* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document
Internal energy wikipedia , lookup
N-body problem wikipedia , lookup
Quantum vacuum thruster wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Equations of motion wikipedia , lookup
Eigenstate thermalization hypothesis wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Routhian mechanics wikipedia , lookup
Hunting oscillation wikipedia , lookup
Work (physics) wikipedia , lookup
Moment of inertia wikipedia , lookup
Symmetry in quantum mechanics wikipedia , lookup
Tensor operator wikipedia , lookup
Old quantum theory wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Classical central-force problem wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Angular momentum wikipedia , lookup
Photon polarization wikipedia , lookup
Angular momentum operator wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
68 •• Picture the Problem. Because there are no external forces or torques acting on the system defined in the statement of Problem 67, both linear and angular momentum are conserved in the collision and the velocity of the center of mass after the collision is the same as before the collision. Kinetic energy is also conserved as the collision of the hard sphere with the bar is elastic. Let the direction the sphere is moving initially be the positive x direction and toward the top of the page in the figure be the positive y direction and v and V be the final velocities of the objects whose masses are m and M, respectively. Apply conservation of linear momentum to obtain: Apply conservation of angular momentum to obtain: pi pf or mv mv'MV ' (1) Li Lf or mvd mv' d 121 ML2 (2) Set v = 0 in equation (1) and solve for V : Use conservation of mechanical energy to relate the kinetic energies of translation and rotation before and after the elastic collision: Substitute (2) and (3) in (4) and simplify to obtain: Solve for d: mv V' M (3) Ki Kf or 2 2 1 1 mv MV ' 12 2 2 (4) m 12m d 2 2 1 M M L d L M m 12m 1 12 ML2 2 70 •• Picture the Problem Let the zero of gravitational potential energy be a distance x below the pivot as shown in the diagram. Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. We can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing. Using conservation of mechanical energy, relate the rotational kinetic energy of the system just after the collision to its gravitational potential energy when it has swung through an angle : Kf Ki Uf Ui 0 or, because Kf = Ui = 0, Ki U f 0 and d 2 1 I Mg mgx 1 cos 2 2 (1) Note: They are using ½ I 2 instead of ½ mv2 because this time, it’s not just a pt mass, it’s a rod, it has an “I” (I is different than for a point mass) due to the fact that the distance is different PE is different. Note: They are using d/2 because that is the com for the uniform rod. What is cos 600? (0.5 ) Note: 1 – cos comes from the triangle drawn as shown: x is the radius, thus, both the original vertical length and the hypotenuse are = x. The length of the dashed line will then be x – x cos . When you later factor out x, you get 1 – cos . Apply conservation of momentum to the collision: Li Lf or 0.8dmv I 13 Md 2 0.8d m 2 Solve for to obtain: 0.8dmv 1 2 2 Md 0 . 64 md 2 (2) Express the moment of inertia of the system about the pivot: I m0.8d 13 Md 2 2 0.64m 13 M d 2 0.640.3 kg 13 0.8 kg 1.2 m 2 0.660 kg m 2 Substitute equation (2) in equation (1) and simplify to obtain: Solve for v: d Mg 0.8dmg 1 cos 2 0.32dmv I 2 v g 0.5 M 0.8m1 cos I 0.32dm2 Substitute numerical values and evaluate v for = 60 to obtain: v 9.81m/s 0.5 0.8 kg 0.8 0.3 kg 0.50.660 kg m 2 2 0.32 1.2 m0.3 kg 2 72 •• Picture the Problem. Because the net external 7.74 m/s torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. (a) Use its definition to express the total angular momentum of the disk and projectile just before impact: L0 mp v0 b (b) Use conservation of angular momentum to relate the angular momenta just before and just after the collision: L0 L I Express the moment of inertia of the disk + projectile: Substitute for I in the expression for to obtain: and L0 I I 12 MR 2 mp b 2 2m p v 0 b MR 2 2mp b 2 (c) Express the kinetic energy of the system after impact in terms of its angular momentum: m v b L2 Kf 1 p 20 2 I 2 2 MR mpb 2 (d) Express the difference between the initial and final kinetic energies, substitute, and simplify to obtain: E K i K f 2 m v b 2 p 0 MR 2 2mpb 2 12 mp v02 m v b 2 p 0 MR 2 2mpb 2 2 m b p 12 m v 1 MR 2 2m b 2 p 2 p 0 80 • Picture the Problem The angular momentum of the particle is given by L r p where r is its position vector and p is its linear momentum. The torque acting on the particle is given by τ dL dt . Express the angular momentum of the particle: Evaluate dr dt : L r p r mv mr v dr mr dt dr 6tˆj dt Substitute and simplify to find L : Find the torque due to the force: L 3 kg 4 m iˆ 3t 2 m/s 2 ˆj 6t m/s ˆj 72.0t J s kˆ dL d 72.0t J s kˆ τ dt dt 72.0 N m kˆ 90 •• #90 Explanation / justification. You cannot just treat the disks inside of the cylinder as point masses. Each point in / on the disk is a different distance away from the axis of rotation. You can, however, use calc to solve a problem like this. Think of it as r m, for each of the r’s in / on the disk. OR You can also think of the resultant I as Icm + other I’s (specifically, the disks), which is the justification for use of the parallel axis theorem. Picture the Problem. Let x be the radial distance each disk moves outward. Because the net torque acting on the system is zero, we can use conservation of angular momentum to relate the initial and final angular velocities to the initial and final moments of inertia. We’ll assume that the disks are thin enough so that we can ignore their lengths in expressing their moments of inertia. Use conservation of angular momentum to relate the initial and final angular velocities of the disks: Solve for f: Li Lf or I i i I f f f Ii i If (1) Express the initial moment of inertia of the system: I i I cyl 2I disk Express the moment of inertia of the cylinder: I cyl 121 ML2 12 MR 2 121 M L2 6 R 2 121 0.8 kg 1.8 m 60.2 m 0.232 kg m 2 2 2 Letting represent the distance of the clamped disks from the center of rotation and ignoring the thickness of each disk (we’re told they are thin), use the parallel-axis theorem to express the moment of inertia of each disk: With the disks clamped: I disk 14 mr 2 m 2 14 m r 2 4 2 1 4 0.2 kg 0.2 m 2 40.4 m 2 0.0340 kg m 2 I i I cyl 2 I disk 0.232 kg m 2 2 0.0340 kg m 2 0.300 kg m 2 With the disks unclamped, = 0.6 m and: Express and evaluate the final moment of inertia of the system: I disk 14 m r 2 4 2 1 4 0.2 kg 0.2 m2 40.6 m2 0.0740 kg m 2 I f I cyl 2 I disk 0.232 kg m 2 2 0.0740 kg m 2 0.380 kg m 2 Substitute in equation (1) to determine f: 0.300 kg m 2 8 rad/s f 2 0.380 kg m Express the energy dissipated in friction: E Ei Ef Apply Newton’s 2nd law to each disk when they are in their final positions: 2 F kx mr radial 6.32 rad/s 12 I i i2 1 2 I f f2 12 kx2 Note: This should really say F radially is caused by the force on the spring. That part is from Hooke’s Law. You are using N2 on the RHS only. They then use the eq to solve for k. Solve for k: mr 2 k x Substitute numerical values and evaluate k: 2 0.2 kg 0.6 m 6.32 rad/s k 0.2 m 24.0 N/m Express the energy Wfr Ei Ef dissipated in friction: 2 12 I ii 1 2 I f f2 12 kx2 Substitute numerical values and evaluate Wfr: Wfr 1 2 0.300 kg m 8 rad/s 2 2 12 0.380 kg m2 6.32 rad/s 12 24 N/m0.2 m 2 2 1.53 J 84 •• Picture the Problem Because the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular velocities of the system. Using conservation of angular momentum, relate the initial and final angular velocities to the initial and final moments of inertia: Li Lf or I i i I f f Solve for i: f Express Ii: I i 101 ML2 2 14 m 2 Express If: I f 101 ML2 2 14 mL2 Substitute to express f in terms of : f Express the initial kinetic energy of the system: Ii I i i If If 1 10 1 10 2 mL ML2 2 14 m 2 2 ML 2 L2 M 5m M 5m Ki 12 I i 2 1 4 2 1 20 ML 2 1 1 2 10 ML2 2 14 m 2 2 5m 2 2 Express the final kinetic energy of the system and simplify to obtain: K f 12 I f f2 1 1 2 10 ML2 2 14 mL2 f2 2 1 20 1 20 2 M 5m 2 L ML2 5mL2 M 5m 1 20 ML 2 5mL2 f2 2 2 ML 5m L 2 1 20 M 5m ML2 5m 2 2 2 2 2 ML 5mL 85 •• Determine the Concept Yes. The net external torque is zero and angular momentum is conserved as the system evolves from its initial to its final state. Because the disks come to the same final position, the initial and final configurations are the same as in Problem 84. Therefore, the answers are the same as for Problem 84.