Phy2048 3


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Phy2048 3

  1. 1. Center of Mass and Linear Momentum Note: The center of mass of an object is a point where the mass of the object appears to be concentrated. • We classify objects onto point objects and rigid objects. For point objects, the mass is assumed to be concentrated at a point, whereas for a rigid object the mass is distributed throughout the object. • For a system of point particles m1, m2,m3,.……, each located at (xi,yi,zi) the coordinates of the center of mass are obtained by using the following equations.
  2. 2. By using summations we can write the center of mass equations as follows: where M is the total mass, i.e similarly and
  3. 3. For a rigid body we partition the mass into differential elements of mass dm and then replace the summation by integration and write where dm is a differential element of mass Once we obtain each coordinate of the center of mass , then we write the center of mass as:
  4. 4. In order to perform the integration we need to know the type of distribution of the mass. There are three possible distributions of mass on on a rigid. These are: i) linear distribution ii) surface distribution iii) volume distribution When mass is linearly distributed then dm is given by Where λ is mass per unit length. λ is called linear mass density. If the mass is uniformly distributed, then λ is constant. • When mass is distribution over an area then dm is given by Where ς is surface charge density
  5. 5. Example: Find the center of mass of a thin uniform rod of mass M and total length L. • When mass is distribution over a volume then dm is given by where ρ is volume mass density or simply density Solution: Note. From our experience with physical objects we expect the center of mass of a uniform rod should coincide with its geometric center. Lets consider a differential element of mass dm distributed over a differential element of length dx The linear density λ =M/L or λ= dm/dx is constant Let the rod lay on the x axis as shown below.
  6. 6. • We start from the following equation for center of mass • Note: Since we have two variables under the integrand, we can not perform the integration right away. We must have only one variable • The way we can reduce the number of variables is by relating the variables to each other • In the present problem we use the linear density to relate dm and dx. i.e Thus • Upon integrating applying the limits we obtain • Note: This was what we expecting at the beginning
  7. 7. 1. In the figure below two blocks are connected over a massless and frictionless pulley. The mass of block A is 10kg and the coefficient of kinetic friction between A and the incline is 0.2 and =300. If Block A slides down at constant speed, Review Outline of the solution i. Do a free body diagram for each object ii. Find the net force on each block iii. Apply Newton’s second law
  8. 8. 2. If a machine moves a package from its initial position (ri) to a new position (rf) in 4.0 seconds by applying a force given by to i. Compute the amount of work done by the force. ii. Compute the amount of work done by the force.
  9. 9. Momentum • Definition: The momentum of an object is a physical quantity whose rate of change gives us the net force on the object. Note: Momentum denoted by the letter P Thus we can write the above definition of the momentum mathematically we obtain The momentum is thus obtained by integration Replacing F by We get
  10. 10. Completing the integration yields the following expression for the momentum of an object. Lets revisit the following equation we had considered earlier • If the net force F is zero, then This means P is constant. This is a statement of conservation of momentum. • Which means • conclusion.: In the absence of net external force momentum is always conserved
  11. 11. • Note: Conservation of momentum has been a powerful tool in the discovery of new particles. It has also been used solve collision problems. • There are two types of collisions. These are: • i) Inelastic collision • ii) Elastic collision • In an inelastic collision momentum is conserved but kinetic energy is not conserved. i.e • In an elastic collision both momentum and kinetic energy are conserved. i.e
  12. 12. Example: Two identical balls are on a frictionless, horizontal tabletop. Ball X initially moves at 10 meters per second, as shown in figure on the left- hand side. It then collides elastically with ball Y , which is initially at rest. After the collision, ball X moves at 6 meters per second along a path at 530 to its original direction, as shown in on the right-hand side. • Find the direction and speed of the second object. Since momentum must be conserved, we can say the second object would scatter with velocity v at an angle θ below the x- axis after the collision. Solution: Object x- component of Initial momentum y- component of initial momentum m1=m 10m 0 m2=m 0 0
  13. 13. Object x- component of final momentum y- component of final momentum m1=m 6mcos53 6msin53 m2=m mvcosθ -mvsinθ • The total initial momentum is thus given by • Note: The initial momentum in the y-direction is zero. • The total final momentum is thus given by Now we can apply the rules of the conservation of momentum,.i.e From the first equation we get
  14. 14. From the second equation we get Dividing by m and rearranging the last two equations we get and Note: Since we have three unknowns, we can reduce the number of unknowns by dividing the second equation by the first equation as follows. Which results in Hence θ=370 Once we obtain θ, we can find v by using any of the above equations and obtain v=6m/s
  15. 15. Collision and Impulse In the absence of a net external force momentum is conserved. However when a nonzero net force acts on an object, the momentum of the object changes. The change in momentum is called the impulse of the object and it is denoted by the letter J, i.e or or or where J is the impulse of a collision
  16. 16. • In order to solve this problem with start from Newton’s second law, i.e • Thus to complete the solution we need to find the change in momentum. Example: A 5 kg steel ball strikes a wall with a speed of 20 m/s at an angle of θ=530 with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure below. If the ball is in contact with the wall for 0.250 s, what is the magnitude of the average force exerted on the ball by the wall? Solution:
  17. 17. From the figure we see that and Thus the magnitude of the change in momentum is: Hence Practice different problems involving collsions F= 2(5kg)(20m/s)cos53/0.25s)=480N
  18. 18. Chapter 10 Rotation Now lets revisit circular motion and consider an object moving on a circle with a radius r where the speed is not constant as shown below, i.e. v1≠v2. • Suppose the car moves from p1 to p2 θ is called angular displacement How fast the angular displacement is called angular velocity and it is denoted by ω, i.e There are two types of motions i. Translational motion ii. Rotational motion
  19. 19. α is measured in rad/s2 Note: All the equations of motion we discussed in translational motion can be applied for rotational motion by exchanging the translational variables with the rotational variables. • When θ is measured in radians, ω will be in radians/second • Angular acceleration which is denoted by α is defined as the rate of change of angular velocity, i.e.
  20. 20. Example1. A wheel rotating with a constant angular acceleration turns through 20 revolutions during a 6 s time interval. Its angular velocity at the end of this interval is 13 rad/s. What is the angular acceleration of the wheel? • Solution: We always start from the definition of what is being asked to find out. Recall that angular acceleration is the rate of change of angular velocity, i.e According to the above equation, we need to know the initial and final angular velocities as well as the duration t. (1)
  21. 21. • Solution: We start by listing the data given in the problem. • Angular displacement θ=20 revolutions=20x2π radians=40π radians • Time=t=6.0s • Final angular velocity=ω=13 rad/s • Initial angular velocity ω0=? • Angular acceleration =α=? • Thus we start by listing the data given in the problem. • According to the information we have from the data there are two unknowns ω0 and α. One has to look for ways to find out ω0 before proceeding. • Lets use the alternate definition of angular displacement θ, i.e • In the new equation, one sees that only ω0 is unknown. Thus we use this equation and solve for ω0 first and obtain (2)
  22. 22. • Now we are in a position to solve for the angular acceleration, i.e Example 2: The turntable of a record player rotates initially at 40.3 rev/min and takes 14.9 s to come to rest. i. What is the angular acceleration of the turntable, assuming it is uniform? Ii. How many rotations does the turntable make before coming to rest? Iii. If the radius of the turntable is 0.109 m, what is the initial linear speed of a bug riding on the rim? • Note we solve example 2 in the same approach we solved example 1, i.e
  23. 23. List all data: • Angular displacement θ=? • Time=t=14.9s• Final angular velocity=ω=0 • Initial angular velocity ω0=40.3 rev/min=4.22rad/s • Angular acceleration =α=? (2)
  24. 24. Acceleration Consider a particle on the rim of circle that is rotating as shown • Note: When the particle is moving on the rim there will always be a nonzero centripetal (radial) component of acceleration. • There may also be a tangential component if the velocity changes in magnitude as well, i.e if ω is not constant • These two components are given by the following equation.
  25. 25. • The the speed is calculated from • The last step is an application of the product rule of differentiation • If an object is moving a when a circular path the radius is constant • This results in: Suppose the particle moves a distance s on the rim from one point to another point through an angle θ. If θ is measured in radians then one we have the following relation.
  26. 26. Moment Inertia (I) • The moment inertia of an object is a physical quantity that tells us the degree of difficulty of the object to be rotated. • The moment of an inertia of an object depends on the mass of the object and the distance of the object from the axis of rotation. • For a point object of mass m the moment of inertia is given by • For a rigid (extended) object the moment of inertia is calculated by • Note: Moment inertia is a scalar quantity • Example: Calculate the moment of inertia of a uniform rod of mass m and length L about an axis passing through its center.
  27. 27. • Solution: For the sake of convenience lets assume the rod is laying on the x-axis as shown below. • The problem is solved by partitioning the rod into differential (small) elements of the mass dm distributed over a differential element of length dx and located at distance x from the center of mass, cm. • Note: x varies from x=-L/2 to x=L/2 x-axis x=-L/2 cm x dx dm x=L/2 • Note: Since we have two variables under the integrand sign we have to reduce that into one variable by relating the variables to each other by using the following relations, i.e dm=λdx r=x λ=mass density=M/L
  28. 28. • Note: The linear density is constant because the rod is uniform The Parallel Axis Theorem • This is a law that enables one to determine the moment inertia of an object about an axis passing through a point p that is a distance h away from the center provided Icm is known by using the equation below. • Completing the integration we get Example: Find the moment of inertia of a thin uniform rod about an axis passing through one of its ends.
  29. 29. Solution: Lets first list what data we have, i.e. h=L/2 Hence Ip=(1/12)ML2+M(L/2)2=(1/3)ML2 Kinetic energy of rotation • An object rotating with an angular velocity ω posses a rotational energy posses a rotational kinetic energy given by • Note: The above equation is analogous to the translational kinetic energy we discussed before, i.e
  30. 30. Torque (τ) • Torque is power of a force to rotate an object. The power of a force to rotate an object depends on the magnitude, direction and location of the force from the axis of rotation. • In general the torque is given as the cross product of the force and the distance from the axis rotation, i.e. where r is the vector from the axis of rotation to the point where the force is applied. rsinθ is called the lever arm. Consider the following situation and discuss qualitatively the torque generated by each force. F1 F2 F3 F4F5
  31. 31. • Suppose all the forces have the same magnitude of 20N • F1 generates a clockwise torque • F2 generates no torque because r=0 • F3 generates a counter clockwise torque • F4 generates no torque because θ=0, because r and F have the same direction • F5 generates a clockwise torque Newton’s Second Law for Rotation For translational motion the net force on an object is related to the acceleration by Newton’s second law, i.e
  32. 32. • Similarly for a rotating object, the net torque is given by • where α is the angular acceleration • Example: Consider a uniform disk with mass M=5kg and R=20cm mounted on a fixed horizontal axle. A block with mass m =1.5kg hangs from a massless cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the tension in the cord, the angular acceleration of the disk. Solution: One has to do the free body diagram of each object as shown here.
  33. 33. Object= block Fnet=T-mg=-ma (1) Object =disk τnet==Iα=TR (2) For a disk =(1/2)MR2 To reduce the number of variables use a=αR Solving for T from equation (1) and substituting that into equation (2) one obtains
  34. 34. Angular Momentum The angular momentum is a physical quantity of rotating object whose rate of change gives the net torque causing the rotation. Angular momentum is denoted by L substituting and writing the force as
  35. 35. Note: Angular momentum is sometimes called the momentum of momentum . The magnitude of the angular momentum Since the velocity is tangential, the angle θ=900 Hence but Thus the angular momentum can be written as But mr2=I, the moment inertia of the rotating object And hence
  36. 36. Conservation of Angular Momentum Question: What is the condition for the conservation of angular momentum? To answer this question we need to revisit the definition of angular momentum earlier, i.e If angular momentum of a system of particles is conserved then its value does not change, hence its derivative is zero. This will happen provided the net torque is zero. Thus the condition for angular momentum conservation is the absence of net external torque. When angular momentum is conserved we have: Lf=Li, or written in a different form
  37. 37. Summary Translational motion Rotational Motion 1. Displacement, x 1. Angular displacement, θ 2. Velocity, v=dx/dt 2. Angular velocity, ω=dθ/dt 3. Acceleration, a=dv/dt 3. angular acceleration, α=dω/dt 4. Mass, m 4. Inertia, I 5. Momentum, P=mv 5. Angular momentum, L=Iω 6. Force, F=ma 6. Torque, τ=Iα 7. Kinetic energy Kt=(1/2)mv2 7. Rotational KE=Kr= (1/2)Iω2 Example: A man stands on a platform that is rotating (without friction) with the angular speed of 1.5 rev/s, his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks and platform about the central vertical axis of the platform is 16.0kg.m2. If by moving the bricks the man decreases the rotational inertia of the system to 4.0 kg.m2, a) what is the physical basis of the situation?, b) what is the angular speed of the platform?, c) what is the ratio of the new kinetic energy of the system to the original kinetic energy?, d) what provided the added kinetic energy?
  38. 38. Solution: a) Since no external torques are involved, this is an internal interaction which conserves angular momentum. b. We solve for the new angular velocity by applying the conservation of angular momentum, i.e Lf=Li , where and From the problem we know that, Ii=16.0kg.m2, ωi=1.5 rev/s, and If=4kg.m2, hence Which results in
  39. 39. Example: Consider two spheres of the same mass 2kg and radius 10cm on an inclined planes of different surfaces. If both spheres starts from rest at the top of the incline of height 10.0m, one sliding without friction while the other rolling down how fast will each be travelling at the bottom of the incline? Solution: This problem is best solved by applying the conservation of mechanical energy in both situations, i.e. Recall that Which translates to Object= sliding sphere: In this case the sphere is executing only translational motion. Lets also take our reference for the potential energy at the bottom, which means Uf=0 and Ui=Mgh for both spheres. Lets consider each object separately
  40. 40. This results in Hence, the speed Object=rolling sphere: In this case there are two parts to the kinetic energy, i.e, translational part and rotational part thus, Recall the relation between angular and translational velocity We also use I=(2/5)mr2 for a sphere Thus:
  41. 41. Equating the initial potential energy with the final kinetic energy yields Hence
  42. 42. Equilibrium • a. An object is said to be in translational equilibrium if the net force (vector sum of all forces) on the object is there. In other words this means the forces are balanced. Mathematically this is written as follows. • b. An object is said to be in rotational equilibrium if the net torque (vector sum of all torques) on the object is zero. In other words this means the total clockwise torques are balanced by the total counter clockwise torques. Mathematically this is written as follows. • When both conditions (a) and (b) are satisfied the object is said to be in complete equilibrium.
  43. 43. • Question: Can a moving object be in equilibrium? • Answer: Yes, but it is moving with no acceleration. Note: In order to calculate torque one has to specify the axis of rotation. Any force located at the axis rotation does not generate any torque. Example: The figure below shows a safe (mass M=430kg) hanging by a rope (negligible mass) from a boom (a=1.9m, and b=2.5m) that consists of a uniform hinged beam (m=85kg) and horizontal cable (negligible mass) a) Find the tension Tc in the cable b) Find the magnitude of the net force on the beam from the hinge? Solution: the first thing to do is to do a free body diagram on the beam
  44. 44. • As we see in the free body diagram there are four forces acting on the beam • The beam is in equilibrium under the action of all these forces • Lets resolve these forces into vertical and horizontal components. Force horizontal component Vertical component Tc -Tc 0 F Fh Fv mg 0 -mg Mg 0 -Mg • Note: Since the beam is in equilibrium net force on the beam is zero. Fh Fv
  45. 45. • Which means and • Examining equation (1) one sees that there are two unknowns, namely, Fh, Tc. • This means one needs to obtain an additional equation. To do so one applies the second condition of equilibrium, i.e, the fact that the net torque is zero. • To calculate the net torque one has to choose an axis of rotation. The axis of rotation can be chosen anywhere. However it is recommended that one choose where it is convenient and advantageous. • In the present problem the best choice is to consider rotation about the hinge. In this case both Fh and and Fv do not generate torque, while Tc generates a counter clock wise torque, where as mg and Mg generate a clockwise torque. • From equation (2) one can solve for Fv, where
  46. 46. Counter clockwise torque is: Clockwise torque is: Equating equations (3) and (4) and solving for Tc one gets • Once Tc is found we can use equations (1) to solve for Fh and Fv , i.e, And hence
  47. 47. Gravitation • Newton’s law of gravitation states that any two masses of mass m1 and m2 separated by a distance r exert gravitational force on each other. • By doing an experiment one can find out that the force is directly proportional to the masses of the two objects and inversely proportional to the square of the separation. • The gravitational force between these two objects is always attractive • The magnitude of the gravitational force is given by • Where G is called universal gravitational constant=6.67x10-11 Nm2/kg2 Equation (1) is called Newton’s law of gravitation.
  48. 48. • Note: The gravitational force between the Sun and the Earth is responsible for keeping the Earth on its orbit. Similarly the gravitational force between the Earth and the Moon keeps the moon moving around the Earth. • Note: Each object exerts gravitational force on another object independent of the presence of other objects. This is called the superposition principle. Gravitational Potential Energy • It takes energy to assemble objects in a certain order. This energy is called the gravitational energy of the system of particles. For two particles of mass M and m that are separated by a distance r is given by
  49. 49. Escape Speed • Note: If an object is fired straight up with an initial velocity, the force of gravity exerted by the Earth slows it down to a stop and the object returns to the Earth with the same initial velocity. • However when the initial speed is increased gradually the object would rise higher and higher before it returns to the Earth. When the initial speed reaches a certain velocity the object would reach high enough and the influence of the force gravity will be diminished to zero and the object will not return to the Earth anymore (i.e it escapes from the Earth). • The initial velocity that resulted in the escape of the object is called escape speed. One can obtain the escape speed by applying the conservation of mechanical energy. • Let Ei be the mechanical energy of the object where,
  50. 50. where and Note: R is the radius of the Earth= 6.37x106m, M is mass of the Earth=5.98x1024kg, and G=6.67x10-11 Nm2/kg2 Hence the initial energy is written as: Similarly the final energy is written as: Applying conservation of energy results in:
  51. 51. • As the objects leaves the influence of the Earth r approaches infinity, which means Uf approaches zero. At minimum v0, the final velocity is also zero. Thus after rearranging we get the following equation for the escape speed. Note. Thus if an object is fired at 11.2x103m/s or greater the object will be gone forever.
  52. 52. Kepler’s Laws • Kepler’s laws deal with the motion of planets or satellites. These are the law of orbits, the law of areas and the law of periods. 1. The law of orbits states that all planets move in elliptical orbits with the sun at one focus. 2. The law of areas states that a line joining any planet to the Sun sweeps out equal areas in equal time intervals. (This statement is equivalent to the conservation of angular momentum). 3. The law of period states that the square of the period T of any planet is proportional to the cube of the semimajor axis a of its orbit. For circular orbits with radius r, this states: Where M is the mass of the attracting body-the Sun in the case of the solar system.
  53. 53. Energy In Planetary Motion When a planet or satellite with mass m moves in a circular orbit o radius r, around the Earth or a planet with mass M, The gravitationa force is responsible for keeping the satellite in its orbit, i.e. • Since F is a centripetal force one can also write: Equating these two equation yields Hence the kinetic energy of the satellite is
  54. 54. Thus the total energy will be given by