#### Transcript Chapter 5: Circular Motion

Chapter 5 Circular Motion Circular Motion • Uniform Circular Motion • Radial Acceleration • Banked and Unbanked Curves • Circular Orbits • Nonuniform Circular Motion • Tangential and Angular Acceleration • Artificial Gravity MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 2 Angular Displacement y is the angular position. f i Angular displacement: x f i Note: angles measured CW are negative and angles measured CCW are positive. is measured in radians. 2 radians = 360 = 1 revolution MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 3 Arc Length y arc length = s = r f r i x s r MFMcGraw-PHY 1401 is a ratio of two lengths; it is a dimensionless ratio! Ch5b-Circular Motion-Revised 6/21/2010 4 Angular Speed The average and instantaneous angular speeds are: av and lim t 0 t t is measured in rads/sec. MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 5 Angular Velocity The average and instantaneous angular speeds are: av and lim t 0 t t is actually a vector quantity. The direction of is along the axis of rotation. Since we are concerned primarily with motion in a plane we will ignore the vector nature until we get to angular momentum when we will have to deal with it. MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 6 Linear Speed y An object moves along a circular path of radius r; what is its average linear speed? f r i x Note: Unfortunately the linear speed is also called the tangential speed. This can cause confusion. total distance r vav r r av total time t t Also, MFMcGraw-PHY 1401 v r (instantaneous values). Ch5b-Circular Motion-Revised 6/21/2010 7 Period and Frequency The time it takes to go one time around a closed path is called the period (T). total distance 2r vav total time T 2 Comparing to v = r: 2f T f is called the frequency, the number of revolutions (or cycles) per second. MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 8 Linear and Angular Speed The linear speed depends on the radius r at which the object is located. y v r v1 Angular speed is independent of position. r1 v1 = ωr1 v2 v2 = ωr2 v1 v2 = =ω r1 r2 r2 x v1 Everyone sees the same ω, because they all experience the same rpms. MFMcGraw-PHY 1401 v1 Ch5b-Circular Motion-Revised 6/21/2010 v1 9 Centripetal Acceleration Consider an object moving in a circular path of radius r at constant speed. y v Here, v 0. The direction of v is changing. v x If v 0, then a 0. Then there is a net force acting on the object. v v MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 10 Centripetal Acceleration Conclusion: with no net force acting on the object it would travel in a straight line at constant speed It is still true that F = ma. But what acceleration do we use? MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 11 Centripetal Acceleration The velocity of a particle is tangent to its path. For an object moving in uniform circular motion, the acceleration is radially inward. MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 12 Centripetal Acceleration The magnitude of the radial acceleration is: 2 v 2 ar r v r MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 13 Rotor Ride Example The rotor is an amusement park ride where people stand against the inside of a cylinder. Once the cylinder is spinning fast enough, the floor drops out. (a) What force keeps the people from falling out the bottom of the cylinder? y fs N Draw an FBD for a person with their back to the wall: x w It is the force of static friction. MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 14 Rotor Ride Example (b) If s = 0.40 and the cylinder has r = 2.5 m, what is the minimum angular speed of the cylinder so that the people don’t fall out? Apply Newton’s 2nd Law: From (2): fs w 1 Fx N mar m 2 r 2 Fy f s w 0 s N s m 2 r mg From (1) 9.8 m/s 2 3.13 rad/s 0.402.5 m s r g MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 15 Unbanked Curve A coin is placed on a record that is rotating at 33.3 rpm. If s = 0.1, how far from the center of the record can the coin be placed without having it slip off? y We’re looking for r. N Draw an FBD for the coin: Apply Newton’s 2nd Law: fs x 1 Fx f s mar m 2 r 2 Fy N w 0 MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 w 16 Unbanked Curve From (2) From 1 : f s m 2 r f s s N s mg m 2 r s g Solving for r: r 2 What is ? rev 2 rad 1 min 33.3 3.5 rad/s min 1 rev 60 sec s g 0.19.8 m/s 2 r 2 0.08 m 2 3.50 rad/s MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 17 Banked Curves A highway curve has a radius of 825 m. At what angle should the road be banked so that a car traveling at 26.8 m/s has no tendency to skid sideways on the road? (Hint: No tendency to skid means the frictional force is zero.) R Take the car’s motion to be into the page. MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 18 Banked Curves The normal force is the cause of the centripetal acceleration. Nsinθ. We need the radial position of the car itself, not the radius of the track. In either case the radius is not a uniquely determined quantity. y This is the only path by which to traverse the track as there is no friction in the problem. N x w MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 19 Banked Curves y FBD for the car: N x w Apply Newton’s Second Law: v2 1 Fx N sin mar m r 2 Fy N cos w 0 MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 20 Banked Curves Rewrite (1) and (2): v2 1 N sin m r 2 N cos mg Divide (1) by (2): v 26.8 m/s tanθ = = = 0.089 2 gr 9.8 m/s 825 m 2 2 θ = 5.1° MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 21 Circular Orbits Consider an object of mass m in a circular orbit about the Earth. v r Earth The only force on the satellite is the force of gravity. That is the cause of the centripetal force. Gms M e v2 F Fg r 2 ms ar ms r Solve for the speed of the satellite: MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 Gms M e v2 ms 2 r r GM e v r 22 Circular Orbits Consider an object of mass m in a circular orbit about the Earth. v r Earth v= GM e r GMe is fixed - v is proportional to 1 r v and r are tied together. They are not independent for orbital motions. MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 23 Circular Orbits Example: How high above the surface of the Earth does a satellite need to be so that it has an orbit period of 24 hours? GM e v r From previous slide: Combine these expressions and solve for r: v Also need, GM e 2 r T 2 4 6.67 10 Nm /kg 5.98 10 kg 2 r 86400 s 2 4 4.225 107 m 11 2 2 24 2r T 1 1 3 3 r Re h h r Re 35,000 km MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 24 Circular Orbits Kepler’s Third Law GM e 2 r T 2 4 GM 2 r It can be generalized to: 2T 4 1 1 3 3 Where M is the mass of the central body. For example, it would be Msun if speaking of the planets in the solar system. For non-circular orbits (elliptical) the mean radius is used and so the mean velocity is obtained. MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 25 Nonuniform Circular Motion Nonuniform means the speed (magnitude of velocity) is changing. a at There is now an acceleration tangent to the path of the particle. ar v The net acceleration of the body is a ar at 2 2 This is true but useless! MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 26 Nonuniform Circular Motion a at at changes the magnitude of v. ar Changes energy - does work ar changes the direction of v. Doesn’t change energy does NO WORK Can write: MFMcGraw-PHY 1401 F ma F ma r r t t The accelerations are only useful when separated into perpendicular and parallel components. Ch5b-Circular Motion-Revised 6/21/2010 27 Loop Ride Example: What is the minimum speed for the car so that it maintains contact with the loop when it is in the pictured position? FBD for the car at the top of the loop: r y Apply Newton’s 2nd Law: x N w MFMcGraw-PHY 1401 v2 Fy N w mar m r v2 N wm r Ch5b-Circular Motion-Revised 6/21/2010 28 Loop Ride The apparent weight at the top of loop is: You lose contact when N = 0 N = 0 when v2 N wm r v2 N m g r v2 N m g 0 r v gr This is the minimum speed needed to make it around the loop. MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 29 Loop Ride Consider the car at the bottom of the loop; how does the apparent weight compare to the true weight? FBD for the car at the bottom of the loop: y N x w MFMcGraw-PHY 1401 Apply Newton’s 2nd Law: v2 Fy N w mac m r v2 N wm r v2 N m g r Here, N mg Ch5b-Circular Motion-Revised 6/21/2010 30 Linear and Angular Acceleration The average and instantaneous angular acceleration are: av and lim t 0 t t is measured in rads/sec2. MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 31 Linear and Angular Acceleration Recalling that the tangential velocity is vt = r means the tangential acceleration is vt r r at t t t MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 32 Linear and Angular Kinematics Angular Linear (Tangential) v v0 at 0 t 1 2 x x0 v0 t at 2 v 2 v02 2ax 1 2 0 0 t t 2 2 02 2 With vt r and at r “a” and “at” are the same thing MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 33 Dental Drill Example A high speed dental drill is rotating at 3.14104 rads/sec. Through how many degrees does the drill rotate in 1.00 sec? Given: = 3.14104 rads/sec; t = 1 sec; = 0 Want . 1 0 0 t t 2 2 0 0 t 0 t 3.14 10 4 rads/sec 1.0 sec 3.14 10 4 rads 1.80 106 degrees MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 34 Dental Drill Example A high speed dental drill is rotating at 3.14104 rads/sec. What is that in rpm’s? rad rev rev 4 rev 1 π × 10 × = 2 10 = 5000 s 2πrad s s rev 60sec 5000 × = 300,000rpm s min 4 MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 35 Car Example Your car’s wheels are 65 cm in diameter and are rotating at = 101 rads/sec. How fast in km/hour is the car traveling, assuming no slipping? v X total distance 2r N 2r v r T N T total time 101 rads/sec 32.5 cm 3.28 103 cm/sec 118 km/hr MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 36 Artificial Gravity A large rotating cylinder in deep space (g0). MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 37 Artificial Gravity y y N x x N Top position Bottom position Apply Newton’s 2nd Law to each: 2 F N ma m r y r 2 F N ma m r y r N is the only force acting. It causes the centripetal acceleration. MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 38 Space Station Example A space station is shaped like a ring and rotates to simulate gravity. If the radius of the space station is 120m, at what frequency must it rotate so that it simulates Earth’s gravity? Using the result from the previous slide: 2 F N ma m r y r N mg mr mr g 0.28 rad/sec r The frequency is f = (/2) = 0.045 Hz (or 2.7 rpm). MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 39 Summary • A net force MUST act on an object that has circular motion. • Radial Acceleration ar=v2/r • Definition of Angular Quantities (, , and ) • The Angular Kinematic Equations • The Relationships Between Linear and Angular Quantities v r and a r t • t Uniform and Nonuniform Circular Motion MFMcGraw-PHY 1401 Ch5b-Circular Motion-Revised 6/21/2010 40