Lecture08

Lecture08

• 1.
  Centripetal Acceleration and Circular Motion Today’s lecture will cover Chapter 5 Physics 101: Lecture 08 Exam I http://www.youtube.com/watch?v=ZyF5WsmXRaI
• 2.
  Circular Motion ActA B C Answer: B v A ball is going around in a circle attached to a string. If the string breaks at the instant shown, which path will the ball follow (demo)?
• 3.
  Acceleration in UniformCircular Motion a ave =  v /  t Acceleration inward Acceleration is due to change in direction, not speed. Since turns “toward” center, must be a force toward center. v 1 v 2  v v 2 v 1 R  R centripetal acceleration
• 4.
  Preflights Consider the following situation: You are driving a car with constant speed around a horizontal circular track. On a piece of paper, draw a Free Body Diagram (FBD) for the car. How many forces are acting on the car? A) 1 B) 2 C) 3 D) 4 E) 5 1% 24% 38% 23% 14% “ Friction, Gravity, & Normal” f W F N correct  F = m a = m v 2 /R a=v 2 /R R
• 5.
  Common Incorrect ForcesAcceleration:  F = ma Centripetal Acceleration Force of Motion (Inertia not a force) Forward Force, Force of velocity Speed Centrifugal Force (No such thing!) Centripetal (really acceleration) Inward force (really friction) Internal Forces (don’t count, cancel) Car Engine
• 6.
  Preflights Consider the following situation: You are driving a car with constant speed around a horizontal circular track. On a piece of paper, draw a Free Body Diagram (FBD) for the car. The net force on the car is A. Zero B. Pointing radially inward C. Pointing radially outward 14% 81% 5% “ Centripetal acceleration (the net force) is always pointing INWARD toward the center of the circle you are moving in..” “ must be a net force drawing the vehicle inward otherwise the cars direction would not change, and so the car would drive off the track and crash and burn” f W F N  F = m a = m v 2 /R a=v 2 /R R correct
• 7.
  ACT Suppose youare driving through a valley whose bottom has a circular shape. If your mass is m , what is the magnitude of the normal force F N exerted on you by the car seat as you drive past the bottom of the hill A. F N < mg B. F N = mg C. F N > mg v  F = m a F N - mg = m v 2 /R F N = mg + m v 2 /R mg F N R a=v 2 /R correct
• 8.
  Roller Coaster ExampleWhat is the minimum speed you must have at the top of a 20 meter roller coaster loop, to keep the wheels on the track. mg Y Direction: F = ma -N – mg = m a Let N = 0, just touching -mg = m a -mg = -m v 2 /R g = v 2 / R v = sqrt(g*R) = 9.9 m/s N
• 9.
  Circular Motion Angulardisplacement  =  2 -  1 How far it has rotated Angular velocity  =  t How fast it is rotating Units radians/second 2  = 1 revolution Period =1/frequency T = 1/f = 2  Time to complete 1 revolution
• 10.
  Circular to LinearDisplacement  s = r    in radians) Speed |v| =  s/  t = r  /  t = r  Direction of v is tangent to circle
• 11.
  Bonnie sits onthe outer rim of a merry-go-round with radius 3 meters, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds (demo). Klyde’s speed is: Merry-Go-Round ACT (a) the same as Bonnie’s (b) twice Bonnie’s (c) half Bonnie’s Klyde Bonnie Bonnie travels 2  R in 2 seconds v B = 2  R / 2 = 9.42 m/s Klyde travels 2  (R/2) in 2 seconds v K = 2  (R/2) / 2 = 4.71 m/s
• 12.
  Merry-Go-Round ACT IIBonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds. Klyde’s angular velocity is: (a) the same as Bonnie’s (b) twice Bonnie’s (c) half Bonnie’s Klyde Bonnie The angular velocity  of any point on a solid object rotating about a fixed axis is the same . Both Bonnie & Klyde go around once (2  radians) every two seconds.
• 13.
  Angular Acceleration Angularacceleration is the change in angular velocity  divided by the change in time. If the speed of a roller coaster car is 15 m/s at the top of a 20 m loop, and 25 m/s at the bottom. What is the cars average angular acceleration if it takes 1.6 seconds to go from the top to the bottom? = 0.64 rad/s 2
• 14.
  Summary (withcomparison to 1-D kinematics) Angular Linear And for a point at a distance R from the rotation axis: x = R  v =  R  a =  R
• 15.
  CD Player ExampleThe CD in your disk player spins at about 20 radians/second. If it accelerates uniformly from rest with angular acceleration of 15 rad/s 2 , how many revolutions does the disk make before it is at the proper speed?  = 13.3 radians 1 Revolutions = 2  radians  = 13.3 radians = 2.12 revolutions
• 16.
  Summary of ConceptsUniform Circular Motion Speed is constant Direction is changing Acceleration toward center a = v 2 / r Newton’s Second Law F = ma Circular Motion  = angular position radians  = angular velocity radians/second  = angular acceleration radians/second 2 Linear to Circular conversions s = r  Uniform Circular Acceleration Kinematics Similar to linear!