Chapter 5, Lect. 10 Additional Applications of Newton’s Laws Today: circular motion, center of mass 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 (!). About Midterm Exam 1 4/24/2017 Phys 201, Spring 2011

impossible to tell from the given information. 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. 4/24/2017 Phys 201, Spring 2011

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 4/24/2017

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: 4/24/2017 Phys 201, Spring 2011

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 4/24/2017 Phys 201, Spring 2011

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 4/24/2017 Phys 201, Spring 2011

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 4/24/2017 Phys 201, Spring 2011

(2/3)ac (4/3)ac (2/9)ac (9/2)ac (3/2)ac 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 4/24/2017 Phys 201, Spring 2011

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 = 4/24/2017 Phys 201, Spring 2011

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 4/24/2017 Phys 201, Spring 2011

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 4/24/2017 Phys 201, Spring 2011

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: 4/24/2017 Phys 201, Spring 2011

The Center of Mass Definition of center of mass: Where For a continuous object (e.g., a solid sphere) 4/24/2017 Phys 201, Spring 2011

CM position for a semicircular hoop where M = λπR, CM position can be outside the body. 4/24/2017 Phys 201, Spring 2011

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. 4/24/2017 Phys 201, Spring 2011

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 . 4/24/2017 Phys 201, Spring 2011