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Transcript
Chapter 5: Circular Motion •Uniform Circular Motion •Radial Radial Acceleration •Banked and Unbanked Curves •Circular Orbits •Nonuniform Circular Motion •Tangential and Angular Acceleration •Artificial Gravity 1 Uniform Circular Motion y θ is the angular position. θf Δθ θi Angular displacement: x Δθ = θ f − θ i Note: angles measured CW are negative and angles measured CCW are p positive. θ is measured in radians. 2π radians = 360° = 1 revolution 2 The average and instantaneous angular velocities are: Δθ Δθ ωav = and ω = lim Δt →0 Δt Δt ω is measured in rads/sec. 3 y arclength = s = rΔθ θf r Δθ θi x s Δθ = r Δθ is a ratio of two lengths; it is a dimensionless ratio! 4 y An object moves along a circular path of radius r; what is its average speed? θf r Δθ θi x total distance rΔθ ⎛ Δθ ⎞ vav = = = r⎜ ⎟ = rω av total time Δt ⎝ Δt ⎠ Also, v = rω (instantaneous values). 5 The time it takes to go one time around a closed path is called the period (T). totall distance di 2πr = vav = total time T 2π Comparing to v = rω: ω = = 2πf T f is called the frequency, the number of revolutions (or cycles) per second. 6 Centripetal Acceleration y Consider an object moving in a circular path of radius r at constant speed speed. v v x Here, Δv ≠ 0. The direction of v is changing. v If Δv ≠ 0, then a ≠ 0. The net force cannot be zero zero. v 7 Conclusion: to move in a circular path, an object must have a nonzero net force acting on itit. It is still true that ΣF = ma, but what acceleration do we use? 8 Th velocity The l i off a particle i l iis tangent to iits path. h For an object moving in uniform circular motion motion, the acceleration is radially inward. 9 The magnitude of the centripedal (or radial) acceleration is: 2 v 2 ar = = rω = ωv r 10 Uniform Circular Motion Slide 6-13 Examples The disk in a hard drive in a desktop computer rotates at 7200 rpm. The disk has a diameter of 5.1 in (13 cm.) What is the angular speed of the disk? The hard drive disk in the previous example rotates at 7200 rpm. The disk has a diameter of 5.1 in (13 cm.) What is the speed d off a point i t 6.0 6 0 cm ffrom th the center t axle? l ? Wh Whatt iis th the acceleration of this point on the disk? Slide 6-14 Quiz 1 For uniform circular motion 1. motion, the acceleration A. is parallel to the velocity. B. is directed toward the center of the circle. C is larger for a larger orbit at the same speed C. speed. D. is always due to gravity. E. is always negative. Slide 6-2 Answer 1 For uniform circular motion 1. motion, the acceleration B. is directed toward the center of the circle. Slide 6-3 Understanding When a ball on the end of a string is swung in a vertical circle: We know that the ball is accelerating because A. the speed is changing. B. the direction is changing. C. the speed and the direction are changing. Slide 6-9 Answer When a ball on the end of a string is swung in a vertical circle: We know that the ball is accelerating because B. the direction is changing. Slide 6-10 Understanding When a ball on the end of a string is swung in a vertical circle: What is the direction of the acceleration of the ball? A. Tangent to the circle, in the direction of the ball’s motion B. Toward the center of the circle Slide 6-11 Answer When a ball on the end of a string is swung in a vertical circle: What is the direction of the acceleration of the ball? B. Toward the center of the circle Slide 6-12 Circular Motion Dynamics When the ball reaches the break in the circle, circle which path will it follow? Slide 6-19 Answer When the ball reaches the break in the circle, circle which path will it follow? Slide 6-20 Forces in Circular Motion v = ωr v2 a = — = ω2 r r → → Fnett = ma = { } mv2 —, toward center of circle r Slide 6-21 Example A level curve on a country road has a radius of 150 m. What is the maximum speed at which this curve can be safely negotiated on a rainy day when the coefficient of friction between th tires the ti on a car and d th the road d iis 0.40? Slide 6-24 Driving over a Rise A car of mass 1500 kg goes over a hill at a speed of 20 m/s. The shape of the hill is approximately circular, with a radius of 60 m, as in the figure at right. When the car is at the highest point of the hill, a What is the force of gravity on a. the car? b. What is the normal force of the road on the car at this point? Slide 6-26 Example: The rotor is an amusement park ride where people stand t d against i t the th inside i id off a cylinder. li d O Once th the cylinder li d iis spinning fast enough, the floor drops out. (a) What force keeps the people from falling out the bottom of the cylinder? y fs Draw an FBD for a person with ith th their i b back k tto th the wall: ll N x w It is the force of static friction. 24 Example continued: (b) If μs = 0.40 and the cylinder has r = 2.5 m, what is the minimum angular speed of the cylinder so that the people don’tt fall out? don Apply pp y Newton’s 2nd Law: From (2): (1) ∑ Fx = N = mar = mω 2 r (2) ∑ Fy = f s − w = 0 From (1) fs = w μ s N = μ s (mω 2 r ) = mg ω= 9.8 m/s 2 = = 3.13 rad/s (0.40)(2.5 m ) μs r g 25 Example (text problem 5.79): A coin is placed on a record that is rotating at 33 33.3 3 rpm rpm. If μs = 0.1, 0 1 how ho far from the center of the record can the coin be placed without having it slip off? y Draw an FBD for the coin: N fs x Apply Newton’s 2nd Law: (1) ∑ Fx = f s = mar = mω (2) ∑ Fy = N − w = 0 2 w r 26 Example continued: From (1) : f s = mω 2 r From (2) f s = μ s N = μ s (mg ) = mω 2 r μs g Solving for r: r = 2 ω What is ω? rev ⎛ 2π rad ⎞⎛ 1 min ⎞ ω = 33.3 ⎟⎜ ⎟ = 3.5 rad/s ⎜ min ⎝ 1 rev ⎠⎝ 60 sec ⎠ μ s g (0.1)(9.8 m/s 2 ) = 0.08 m r= 2 = 2 ω (3.50 rad/s ) 27 Unbanked and Banked Curves Example (text problem 5.20): 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.)) Take the car’s motion to be into the page. θ 28 Example continued: y FBD for the car: θ N x w Apply Newton’s Second Law: v2 (1) ∑ Fx = N sin θ = mar = m r (2) ∑ Fy = N cosθ − w = 0 29 Example continued: Rewrite (1) and (2): v2 (1) N sin θ = m r (2) N cosθ = mg Divide (1) by (2): 2 ( ) 26 . 8 m/s v tan θ = = = 0.089 2 gr (9.8 m/s )(825 m ) 2 θ = 5.1° 30 The Force of Gravity Slide 6-31 Circular Orbits r Earth Consider an object of mass m in a circular orbit about the Earth. The only force on the satellite is the force of gravity: Gms M e v2 ∑ F = Fg = r 2 = ms ar = ms r Solve for the speed p of the satellite: Gms M e v2 = ms 2 r r GM e v= r 32 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 = From previous slide: r Also need need, Combine these expressions and solve for r: ( )( ) 2πr v= T ⎛ GM e 2 ⎞ r =⎜ T ⎟ 2 ⎠ ⎝ 4π ⎛ 6.67 ×10 Nm /kg 5.98 ×10 kg 2⎞ ⎟⎟ ( ) r = ⎜⎜ 86400 s 2 4π ⎝ ⎠ = 4.225 ×107 m −11 2 2 24 1 1 3 3 r = Re + h ⇒ h = r − Re = 35,000 km k 33 ⎛ GM e 2 ⎞ r =⎜ T ⎟ 2 ⎝ 4π ⎠ 1 3 is Kepler’s Third Law. It can be generalized to: ⎛ GM 2 ⎞ r =⎜ 2 T ⎟ ⎝ 4π ⎠ 1 3 Where M is the mass of the central body. y For example, p , it would be Msun if speaking of the planets in the solar system. 34 Example: What is the minimum speed for the car so that it maintains i t i contact t t with ith the th loop l when h it iis iin th the pictured i t d position? FBD ffor the th car att the top of the loop: r y Apply Newton’s 2nd Law: x N w v2 ∑ Fy = − N − w = −mar = −m r v2 N +w=m r 35 Example continued: The apparent weight at the top of loop is: N = 0 when v2 N +w=m 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. 36 Example continued: 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: p y N x w Apply Newton’s 2nd Law: v2 ∑ Fy = N − w = mac = m r v2 N −w=m r ⎛ v2 ⎞ N = m⎜⎜ + g ⎟⎟ ⎝ r ⎠ Here, N > mg 37 Nonuniform Circular Motion Here, the speed is not constant. a at There is now an acceleration tangent g to the p path of the p particle. ar v The net acceleration of the body is a = ar + at 2 2 38 a ar at at changes g the magnitude g of v. ar changes the direction of v. Can write: ∑ F = ma ∑ F = ma r r t t 39 Artificial Gravity A large rotating cylinder in deep space (g≈0). 40 FBD for person at th bottom the b tt position iti FBD for person at the top position y y N x x N Apply Newton’s 2nd Law to each: 2 F = N = ma = m ω r ∑ y r 2 F = − N = − ma = − m ω r ∑ y r 41 Example (text problem 5.56): A space station is shaped like a ring i and d rotates t t tto simulate i l t gravity. it If th the radius di off the th space station is 120m, at what frequency must it rotate so that it simulates Earth’s gravity? g y Using the result from the previous slide: 2 F = N = ma = m ω r ∑ y r N mg ω= = = mr mr g = 0.28 rad/sec r The frequency is f = (ω/2π) = 0.045 Hz (or 2.7 rpm). 42 Tangential and Angular A Acceleration l i The average and instantaneous angular acceleration are: Δω Δω α av = and α = lim Δt →0 Δt Δt α is measured in rads/sec2. 43 Recalling that the tangential velocity is vt = rω means the tangential acceleration is Δvt Δω αt = =r = rα Δt Δt 44 The kinematic equations: Linear Angular v = v0 + aΔt ω = ω0 + αΔt 1 2 x = x0 + v0 Δt + aΔt 2 v 2 = v02 + 2aΔx 1 2 θ = θ 0 + ω0 Δt + αΔt 2 ω 2 = ω02 + 2αΔθ With vt = rω and at = rα 45 Example (text problem 5.66): A high speed dental drill is rotating t ti att 3.14×10 3 14 104 rads/sec. d / Th Through hh how many d degrees does the drill rotate in 1.00 sec? Given: ω = 3.14×104 rads/sec; Δt = 1 sec; α = 0 Want Δθ. 1 θ = θ 0 + ω0 Δt + αΔt 2 2 θ = θ 0 + ω0 Δt ( ) Δθ = ω0 Δt = 3.14 ×10 rads/sec (1.0 sec ) 4 = 3.14 ×10 rads = 1.80 ×10 degrees g 4 6 46