserved. Answer: B Slide 9-7 Conservation of linear momentum
This applies to collision of objects that interact with each other but whose interactions with the rest of the world can be neglected.This the definition of an isolated system. As an example, one can considera collision between two hockey pucks (one larger and the other smaller) that are sliding without friction on a frictionless ice surface. Proof: From Newtons 3d law: If there are external forces like gravity, momentum may not be conserved. In such cases: Conceptual question Which of these systems are isolated?
a) While slipping on a patch of ice (k=0), a car collides totally inelastically with another car. System: both cars b) Same situation as in a). System: the slipping car c) A single car slips on a patch of ice. System: car d) A car makes an emergency stop on a road. System: car e) A ball drops to Earth. System: ball f) A billiard ball collides elastically with another billiard ball on a pool table. System: both balls Principles of collisions
If there are no external forces, the total momentum is always conserved during a collision: In such collisions, however, the mechanical energy may or may not be conserved. We have two important limits: Totally inelastic collisions where the two objects stick together after the collision. Here, largest energy loss possible for an isolated system occurs. Totally elastic collisions where the two objects bounce off each other and the mechanical energy is the same after the collisions as it is before the collision. Inelastic collisions can occur in which the objects do not stick together. The energy loss in such collisions is less than what occurs in totally inelastic collisions where the object do stick together. Totally inelastic collisions
In isolated systems (systems without external forces) momentum is conserved. In totally inelastic collisions, the particles stick together after the collision. (Maximum possible energy loss) Example: A 40 kg skater, sliding to the right without friction with a velocity of 1.5 m/s, suffers a head on collisions with a 30 kg skater who is initially at rest.After the collisions the two cling together. a) 0.17 m/s b) 0.23 m/s c) 0.42 m/s d) 0.86 m/s m1 40 kg m2 30 kg v1,0 1.5 m/s v2,0 vf ? totally inelasstic collision Physics 231 Lecture 12 Main points of this lect. Impulses
Momentum conservation Isolated systems Totally inelastic collisions Main points of this lect. Quiz choose east to be positive
Jack swings at a 0.2 kg ball that is moving west with a velocity of 40 m/s and hits a line drive. The leaves his bat with a velocity of 40 m/sdue east. Assuming the ball is in contact with the bat for s, what is the average impulse force of the bat on the ball? a) 800N east b) 1600 N east c) 1600 N west d) 800 N west m 0.2 kg t 0.01 s v0 -40 m/s vf 40 m/s ? choose east to be positive Reading Quiz The total momentum of a system is conserved always.
if no external forces act on the system. if no internal forces act on the system. never; momentum is only approximately conserved. Answer: B Slide 9-7 Conservation of linear momentum
This applies to collision of objects that interact with each other but whose interactions with the rest of the world can be neglected.This the definition of an isolated system. As an example, one can considera collision between two hockey pucks (one larger and the other smaller) that are sliding without friction on a frictionless ice surface. Proof: From Newtons 3d law: If there are external forces like gravity, momentum may not be conserved. In such cases: Example An astronaut is motionless in outer space. Upon command, his propulsion unit strapped to his back ejects some gas with a velocity of +14 m/s, and the astronaut recoils with a velocity of -0.5m/s. After the gas is ejected, the mass of the astronaut is 160kg. What is the mass of the ejected gas? Conceptual question Which of these systems are isolated?
a) While slipping on a patch of ice (k=0), a car collides totally inelastically with another car. System: both cars b) Same situation as in a). System: the slipping car c) A single car slips on a patch of ice. System: car d) A car makes an emergency stop on a road. System: car e) A ball drops to Earth. System: ball f) A billiard ball collides elastically with another billiard ball on a pool table. System: both balls Principles of collisions
If there are no external forces, the total momentum is always conserved during a collision: In such collisions, however, the mechanical energy may or may not be conserved. We have two important limits: Totally inelastic collisions where the two objects stick together after the collision. Here, largest energy loss possible for an isolated system occurs. Totally elastic collisions where the two objects bounce off each other and the mechanical energy is the same after the collisions as it is before the collision. Inelastic collisions can occur in which the objects do not stick together. The energy loss in such collisions is less than what occurs in totally inelastic collisions where the object do stick together. Reading Quiz In an inelastic collision in an isolated system,
impulse is conserved. momentum is conserved. force is conserved. energy is conserved. elasticity is conserved. Answer: B Slide 9-9 Principles of collisions
If there are no external forces, the total momentum is always conserved during a collision: In such collisions, however, the mechanical energy may or may not be conserved. We have two important limits: Totally inelastic collisions where the two objects stick together after the collision. Here, largest energy loss possible for an isolated system occurs. Totally elastic collisions where the two objects bounce off each other and the mechanical energy is the same after the collisions as it is before the collision. Inelastic collisions can occur in which the objects do not stick together. The energy loss in such collisions is less than what occurs in totally inelastic collisions where the object do stick together. Totally inelastic collisions
In isolated systems (systems without external forces) momentum is conserved. In totally inelastic collisions, the particles stick together after the collision. (Maximum possible energy loss) Example: A 40 kg skater, sliding to the right without friction with a velocity of 1.5 m/s along the x axis, suffers a head on collisions with a 30 kg skater who is initially at rest.After the collisions the two cling together. What is the final velocity? a) 0.17 m/s b) 0.23 m/s c) 0.42 m/s d) 0.86 m/s m1 40 kg m2 30 kg v1,0 1.5 m/s v2,0 vf ? totally inelasstic collision Quiz A 40 kg skater, sliding to the right without friction with a velocity of 1.5 m/s, suffers a head on collisions with another skater, moving to the left with a velocity of 2 m/s. After the collision, the two skaters come to rest. The mass of the second skater is: a) 120 kg b) 30 kg c) 13.3 kg d) 53.3 kg m1 40 kg v1 1.5 m/s v2 -2 m/s vf m2 ? Conceptual question If all three collisions in the figure shown here are totally inelastic, which bring(s) the car on the left to a halt? Assume these systems to isolated. a) I b) II c) III d) I, II e) I, III f) II, III g) all three Conceptual quiz A compact car and a large truck collide head on and stick together. Which undergoes a larger change in the magnitude of themomentum? Assume this system of car plus truck to be isolated a) car b) truck c) The magnitude of the momentum change is the same for both vehicles. d) Cant tell without knowing the final velocity of combined mass. Energy loss in totally inelastic collisions
Example: A 40 kg skater, sliding to the right without friction with a velocity of 1.5 m/s, suffers a head on collisions with a 30 kg skater who is initially at rest. How much energy is lost in this collision? a) 19 J b) 22 J c) 25 J d) 28 J m1 40 kg m2 30 kg v1,0 1.5 m/s v2,0 KEloss ? Expressing Kinetic energy in terms of momentum
A car has momentum and mass m. In terms of these two quantities, express the kinetic energy. Additional Questions In the demonstration, two cars start from rest. One car is heavier than the other, but both experience the same force and both run for the same time. Which car has the greater final momentum? The lighter car. The heavier car. They have the same momentum. Answer: C Slide 9-31 Review Conceptual Question
In the demonstration, one car is heavier than the other, but both experience the same force and both start from rest and run until they achieve the same displacement in the positive direction. Which car has the greater final momentum? Hint: express momentum in terms of work The lighter car. The heavier car. They have the same momentum. Force does the same work on each car. By the work energy theorem: Answer: B Slide 9-33 Totally elastic collisions
To calculate the result of an elastic collision in one dimension, we considered the constraints of total momentum and energy conservation: We rearrange both equations to get object 1 on the left and object 2 on the right: totally elastic demo vary masses Combining the last equation and the rearranged Equation 1, we have two equations and 2 unknowns which we can solve to get v1,f and v2,f : Example A 2 kg cart moves with a velocity of 3 m/s to the right on a frictionless track. It collides elastically with a stationary 1 kg cart. a) What are the final velocities of the two carts? b) In a frame moving with the initial velocity of the first cart, what are the initial and final velocities of the two carts? Physics 231 Lecture 13 Main points of todays lecture:
Elastic collisions in one dimension: Multiple impulses and rocket propulsion. Center of Mass 1st Midterm Exam Exam is Wednesday October 5th.
If you will be away on a University sponsored trip, you need to make alternative testing arrangements with me by the end of today (Friday). Problems will be similar to Lon-Capa, but somewhat simpler on the average. Answers will be Multiple Choice. You should prepare one 8.5 by 11 sheet of formulae. You can use both sides of the sheet. We will have a review on Monday. There are a set of practice problems on course web site. I have also distributed bysome suggestions about how you can prepare for the exam. You should review the homework, lectures, book and then prepare your formula sheet. You should then attempt the practice problems with only the formula sheet as a reference. Homework is not due next week. Instead, you should the do the corrections set, which will be available on LONCAPA as a regular homework assignment on Thursday evening next week. The corrections set will consist of the same problems as on the midterm exam.I have already sent anabout how the corrections set can improve your midterm score. Conceptual quiz A compact car and a large truck collide head on and stick together. Which undergoes a larger change in the magnitude of themomentum? Assume this system of car plus truck to be isolated a) car b) truck c) The magnitude of the momentum change is the same for both vehicles. d) Cant tell without knowing the final velocity of combined mass. Energy loss in totally inelastic collisions
Example: A 40 kg skater, sliding to the right without friction with a velocity of 1.5 m/s, suffers a head on collisions with a 30 kg skater who is initially at rest. How much energy is lost in this collision? a) 19 J b) 22 J c) 25 J d) 28 J m1 40 kg m2 30 kg v1,0 1.5 m/s v2,0 KEloss ? Expressing Kinetic energy in terms of momentum
A car has momentum and mass m. In terms of these two quantities, express the kinetic energy. Review Conceptual Question
In the demonstration, one car is heavier than the other, but both experience the same force and both start from rest and run until they achieve the same displacement in the positive direction. Which car has the greater final momentum? Hint: express momentum in terms of work The lighter car. The heavier car. They have the same momentum. Force does the same work on each car. By the work energy theorem: Answer: B Slide 9-33 Additional Questions In the demonstration, two cars start from rest. One car is heavier than the other, but both experience the same force and both run for the same time. Which car has the greater final momentum? The lighter car. The heavier car. They have the same momentum. Answer: C Slide 9-31 Totally elastic collisions
Proof: To calculate the result of an elastic collision in one dimension, we considered the constraints of total momentum and energy conservation: We rearrange both equations to get object 1 on the left and object 2 on the right: Combining the last equation and the rearranged Equation 1, we have two equations and 2 unknowns which we can solve to get v1,f and v2,f : In one dimension the result for the collisions oftwo masses is: Example A 2 kg cart moves with a velocity of 3 m/s to the right on a frictionless track. It collides elastically with a stationary 1 kg cart. a) What are the final velocities of the two carts? b) In a frame moving with the initial velocity of the first cart, what are the initial and final velocities of the two carts? m1 2 kg m2 1 kg v1,0 3 m/s v2,0 v1,f ? v2,f Conceptual question v2,0=0 If m2>>m1
A golf ball (mass 1) is fired at a bowling ball (mass 2) initially at rest and bounces back elastically. Compared to the bowling ball, the golf ball after the collision has a) more momentum but less kinetic energy. b) more momentum and more kinetic energy. c) less momentum and less kinetic energy. d) less momentum but more kinetic energy. e) none of the above v2,0=0 If m2>>m1 Example The two carts exchange velocities if they have equal masses.
A 3 kg cart (cart 1) moving with a velocity of +2 m/s collides with a 3 kg cart (cart 2) moving with a velocity of -3 m/s.What are the final velocities of cart 1 and cart 2? a) v1= 2 m/s, v2= -3 m/s b) v1= -3 m/s, v2= 2 m/s c) v1= 1 m/s, v2= -1.5 m/s d) v1= -1.5 m/s, v2= 1 m/s m1 3 kg m2 v1,0 2 m/s v2,0 -3 m/s v1,f ? v2,f The two carts exchange velocities if they have equal masses. Conceptual problem Suppose you are on a cart, initially at rest on a track with very little friction.You throw balls at a partition that is rigidly mounted on the cart. If the balls bounce straight back as shown in the figure, is the cart put in motion? a) Yes, it moves to the right. b) Yes, it moves to the left. c) No, it remains in place. Center of Mass The center of mass of a system is a mass weighted average over the positions of the various masses. Example Three masses are lined up along the x axis, with y=0, and z=0 for all three masses. Mass 1 has m1=2 kg is at x=1 m. Mass 2 has m2=0.5 kg and is at x=3 m. Mass 3 has m3=1.5 kg and is at x=4 m. Where is the center of mass? Properties of the center of mass
The velocity of the center of mass is given by the total momentum divided by the total mass. Therefore, the center of mass velocity of an isolated system is constant. Movie If an external force acts on a system of particles, the center of mass follows a trajectory that this the same as would be followed by a single particlewith mass Mtotal. Movie Physics 231 Lecture 14 Main points of todays lecture:
Multiple impulses and rocket propulsion. Things that oscillate Mass and Spring Circular motion Simple harmonic motion Mid-term correction set open
Final Mid-term exam scores will be sent byon Monday after corrections set closes. Did corrections set Did not do corrections In the past, most students did the corrections set. Allows you to learn how to do the problems you may have gotten wrong. Improves your score. Some students didnt do the corrections set. Why? Conceptual question Example of slow accumulation of mass by
Suppose rain falls vertically into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating water, the speed of the cart a) increases. b) does not change. c) decreases. Example of slow accumulation of mass by a continuous sequence of totally inelastic collisions. Rockets Rocket propulsion is an example of conservation of momentum:
The rocket doesnt push on the environment, as in propulsion. It pushes on the exhaust gas, and the exhaust gas pushes the rocket forward. It is a continuous sequence of small explosions. Newtons third law, but seen more easily from the perspective of conservation of momentum. Slide 9-24 Example An astronaut is motionless in outer space. Upon command, his propulsion unit strapped to his back ejects some gas with a velocity of +14 m/s, and the astronaut recoils with a velocity of -0.5m/s. After the gas is ejected, the mass of the astronaut is 160kg. What is the mass of the ejected gas? If the mass is ejected in 2 s, what is the thrust force and the average rate of emitted gas m/t? (Hint treat thrust as the average force acting over t.) Rocket propulsion The thrust force on a rocket can be computed using the impulse momentum theorem knowing the rate of mass m/t of propellant that is being emitted per second and its emission velocity relative t the rocketvf- v0. Periodic motion Periodic or oscillatory motion is motion that exactly repeats itself after a certain amount of time T has elapsed. This time T is called the period. The number of times the motion repeats itself in a second is call the frequency. Examples of systems that display periodic motion. a system of masses and springs a pendulum or swing. the second hand on a clock, which repeats its motion in exactly 60 seconds. circular motion with constant angular velocity. The vibrating surface that makes the sound in a musical instrument. Piano string Bell. Simple Harmonic Motion is a common form of periodic motion. Springs and simple harmonic motion
Consider the mass and spring system to the right. The system is at equilibrium at x=0. This is an example of stable equilibrium. If an object is in equilibrium then the net force on it is zero. If you displace an object away from a stable equilibrium there will be a force pushing it back in the opposite direction of the displacement. When the equilibrium is stable, the force around equilibrium often depends linearly on displacement. For the spring, this gives rise to the Hookes law force: mass with two springs on air track mass hanging on spring If you displace the mass away from equilibrium by an amount x0 =A, the mass will oscillate about the minimum. This oscillation is an example of Simple Harmonic Motion (SHM) has units rad/s or Hz 1/s is sometimes called Hz Review problem: Hookes law
In a room that is 2.44 m high, a spring (unstrained length=0.30 m) hangs from the ceiling. From this spring, a board hangs so that its 1.98 m length is perpendicular to the floor, the lower end just extending to, but not touching, the floor. The board weighs 102 N. What is the spring constant of the spring? d=( )m mass hanging on spring If displaced from d and released, the board will oscillate up and down (if the floor were not in the way). Vertical vs. horizontal oscillations
Whether the motion is vertical or horizontal: The only thing gravity does is displace the equilibrium point a distance L=mg/k downwards. Mass moves with Simple Harmonic Motion: Not that increasing the amplitude A only increases the displacement from the equilibrium point at x=0. Conceptual question An object can have simple harmonic motion (oscillation) about a. any equilibrium point. b. any stable equilibrium point. c. around certain stable equilibrium points where Hookes law is obeyed. d. any point, provided the forces exerted on it obey Hookes law. e. any point. What is the acceleration in Simple Harmonic Motion (SHM)
From Hookes law and from Newtons second law: Putting it together Simple harmonic motion means Physics 231 Lecture 15 Main points of todays lecture:
Simple harmonic motion Mass and Spring Pendulum Circular motion Graphical Representation of Simple Harmonic Motion
x T When does x resume its maximum? cos(0)=1; cos(2)=1; cos(4)=1, etc T=; T=()/; f=1/T; f=2/ T is the period. It is the time for the motion to repeat itself f is the frequency. It is the number of times the motion repeats itself per second.Units are Hertz (Hz) or s-1. is the angular frequency. The period does not depend on the amplitude A When x is a maximum or minimum, velocity is zero When x is zero, the speed is a maximum When x achieves its most positive value, a is at its most negative value. v a Conceptual question A mass attached to a spring oscillates back and forth as indicated in the position vs. time plot below. At point P, the mass has a. positive velocity and positive acceleration. b. positive velocity and negative acceleration. c. positive velocity and zero acceleration. d. negative velocity and positive acceleration. e. negative velocity and negative acceleration. x is increasing with time, therefore the velocity is positive. x is positive, a=-kx/m, therefore the accleration is negative. Example a) c) Just plug the time into the equations above
The motion of an object is described by the equation x = (0.30 m) cos(t/3), where t is assumed to be in seconds. Find (a) the position, (b) velocity and (c) acceleration of the object at t = 0 and t = 0.60 s, (d) the amplitude of the motion, (e) the frequency of the motion, and (f) the period of the motion. a) c) Just plug the time into the equations above Example A 50 coil spring has a spring constant of 860 N/m. One end of the 50-coil spring is attached to a wall. An object of mass 45 kg is attached to the other end of the spring and the system is set in horizontal oscillation. What is the angular frequency of the motion? a) 2.39 Hz b) 4.37 Hz c) 5.21 Hz d) 6.85 Hz e) 9.22 Hz Reading quiz If mass spring system was arranged vertically with the mass suspended from the 50 coil spring, how would the frequency change? a) it would be smaller because gravity subtracts from the spring force at the bottom of its motion. b) it would be larger because gravity adds to the spring force at the top of its motion. c) it would be exactly the same. Gravity only displaces the equilibrium point so that the equilibrium length is greater. hanging mass on spring Example If a mass of 0.4 kg is suspended vertically by a spring, it stretches the spring by 2 m. Assume the spring is stretched further and released, and the mass plus spring system undergoes vertical oscillations. Calculate the angular frequency of the oscillatory motion.(Hint: Solve the static equilibrium to get k and then solve for ) a) 0.7 rad/s= 0.7 Hz b) 4.5 rad/s=4.5 Hz c) 2.9 rad/s=2.9 Hz d) 13.8 rad/s=13.8 Hz e) 2.2 rad/s=2.2 Hz d=2 m 0.4 kg Verification of Sinusoidal Nature
This experiment shows the sinusoidal nature of simple harmonic motion The spring mass system oscillates in simple harmonic motion The attached pen traces out the sinusoidal motion With no friction or viscosity, the amplitude of the oscillation remains the same. With damping, the amplitude decreases: Simple Pendulum d=Lsin
The simple pendulum is another example of simple harmonic motion The torque is given by the gravitational force times the moment arm d=Lsin = -mgd=- m gL sin Newtons second law states: = I - m gL sin =mL2 = - g/L sin For small angles < 15 : L sin L sin (in radians) = - g/L This is similar to the equation ax= - k/m x which describes the motion of mass plus spring. We therefore expect simple harmonic motion with: d=Lsin note:I=mL2 do demo of pendulum first T depends on L and g not on max Comparison of simple pendulum to a spring-mass system Checking Understanding
A series of pendulums with different length strings and different masses is shown below. Each pendulum is pulled to the side by the same (small) angle, the pendulums are released, and they begin to swing from side to side. 20 cm Rank the frequencies of the five pendulums, from highest to lowest. A E B D C D A C B E A B C D E B E C A D do several differing pendula with different lengths Slide 14-17 Conceptual question A person swings on a swing.When the person sits still, the swing oscillates back and forth at its natural frequency. If, instead, two people sit on the swing, the natural frequency of the swing is a. greater. b. the same. c. smaller. Quiz Two playground swings start out together. During the time that swing 1 makes 10 complete cycles, swing 2 makes only 8.5 cycles. What is the ratios L1/L2 of the lengths of the swings? (Hint: use ratio technique) a) .32 b) .42 c) .52 d) .62 e) .72 Example The period of a simple pendulum is 0.2% longer at location A than it is a location B. Find the ratio gA/gB of the acceleration due to gravity at these two locations. Example An object is attached to the lower end of a 100 coil spring that is hanging from the ceiling. The spring stretches by 0.16 m. The spring is then cut into two identical springs of 50 coils each. As the drawing shows, each spring is attached between the ceiling and the object. By how much does each string stretch? Physics 231 Lecture 16 Main new points of todays lecture:
Description of circular motion in cylindrical coordinates (r,): First equations for circular motion: Centripetal acceleration: vt ac Rotation as periodic motion: Polar coordinates
When an object rotates about a fixed axis, the displacement of any point P in the object can be economically described by the angular displacement . Here P originally at (x0,y0) = (r,0). The rotation moves it to (x,y)=(rcos(), rsin()). Three different units are often used to quantify: degrees, radians, and revolutions. degrees radians 1 1o= rad 1 rad=57.30 revolutions 1 rev=3600 1 rev=2 rad One radian is the angle change for which the distance s traveled equals the radius r of the circle; i.e. s=r (when is in radians.) Example A wheel has a radius of 4.1 m. How far (path length) does a point on the circumference travel if the wheel is rotated through angles of 30, 30 rad, and 30 rev, respectively? Angular velocity The rate of change of the angular displacement is defined to be the angular velocity . In term of the angular displacement, the average angular velocity over a time interval t is: The instantaneous angular velocity can be obtained by making the time interval very short: Note: if measured in radians, s=r is the arc length covered in time t. Thus: is the average tangential speed and is the instantaneous tangential speed of the object. Example Two people start at the same place and walk around a circular lake in opposite directions. The first has an angular speed of 1.7x10-3 rad/s, while the second has an angular speed of 3.4x10-3 rad/s. How long will it be before they meet? If first person is walking at a speed of 0.25 m/s, what is the approximate radius of the lake? a) 150 m b) 15 m c) 45 m d) 1200 m 1 2 Reading Quiz For uniform circular motion, the acceleration
is parallel to the velocity. is directed toward the center of the circle. is larger for a larger orbit at the same speed. is always due to gravity. is always negative. Answer: B Slide 6-6 Reading question When a car turns a corner on a level road, which force provides the necessary centripetal acceleration? Friction Tension Normal force Air resistance Gravity Answer: A Slide 6-9 Checking Understanding
When a ball on the end of a string is swung in a horizontal circle: What is the direction of the acceleration of the ball? Tangent to the circle, in the direction of the balls motion Toward the center of the circle Answer: B Slide 6-15 Checking Understanding: Circular Motion Dynamics
When the ball reaches the break in the circle, which path will it follow? Answer: C Do demo of a object on the uverhead Slide 6-21 Forces and uniform circular motion
vt ac Consider the mass undergoing horizontal circular motion at the right. At any point, the instantaneous velocity is tangential to the circle. Because the direction of the tangent changes, however, the direction of the velocity changes. This means there is an acceleration pointing inward towards the center of the circle. If we look at the component of the acceleration in this direction: Show slide from work and do the demo with howler on string Here we have used: Motion in a horizontal circle: A car going through a bend
A car is going down a winding road with a speed of 22 m/s and is going around a curve with a radius r = 150 m. What is the centripetal acceleration? a) 0.16 m/s2 b) 3.23 m/s2 c) 9.8 m/s2 The coefficient of static friction between road and car is 0.5. What is the minimum radius thebend can have without the car sliding of the road? ac =mv2/r v 22 m/s r 150 m ac ? =(22m/s)2/150 m =3.23 m/s2 Do demo ofchips on wheel v 22 m/s r ? s 0.5 fs = mac= mv2/r also fs =smg sg =v2/r r= v2/(sg )=(22m/s2)2/(0.59.81m/s)=99m Conceptual question You are a passenger in a car and not wearing your seat belt.Without increasing or decreasing its speed, the car makes a sharp left turn, and you find yourself colliding with the right-hand door. Which is the correct analysis of the situation? a) Before and after the collision, there is a rightward force pushing you into the door. b) Starting at the time of collision, the door exerts a leftward force on you. c) both of the above d) neither of the above Conceptual question a) b) c) d) e)
A rider in a barrel of fun finds herself stuck with her back to the wall.Which diagram correctly shows the forces acting on her? a) b) c) d) e) Banked Turns ( in Lon-capa homework)
y x An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. She does so by banking the road in such a way that the force causing the centripetal acceleration will be supplied by the component of the normal force toward the center of the circular path. Strategy: Use Newton's second law : Example Consider the motion of a 1 kg mass constrained to move in a vertical circle of radius r=2 m by a massless wire. What is the tension in this wire if the mass is at the side of the circle (wire) is horizontal and is moving with an instantaneous speed of 3 m/s? a) 1.5 N b) 2.3 N c) 4.5 N d) 0 vt ac v 3 m/s r 2 m m 1 kg T ? do howler on vertical string Example ac vt Hint: there are two forces in the y direction.
Consider the motion of a 1 kg mass constrained to move in a vertical circle of radius r=2 m by a massless wire. a) What is the tension in this wire if the mass is at the bottom of the circle and is moving with a instantaneous speed of 3 m/s? a) 4.5N b) 9.8 N c) 14 .3N d) 24 .3N vt ac v 3 m/s r 2 m m 1 kg T ? Hint: there are two forces in the y direction. Example Consider the motion of a 1 kg mass constrained to move in a vertical circle of radius r=2 m by a massless wire. What is the tension in this wire if the mass is at the top of the circle and is moving with an instantaneous speed of 3 m/s? a) 5.3N b) 9.8 N c) 5.3N d) 0 vt ac now do wheel on sttrack at two heights v 3 m/s r 2 m m 1 kg T ? Physics 231 Lecture 17 Main points of todays lecture:
Rotational motion definitions: Rotational kinematics equations: Rolling motion: Conceptual question ) Angles for ladybug and
A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second.The gentleman bugs angular speed is a) half the ladybugs. b) the same as the ladybugs. c) twice the ladybugs. d) impossible to determine ) you put two chips on the platter Angles for ladybug and gentleman bug are the same, so the angular velocities are the same as well. 12 Angular acceleration Many times, the angular velocity changes with time. We quantify this by the angular acceleration . In term of the angular velocity , the average angular acceleration over a time interval t is: Note: if measured in rad/s, vt=r is the change in tangential speed during time t. Thus: The instantaneous angular acceleration can be obtained by making the time interval very short: is the average tangential acceleration. And at=ris the instantaneous tangential speed of the object. 13 Analogy between linear and angular kinematics for constant acceleration
If the acceleration is constant, we obtained a set of equation on the left side of the table relating x, v, a and t. If the angular acceleration is constant and we go through the same steps, we obtain the analogous set of equation on the right side of the table relating , , , and t. 14 Example After 10 s, a spinning roulette wheel has slowed to an angular speed of 1.88 rad/s. During this time, the wheel rotates through an angle of 44.0 rad. Determine the angular acceleration of the wheel. 1.88 rad/s t 10 s 44.0 rad 15 Quiz A centrifuge decelerates to rest over a time interval of 10 seconds from an initial angular velocity of 1000 rad/s. Over this time interval, the angular displacement is about: a) 5 rad b) 5000 rad c) 10 rad d) rad 16 Rolling motion Consider a wheel of radius r rolling on the ground. The relationship between the distance traveled and the angle is obtained by considering the arc length s: Similarly there are corresponding relationships between v, a and , . show rolling wheel. 17 Quiz A ball of radius 0.2 m rolls without slipping at a velocity of 5 m/s. What is the angular velocity of the ball? a) 25 rad/s b) 250 rad/s c) 50 rad/s d) 5 rad/s 18 Physic 231 Lecture 16 Main points of last lecture:
Description of circular motion in cylindrical coordinates (r,): Main points of todays lecture: Rotational motion definitions: Rotational kinematics equations: Rolling motion: Physic 231 Lecture 11 Main points of todays lecture:
Centripetal acceleration: Newtons law of universal gravitation: Keplers laws and the relation between the orbital period and orbital radius. vt ac Newtons law of universal gravitation
All objects (even light photons) feel a gravitational force attracting them to other objects. This force is proportional to the two masses and inversely proportional to the square of the distance between them. Show the cavendish balance G = x N m /kg A spaceship is on a journey to the moon
A spaceship is on a journey to the moon. The masses of the earth and moon are, respectively, 5.98x1024 kg and 7.36x1022 kg. The distance between the centers of the earth and the moon is 3.85x106m. At what point, as measured from the center of the earth, does the gravitational force exerted on the craft by the earth balance the gravitational force exerted by the moon? This point lies on a line between the centers of the earth and the moon. Moon Earth Example Hint: This is a ratio problem.
George weighs 100 N on the Earth. What would his weight be on the surface of another planet that has a planetary mass, which is 3 times the mass of Earth and mass and a planetary radius, which is 2 times the radius of Earth? a) 75 N b) 100 N c) 125 N d) 150 N Mp 3ME Rp 2Re WE 100 N Wp ? Hint: This is a ratio problem. Physic 231 Lecture 12 1 Main new points of todays lecture:
Kinematic equations for circular motion: Torque 1 midterm 1 grades You should have received a message with the subject: LON-CAPA] 'New'critical message from William Lynch At the end of the message there should be a line like: The midterm 1 score for"Lynch, William" is 48 out of a total possible score of48 points This is not the total score. It is the in class score. The total is computed as follows: Total score = in class score +0.3(corrections score- in class score) If corrections is less than in class, then total = in class score Some examples: in class = 30, corrections =48, total=30+0.3(48-30)=35.5 in class = 15, corrections =40, total=15+0.3(40-15)=22.5 in class = 40, corrections =9, total=40 2 Conceptual question An astronaut floating weightlessly in orbit shakes a large iron anvil rapidly back and forth. She reports back to Earth that a) the shaking costs her no effort because the anvil has no inertial mass in space. b) the shaking costs her some effort but considerably less than on Earth. c) although weightless, the inertial mass of the anvil is the same as on Earth. 3 Keplers laws Johannes Kepler proposed three laws of planetary motion: All planets moved in elliptical orbits with the Sun at one of the focal points. Planetary orbits sweep out equal areas in equal times. (This is a consequence of angular momentum conservation.) 2a r The square of the orbital period of any planet is proportional to the cube of the average distance from the planet to the Sun. For a circular orbit: For an elliptical orbit: / 4 Conceptual question Hint: This is a ratio problem. 5
Two satellites A and B of the same mass are going around Earth in concentric circular orbits.The distance of satellite B from Earths center is twice that of satellite A.What is the ratio of the centripetal force acting on B to that acting on A? a) 1/8 b) 1/4 c) 1/2 d) e) 1 RB 2RA Fc,B/Fc,A ? Hint: This is a ratio problem. 5 Example Hint: this can also be solved as a ratio problem 6
One satellite is in an orbit about Jupiter of radius r1 and a period of 100 days. If a second satellite is placed in an orbit with 4 times the radius (i.e. r2=4r1), what is the period for the orbit of the second satellite? a) 2 b) 4 c) 16 d) 64 r2 4r1 T1 100 d T2 ? Hint: this can also be solved as a ratio problem 6 Conceptual quiz Note: someone in a space ship in the same orbit would
Suppose Earth had no atmosphere and a ball were fired from the top of Mt. Everest in a direction tangent to the ground. If the initial speed were high enough to cause the ball to travel in a circular trajectory around Earth, the balls acceleration would a) be much less than g (because the ball doesnt fall to the ground). b) be approximately g. c) depend on the balls speed. Note: someone in a space ship in the same orbit would feel weightless and many would call this a zero g environment. weightlessness is a sensation that occurs when one feels no effects of gravity. It happens when g=0 or when someone is falling freely. 7 Example The Earth orbits the sun in an circular orbit of radius rE=1.5x1011 m. What is the mass of the sun? 8 Gravitational Potential Energy
PE = mgy is valid only near the earths surface For objects high above the earths surface, an alternate expression is needed Zero potential energy is defined to be the value infinitely far from the earth If the total mechanical energy of an object exceeds zero, the object can escape the gravitation field of the earth.The escape velocity is when the Emech=0. you will have loncapa problems on this. 9 Conceptual question A rock, initially at rest with respect to Earth and located an infinite distance away is released and accelerates toward Earth. An observation tower is built 3 Earth-radii high to observe the rock as it plummets to Earth. Neglecting friction, the rocks speed when it hits the ground is a) twice b) three times c) four times d) six times e) eight times its speed at the top of the tower. 10 Conceptual question ) Angles for ladybug and
A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second.The gentleman bugs angular speed is a) half the ladybugs. b) the same as the ladybugs. c) twice the ladybugs. d) impossible to determine ) you put two chips on the platter Angles for ladybug and gentleman bug are the same, so the angular velocities are the same as well. 12 Angular acceleration Many times, the angular velocity changes with time. We quantify this by the angular acceleration . In term of the angular velocity , the average angular acceleration over a time interval t is: Note: if measured in rad/s, vt=r is the change in tangential speed during time t. Thus: The instantaneous angular acceleration can be obtained by making the time interval very short: is the average tangential acceleration. And at=ris the instantaneous tangential speed of the object. 13 Analogy between linear and angular kinematics for constant acceleration
If the acceleration is constant, we obtained a set of equation on the left side of the table relating x, v, a and t. If the angular acceleration is constant and we go through the same steps, we obtain the analogous set of equation on the right side of the table relating , , , and t. 14 Example After 10 s, a spinning roulette wheel has slowed to an angular speed of 1.88 rad/s. During this time, the wheel rotates through an angle of 44.0 rad. Determine the angular acceleration of the wheel. 1.88 rad/s t 10 s 44.0 rad 15 Quiz A centrifuge decelerates to rest over a time interval of 10 seconds from an initial angular velocity of 1000 rad/s. Over this time interval, the angular displacement is about: a) 5 rad b) 5000 rad c) 10 rad d) rad 16 Rolling motion Consider a wheel of radius r rolling on the ground. The relationship between the distance traveled and the angle is obtained by considering the arc length s: Similarly there are corresponding relationships between v, a and , . show rolling wheel. 17 Quiz A ball of radius 0.2 m rolls without slipping at a velocity of 5 m/s. What is the angular velocity of the ball? a) 25 rad/s b) 250 rad/s c) 50 rad/s d) 5 rad/s 18 Archimedes: GIVE ME A PLACE TO STAND AND I WILL MOVE THE EARTH
19 Torques The direction that you make an object turn about an axis depends not only on the force, but the point at which you apply the force. The relevant quantity that governs rotation is not the force, but is the torque instead: movie 20 Torques r1 r2 d1 d2 The direction that you make an object turn about an axis depends not only on the force, but the point at which you apply the force. The relevant quantity that governs rotation is not the force, but is the torque instead: 1 2 To calculate the torque about an axis: Chose your axis Draw your force. Draw the line, which is perpendicular to the force and goes from the line of force to the axis. This line is your lever arm d. Torque has the units Nm or Joules. do demo of wrench do demo of door with newton scale. 21 Example Find the net torque (magnitude and direction) produced by the forces F1 and F2 about the rotational axis shown in the drawing. The forces are acting on a thin rigid rod, and the axis is perpendicular to the page. d2 300 600 22 Conceptual quiz a) b) d) c) 23
You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is least effective in loosening the nut? a) b) d) c) 23 Example One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. One force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The other force has a magnitude of 6.00 N and acts at a 300 angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end that is pinned. d h 6.00 N 30o 2.00 N 24 Rotational Equilibrium I
Newtons second law for linear motion states that the net force on an object is equal to its mass times its acceleration. If the net force is zero, the acceleration of the object is zero. It could be at rest or moving with constant non-zero velocity. But is the object necessarily in rotational equilibrium? Consider the case of a merry-go-round sitting on a frictionless surface. The only forces parallel to the surface are shown in the Figure. Obviously, the net horizontal force vanishes. What can you say about the angular acceleration around the vertical axis shown? a) It is zero. The two torques cancel out. b) It is non-zero and counterclockwise. c) It is non-zero and clockwise 100 N Rotational equilibrium II:
Newtons 2nd law for rotational motion states that the angular acceleration is given by the net torque. It states that the net torque, i.e. the sum of all torques about an axis, on a body is equal to the moment of inertial I times the angular acceleration about that axis: Here, I is the moment of inertia. It plays the same role as the mass in Newtons law for linear motion. We discuss I later. We define equilibrium to be the situation when both the linear acceleration and angular acceleration vanish. This occurs when both the net force and the net torque vanish. In other words: Here, torques must vanish around any axis. Axes can be chosen so as to make the solutions of problems easier. Example A 71 kg boulder is placed on a 2.00 m long board at a point that is 1.4 m from one end. Cliff and Will support the board at each end so that it is horizontal. Cliff is nearest the boulder. If the weight of the board is negligible, what force does each apply to the board? 2.00m 1.4m Center of mass What if the weight of the board, or any other extended object, is not zero? How do we calculate the torque due to the weight? We can break up the object into little mass elements mi. When these mass elements are added, i mi = m, where m is the total mass of the object. When this is done, the torque due to gravity around axis O can be calculated as follows: Thus, we put the entire weight of the object at For a uniform rod or some other uniform object,is at the center of the object. In general, however, may not even be within the object itself. Example A uniform steel beam of length 5.00 m has a weight of 4.5x103 N. One end of the beam is bolted to a vertical wall. The beam is held in a horizontal position by a cable attached between the other end of the beam and a point on the wall. The cable makes an angle of 25o above the horizontal. A load whose weight is 12.0x103 N is hung from the beam at a point that is 3.5 m from the wall. Find (a) the magnitude of the tension in the supporting cable and (b) the magnitude of the force exerted on the end of the beam by the bolt that attaches to the wall. 25o load Example A uniform steel beam of length 5.00 m has a weight of 4.5x103 N. One end of the beam is bolted to a vertical wall. The beam is held in a horizontal position by a cable attached between the other end of the beam and a point on the wall. The cable makes an angle of 25o above the horizontal. A load whose weight is 12.0x103 N is hung from the beam at a point that is 3.5 m from the wall. Find (a) the magnitude of the tension in the supporting cable and (b) the magnitude of the force exerted on the end of the beam by the bolt that attaches to the wall. 25o load Quiz A uniform beam of length 5 m and mass 100kg is supported at the middle by a pivot and is clamped at the right end as shown below. A 80 kg man stands on the beam 0.5 m from the left end. What is the force exerted by the clamp on the beam? a) 1760 N up b) 627 N down c) 1411 N up d) 830 N down clamp 0.5 m Planetary motion Example Problem A 500 kg rocket sled is coasting on a frictionless horizontal surface at 20 m/s. It then turns on its rocket engines for 5.0 s, with a thrust of 1000 N.What is its final speed? Slide 9-36 Conceptual question Consider two people on opposite sides of a rotating merry-go-round. One of them throws a ball toward the other. In which frame of reference is the path of the ball straight when viewed from above: (1) the frame of the merry-go-round or (2) that of Earth? a)(1) only b) (1) and (2)although the paths appear to curve c) (2) only d) neither; because its thrown while in circular motion, the ball travels along a curved path. show frames of refference video Physic 231 Lecture 2 Main points of todays lecture Lecture 3 Some points of vertical motion:
Here we assumed t0=0 and the y axis to be vertical. Vectors andcomponents of vectors Addition and subtraction of vectors do demos Physic 231 Lecture 4 Main points of todays lecture:
Example: addition of velocities Trajectories of objects in 2 dimensions: Physic 231 Lecture 5 Main points of todays lecture: Newtons 1st law:
If there is no net force, the velocity of a mass remains constant (neither the magnitude nor the direction of the velocity changes). Newtons 2nd law: m is the mass of the object. It is proportional to the number of nucleons (neutrons and protons) in an object. is the net (total) force acting on the object. It is the vector sum of all forces acting on the object. read it Physic 231 Lecture 6 Main points of todays lecture: Normal force
Newtons 3d Law Frictional forces: kinetic friction: static friction Examples. Physic 231 Lecture 7 Shelf-life of ideas at the Bookstore
Main points of todays lecture: Work: Kinetic energy: Work-energy theorem: Potential energy for gravity: Conservation of energy x mass for lifting mass for sliding mass for holding ball for twirling inclinded plane with cart: potential energypil various toys pile driver Physic 231 Lecture 8 Main points of todays lecture:
Cons. of Energy with Gravity Potential energy of spring Work, energy and non-conservative and external forces Power Physic 231 Lecture 17 Main points of last lecture:
Rotation kinematic equations Rolling motion: Main points of todays lecture: Centripetal acceleration: Newtons law of universal gravitation: Keplers laws and the relation between the orbital period and orbital radius. vt ac Review session The review will occur during the last 50 minutes of Tuesdays class period. I have attached a number of sample problems that illustrate the concepts we have covered. In preparation for the review, please do as many of these problems as you can. Some of the problems are very similar, so it will be best if you do problems that illustrated a range of physics principles. I will begin by doing a selected set of review problems. After I am done, I will accept suggestions about which additional problems I should do. Some hints for exam preparation
Go through section 2 lecture notes Key Concepts and make sure you understand these in a precise way (check if you are able to describe an equation in your own words) Many of these are contrary to every day language and pre-conceptions you got in high school If you have difficulty, read the relevant sections in the book again, or/and your lecture notes Prepare a smart equation sheet Group equations by classes of problems Focus on basic equations, not the equations for a specific problem Develop your equation sheets throughout the course. Prepare a set of equations before you do homework and then check whether this set is sufficient for you (and expand accordingly) If you havent done this start now, prior to exam preparation Practice with fresh problems and do them completely on your own Use your equation sheet this is a good test for whether you have a goodequation sheet Do only get help if you are really stuck If you need help try a new problem on the same topic until you can do problems on your own A problem you did before is no longer new to you. Redoing old problems is ok, but in addition you should check whether you can do a fresh problem on the same topic without help. Do an example in class:the soccer player from the book: Problem solving strategy
Make a drawing (for some problems a initial and final drawing is helpful, for others draw the moving object at different locations in the same drawing) Introduce a coordinate system and chose t=0 Give names (letters) to all relevant lengths, times, forces, energies and put them into the drawing. For motion do this at all critical points of the motion. Classify the problem (so far we had: 1D Motion, 2D Motion, Projectile Motion, Energy, Power, Force-acceleration, Force-acceleration for connected masses, Equilibrium) - can be more than one class Write down the general equations for this class of problems For some classes of problems there are specific approaches examples: 2D motion/projectile motion: setup equations for x- and y-motion separately Force-acceleration: draw a separate free body force diagram for all masses involved, each diagram gives an equation (or 2 in x- and y-direction). Connected masses can often be treated asa1D problem choosing an axis along the motion of the system (even if it changes direction) Equilibrium: free body force diagram for all critical points (each diagram gives 2 equations, one in x and one in y direction) Adapt the equations to the specifics of the problem Mark knowns and unknowns Develop a strategy to get from the knowns to the unknowns Execute that strategy, ideally using algebra without numbers Convert all quantities in mks units and plug in numbers Do an example in class:the soccer player from the book: Many useful formulae were listed on the first pages of all lectures, which I attach immediately following this page. I dont think multiple g-forces can be achieved on a see-saw, Tommy. Physic 231 Lecture 18 Main points of last lecture:
Newtons law of universal gravitation: Keplers laws and the relation between the orbital period and orbital radius. Main points of todays lecture: Torque Newtons second law for rotations: Equilibrium: Center of mass: Example A person has a mass of 42kg. What is the persons weight on the moon? The mass and radius of the earth and moon are given below. earth moon mass 5.98x1024 kg 7.36x1022 kg radius 6.38x106 m. 1.74x106 m. Additional Questions A satellite orbits the earth. A Space Shuttle crew is sent to boost the satellite into a higher circular orbit. When the satellite is in is final orbit, which of these quantities increases? Speed Angular speed Period Centripetal acceleration Gravitational force of the earth Answer: C Slide 6-43 Reading Quiz If a system is isolated, the total energy of the system increases constantly. decreases constantly. is constant. depends on work into the system. depends on work out of the system. Hint: the total energy isnt just the mechanical energy. It includes thermal energy, chemical energy and all other forms of energy. Answer: C Slide 10-6 Loncapa hints Hints: These points are equally spaced in time and give the displacement vectors during these time intervals The picture records when ball starts, reaches the end of the slide and hits the floor The displacement vectors are proportional to the velocity. The x-component of the velocity clearly doesnt change. We can see where there is acceleration it is where velocity changes.Where is the acceleration largest? Mechanical energy is conserved. The height and the speed are connected by mechanical energy conservation. Example At what rate m/t must a rocket burn propellant to achieve a thrust whose magnitude is 14,000 N, if the speed of the ejected gases is 6400 m/s relative to the rocket? Expressing Kinetic energy in terms of momentum
A car has momentum and mass m. In terms of these two quantities, express the kinetic energy. Review Conceptual Question
In the demonstration, one car is heavier than the other, but both experience the same force and both start from rest and run until they achieve the same displacement in the positive direction. Which car has the greater final momentum? Hint: express momentum in terms of work The lighter car. The heavier car. They have the same momentum. Force does the same work on each car. By the work energy theorem: Answer: B Slide 9-33