1 / 33

Experiment 5: Rotational Dynamics and Angular Momentum 8.01 W10D1 Young and Freedman: 10.5-10.6

Experiment 5: Rotational Dynamics and Angular Momentum 8.01 W10D1 Young and Freedman: 10.5-10.6. Vote Tomorrow Math Review Night Tuesday from 9-11 pm Pset 9 Due Nov 8 at 9 pm No Class Friday Nov 11 Exam 3 Tuesday Nov 22 7:30-9:30 pm W010D2 Reading Assignment Young and Freedman: 10.5-10.6.

bseal
Download Presentation

Experiment 5: Rotational Dynamics and Angular Momentum 8.01 W10D1 Young and Freedman: 10.5-10.6

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Experiment 5: Rotational DynamicsandAngular Momentum8.01W10D1Young and Freedman: 10.5-10.6

  2. Vote Tomorrow Math Review Night Tuesday from 9-11 pm Pset 9 Due Nov 8 at 9 pm No Class Friday Nov 11 Exam 3 Tuesday Nov 22 7:30-9:30 pm W010D2 Reading Assignment Young and Freedman: 10.5-10.6 Announcements

  3. Rotor Moment of Inertia

  4. Review Table Problem: Moment of Inertia Wheel A steel washer is mounted on a cylindrical rotor . The inner radius of the washer is R. A massless string, with an object of mass m attached to the other end, is wrapped around the side of the rotor and passes over a massless pulley. Assume that there is a constant frictional torque about the axis of the rotor. The object is released and falls. As the mass falls, the rotor undergoes an angular acceleration of magnitude a1. After the string detaches from the rotor, the rotor coasts to a stop with an angular acceleration of magnitude a2. Let g denote the gravitational constant. What is the moment of inertia of the rotor assembly (including the washer) about the rotation axis?

  5. Review Solution: Moment of Inertia of Rotor Force and rotational equations while weight is descending: Constraint: Rotational equation while slowing down Solve for moment of inertia: Speeding up Slowing down

  6. Review Worked Example: Change in Rotational Energy and Work While the rotor is slowing down, use work-energy techniques to find frictional torque on the rotor.

  7. Experiment 5: Rotational Dynamics

  8. Review: Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors Direction: determined by the Right-Hand-Rule

  9. Angular Momentum of a Point Particle Point particle of mass m moving with a velocity Momentum Fix a point S Vector from the point S to the location of the object Angular momentum about the point S SI Unit

  10. Cross Product: Angular Momentum of a Point Particle Magnitude: moment arm perpendicular momentum

  11. Angular Momentum of a Point Particle: Direction Direction: Right Hand Rule

  12. Concept Question: • In the above situation where a particle is moving in the x-y plane • with a constant velocity, the magnitude of the angular momentum • about the point S (the origin) • decreases then increases • increases then decreases • is constant • is zero because this is not circular motion

  13. Concept Question: Solution 3. As the particle moves in the positive x-direction, the perpendicular distance from the origin to the line of motion does not change and so the magnitude of the angular momentum about the origin is constant.

  14. Table Problem: Angular Momentum and Cross Product A particle of massm = 2 kg moves with a uniform velocity At timet, the position vector of the particle with respect ot the point S is Find the direction and the magnitude of the angular momentum about the origin, (the pointS) at timet.

  15. Angular Momentum and Circular Motion of a Point Particle: Fixed axis of rotation: z-axis Angular velocity Velocity Angular momentum about the point S

  16. Concept Question A particle of mass m moves in a circle of radius R at an angular speed ω about the z axis in a plane parallel to but above the x-y plane. Relative to the origin 1. is constant. 2. is constant but is not. 3. is constant but is not. 4. has no z-component. .

  17. Concept Question Answer 2. The vector represents a unit vector in the direction of . The angular momentum about the origin is shown in the figure to the right The magnitude is constant. As the particle moves in a circle, the angular momentum sweeps out a cone (shown in the figure below) and so the direction of is changing and hence is not a constant unit vector.

  18. Worked Example : Changing Direction of Angular Momentum A particle of mass m moves in a circle of radius R at an angular speed ω about the z axis in a plane parallel to but a distance h above the x-y plane. Find the magnitude and the direction of the angular momentum relative to the origin. Also find the z component of .

  19. Solution: Changing Direction of Angular Momentum

  20. Solution: Changing Direction of Angular Momentum

  21. Table Problem: Angular Momentum of a Two Particles About Different Points Two point like particles 1 and 2, each of mass m, are rotating at a constant angular speed about point A. How does the angular momentum about the point B compare to the angular momentum about point A? What about at a later time when the particles have rotated by 90 degrees?

  22. Table Problem: Angular Momentum of Two Particles Two identical particles of mass m move in a circle of radius R, 180º out of phase at an angular speed ω about the z axis in a plane parallel to but a distance h above the x-y plane. a) Find the magnitude and the direction of the angular momentum relative to the origin. b) Is this angular momentum relative to the origin constant? If yes, why? If no, why is it not constant?

  23. Table Problem: Angular Momentum of a Ring A circular ring of radius R and mass dm rotates at an angular speed ω about the z-axis in a plane parallel to but a distance h above the x-y plane. a) Find the magnitude and the direction of the angular momentum relative to the origin. b) Is this angular momentum relative to the origin constant? If yes, why? If no, why is it not constant?

  24. Concept Question A non-symmetric body rotates with constant angular speed ω about the z axis. Relative to the origin 1. is constant. 2. is constant but is not. 3. is constant but is not. 4. has no z-component.

  25. Concept Question Answer 2. For this non-symmetric rigid body, the angular momentum about the origin has time varying components in the x-y plane, (where is a radial unit vector pointing outward from the origin) The magnitude of is constant because both and are constant. The direction of in the x-y plane depends on the instantaneous orientation of the body and so the direction of is changing.

  26. Concept Question A rigid body with rotational symmetry body rotates at a constant angular speed ω about it symmetry (z) axis. In this case 1. is constant. 2. is constant but is not. 3. is constant but is not. 4. has no z-component. 5. Two of the above are true.

  27. Concept Question Answer 1. For a symmetric body, all the non-z components of the angular momentum about any point along the z-axis cancel in pairs leaving only a constant non-zero z-component of the angular momentum so is constant.

  28. Angular Momentum of Cylindrically Symmetric Body A cylindrically symmetric rigid body rotating about its symmetry axis at a constant angular velocity with has angular momentum about any point on its axis

  29. Angular Momentumfor Fixed Axis Rotation Angular Momentum about the point S Tangential component of momentum z-component of angular momentum about S:

  30. Concept Question: Angular Momentum of Disk A disk with mass M and radius R is spinning with angular speed ω about an axis that passes through the rim of the disk perpendicular to its plane. The magnitude of its angular momentum is:

  31. Concept Question: Angular Momentum of Disk Answer 6. The moment of inertia of the disk about an axis that passes through the rim of the disk perpendicular to its plane is I = Icm + MR2 = (3/2)MR2 . So the magnitude of its angular momentum is L = (3/2)MR2 ω.

More Related