1 / 16

Chapter 5, Lect. 10 Additional Applications of Newton’s Laws

About Midterm Exam 1. When and where TODAY: 5:45-7:00 pm Rooms: See course webpage. Be sure report to your TA’s room Your TA will give a review during the discussion session next week. Format Closed book, 20 multiple-choices questions (consult with practice exam)

xantha-mccarthy
Download Presentation

Chapter 5, Lect. 10 Additional Applications of Newton’s Laws

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. About Midterm Exam 1 • When and where • TODAY: 5:45-7:00 pm • Rooms: See course webpage. Be sure report to your TA’s room • Your TA will give a review during the discussion session next week. • Format • Closed book, 20 multiple-choices questions (consult with practice exam) • One-page formula sheet allowed, must be self prepared, no photo copying/download-printing of solutions, lecture slides, etc. • Bring a calculator (but no computer). Only basic calculation functionality can be used. Bring a 2B pencil for Scantron. • Fill in your ID and section # ! • Special requests: • One alternative exam all set: • 3:30pm – 4:45pm, Thurs Feb.17, room 5280 Chamberlin (!). Chapter 5, Lect. 10Additional Applications of Newton’s Laws Today: circular motion, center of mass Phys 201, Spring 2011

  2. The figure shows a top view of a ball on the end of a string traveling counterclockwise in a circular path. The speed of the ball is constant. If the string should break at the instant shown, the path that the ball would follow is • 1 • 2 • 3 • 4 • impossible to tell from the given information. Phys 201, Spring 2011

  3. Acceleration on a curved path Instead of considering a = ax i + ay j + az k (time-independent) Decomposed into: a = at + ac Tangential acceleration: at = dv/dt The magnitude change of v. Centripetal acceleration: ac The direction change of v. Phys 201, Spring 2011

  4. Centripetal acceleration is the acceleration perpendicular to the velocity that occurs when a particle is moving on a curved path.  • Centripetal force associated with centripetal acceleration, directed towards the center of the circle: Phys 201, Spring 2011

  5. Uniform Circular Motion If object is moving with constant speed on the circle, v = const. r = const. ac = v2/r = const. Motion in a Horizontal Circle The centripetal force is supplied by the tension Phys 201, Spring 2011

  6. Example: An object of mass m is suspended from a point in the ceiling on a string of length L. The object revolves with constant speed v in a horizontal circle of radius r. (The string makes an angle θ with the vertical). The speed v is given by the expression: θ L y Phys 201, Spring 2011

  7. Horizontal (Flat) Curve • The force of static friction supplies the centripetal force • The maximum speed at which the car can negotiate the curve is Note, this does not depend on the mass of the car Phys 201, Spring 2011

  8. A car going around a curve of radius R at a speed V experiences a centripetal acceleration ac. What is its acceleration if it goes around a curve of radius 3R at a speed of 2V? • (2/3)ac • (4/3)ac • (2/9)ac • (9/2)ac • (3/2)ac Phys 201, Spring 2011

  9. Banked Curve • These are designed to be navigable when there is no friction • There is a component of the normal force that supplies the centripetal force (even μ=0!) nx = ny = Phys 201, Spring 2011

  10. Non-Uniform Circular Motion • The acceleration and force have tangential components • Fr produces the centripetal acceleration (change v in directions) • Ft produces the tangential acceleration (v change in magnitude) • ΣF = ΣFr+ΣFt Phys 201, Spring 2011

  11. Vertical Circle With Non-Uniform Speed • The gravitational force exerts a tangential force on the object • Look at the components of Fg • The tension at any point: • along ac direction: • ac =T - gravity Phys 201, Spring 2011

  12. Top and Bottom of Circle • The tension at the bottom is a maximum: cosθ = +1 • The tension at the top is a minimum: cosθ = -1 • If Ttop = 0, gravity does it: Phys 201, Spring 2011

  13. The Center of Mass • Definition of center of mass: Where For a continuous object (e.g., a solid sphere) Phys 201, Spring 2011

  14. CM position for a semicircular hoop where M = λπR, CM position can be outside the body. Phys 201, Spring 2011

  15. Center of Mass (2) • Newton’s Laws for a collection of objects: The acceleration of the center of mass is determined entirely by the external net force on the objects. Phys 201, Spring 2011

  16. Changing Places in a Rowboat : Fnet Ext = 0  Xcm fixed (initial condition: 0) mP XP + mD XD + mb Xb = 0 mP X’P + mD X’D + mb X’b = 0 mP ΔXP + mD ΔXD + mb ΔXb = 0 with ΔXP = -ΔXD = L Thus, (mP – mD) L = -mb ΔXb ΔXb = L (mD – mP)/mb . Phys 201, Spring 2011

More Related