Purdue University, Physics 220 Feb. 2 Lecture 7 missed Chapter 5 Feb. 4 Recitation 4 HW4 Feb. 7 Lecture 7 Feb. 9 Lecture 8 Feb. 11 Recitation 5 Feb. 14 Lecture 9 Chapter 6 Feb. 16 Lecture 10 Exam 1 Feb. 18 Recitation 6 HW5 /HW6 Feb. 21 Lecture 11 Chapter 7 Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 UNIMPORTABLE: #1D987FFA,2 #1DB9EC48,2 #2238DDC7,4 #DB602B9,2 #DD17CA0,4 #DF9EC18,2 #E316E51,4 Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 Lecture 07 Circular Motion Lecture 7 Purdue University, Physics 220

Examples of Circular Motion Rotate ruler 90 degrees ask students how far rotated. Motivate meters bad unit, use angles. Also w/ record player Lecture 7 Purdue University, Physics 220

Uniform Circular Motion Assume constant speed The direction of the velocity is continually changing The vector is always tangent to the circle Uniform circular motion assumes constant speed period of the motion: T = 2r/v [s]

Purdue University, Physics 220 Angular Variables The motion of objects moving in circular (or nearly circular) paths, is often described by angles measured in radians rather than degrees. The angle  in radians, is defined as: If s = r the angle is 1 rad If s = 2r (the circumference of the circle) the angle is 2 rad. (In other words, 360° = 2 rad.) Rotate ruler 90 degrees ask students how far rotated. Motivate meters bad unit, use angles. Also w/ record player Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 Circular Motion Angular displacement Dq = q2-q1 How far it has rotated Angular velocity wav = Dq/Dt How fast it is rotating Units: radians/second (2p = 1 revolution) Period = 1/frequency T = 1/f = 2p / w Time to complete 1 revolution Rotate ruler 90 degrees ask students how far rotated. Motivate meters bad unit, use angles. Also w/ record player Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 Circular to Linear Displacement Ds = r Dq (q in radians) Speed |v| = |Ds/Dt| = r |Dq/Dt| = r|w |v| = 2rf |v| = 2r/T Direction of v is tangent to circle Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 iClicker Bonnie sits on the 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. Klyde’s speed is: Klyde Bonnie A) the same as Bonnie’s B) twice Bonnie’s C) half Bonnie’s Bonnie travels 2 p R in 2 seconds vB = 2 p R / 2 = 9.42 m/s Klyde travels 2 p (R/2) in 2 seconds vK = 2 p (R/2) / 2 = 4.71 m/s Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 Examples The wheel of a car has a radius of 0.29 m and is being rotated at 830 revolution per minute (rpm) on a tire-balancing machine. Determine the speed (in m/s) at which the outer edge of the wheel is moving: The speed could be obtained by |v| = 2r/T A CD spins with an angular frequency 20 radians/second. What is the linear speed 6 cm from the center of the CD? v = r w = 0.06  20 = 1.2 m/s Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 Circular Motion 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? 1 2 3 v 4 5 Use this to motivate circular motion involves acceleration. Answer: 2 Lecture 7 Purdue University, Physics 220

Centripetal Acceleration Magnitude of the velocity vector is constant, but direction is constantly changing At any instant of time, the direction of the instantaneous velocity is tangent to the path Therefore: nonzero acceleration Lecture 7 Purdue University, Physics 220

Uniform Circular Motion Circular motion with constant speed R centripetal acceleration Recall: v = w R v a Instantaneous velocity is tangent to circle Instantaneous acceleration is radially inward There must be a force to provide the acceleration Lecture 7 Purdue University, Physics 220

Roller Coaster Example What is the minimum speed you must have at the top of a 20 meter diameter roller coaster loop, to keep the wheels on the track. y-direction: F = ma N + mg = m a Let N = 0, just touching mg = m a mg = m v2/R g = v2 / R v = sqrt(g*R) = 10 m/s N y mg Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 Unbanked Curve What force accelerates a car around a turn on a level road at constant speed? A) it is not accelerating B) the road on the tires C) the tires on the road D) the engine on the tires Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 Unbanked Curve What is the maximum velocity a car can go around an unbanked curve in a circle without slipping? The maximum velocity to go around an un-banked curve depends only on s (for a given r) Dry road: s=0.9 Icy road: s=0.1 Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 Banked Curve A car drives around a curve with radius 410 m at a speed of 32 m/s. The road is banked at 5.0°. The mass of the car is 1400 kg. A) What is the frictional force on the car? B) At what speed could you drive around this curve so that the force of friction is zero? Work out on overhead. Use all the standard steps: draw figure, list what is given using symbols, draw a FBD, write 2nd law (component form), write equations that result, give answer (can skip algebra). Algebra is rather complicated and is in the answer book. No need to spend time on that. Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 Banked Curve  x y y-direction (1) (2) N W f x-direction Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 Banked Curve 2 equations and 2 unknown we can solve for N in (1) and substitute in (2) Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 Banked Curve A car drives around a curve with radius 410 m at a speed of 32 m/s. The road is banked at 5.0°. The mass of the car is 1400 kg. A) What is the frictional force on the car? B) At what speed could you drive around this curve so that the force of friction is zero? Like an airplane Work out on overhead. Use all the standard steps: draw figure, list what is given using symbols, draw a FBD, write 2nd law (component form), write equations that result, give answer (can skip algebra). Algebra is rather complicated and is in the answer book. No need to spend time on that. Lecture 7 Purdue University, Physics 220

Purdue University, Physics 220 iClicker Suppose you are driving through a valley whose bottom has a circular shape. If your mass is m, what is the magnitude of the normal force N exerted on you by the car seat as you drive past the bottom of the hill. A) N < mg B) N = mg C) N > mg v mg N R a=v2/R correct Since there is centripetal acceleration, the normal force is greater than simply mg SF = ma N - mg = mv2/R N = mg + mv2/R Lecture 7 Purdue University, Physics 220