Physics 23 Notes Chapter 5

Transcription

1 Physics 23 Notes Chapter 5 Dr. Alward Circular Motion Example A: An object is moving in a circular path of radius r = 1.2 m, and is completing 0.40 revolutions per second. What is the object s centripetal acceleration? ar = v 2 /r Centripetal acceleration always points toward the center of the circle--a direction called radial. The positive r-axis points toward the center of the circle. Wherever the object is at any moment, the direction pointing from the object to the circle s center is the positive r-direction--the radial direction. Newton s Second Law applied to the radial direction is shown below: v = 0.40 rev/s [2π(1.2 m)/rev] = 3.02 m/s a = (3.02 m/s) 2 /1.2 m = 7.60 m/s 2 The centripetal acceleration of an object moving at the end of a string is 4.0 m/s 2. If the string s length (r) is doubled, and the object s speed tripled, what would be the new centripetal acceleration? a1 = v1 2 /r1 = 4.0 m/s 2 a2 = v2 2 /r2 = (3v1) 2 /(2r1) = (9/2) v1 2 /r1 = (9/2) 4.0 = 18.0 m/s 2 Other way: (3) 2 (½) = 4.5 a = 4.5 (4.0) = 18 m/s 2 1

2 An object of mass m = 3.0 kg is moving at 12.0 m/s in a circular path of radius r = 2.4 m in a vertical plane. What is the tension in the string? The object is now at the bottom of the swing. What is the tension in the string? (Positive radial direction is downward.) Tr + wr = mv 2 /r T + mg = mv 2 /r T = 3.0 (12.0) 2 / (9.8) = N (Positive radial direction is upward.) Tr + wr = mv 2 /r T mg = mv 2 /r T = 3.0 (9.8) (12.0) 2 /2.4 = N Example C: In the figure at the right, an object of mass m = 2.0 kg is revolving at the end of the string in a vertical circle of radius r = 1.2 m at a speed of v = 5.0 m/s. The string makes an angle of 60 o relative to the vertical diretion. What is the tension T in the string? mar = Fr mv 2 /r = Tr + wr = T + mgcos 60 T = N 2

3 A race car travels around a banked, frictionless track. The angle the ramp makes with respect to the vertical is the so-called banking angle, θ. Derive an expression for the speed necessary to travel in a circular path of a particular radius, r, and banking angle, θ. may = Fy = Cy + wy The y-coordinate of the car never changes, so ay = 0: 0 = Cy + wy = C cos θ - mg Solve for C: C = mg/cosθ Equation (1) Cr + wr = mar C sin θ + 0 = mv 2 /r Equation (2) Substitute (1) into (2): (mg/cos θ) sinθ = mv 2 /r v = (grtan θ) 1/2 (continued at the right) The radius of curvature of a circular ramp exit from a freeway is 160 meters. What must be the banking angle to ensure a safe exit at 20 m/s, even if there is zero friction? v = (grtan θ) 1/2 20 = [(9.8)160 tan θ] 1/2 θ = 14.3 o 3

4 The water in a bucket being swung in a vertical circle is a distance r = 0.60 m from the center of the circle. What is the least speed v the water could have without the water losing contact with the bottom of the bucket? mar = Fr mv 2 /r = Cr + wr Let Cr = N (virtually zero.) Water is still in contact with the bottom of the bucket, but just barely. mv 2 /r = wr = mg v = (rg) 1/2 = [0.60 (9.8)] 1/2 = 2.42 m/s Above is shown a Ferris Wheel, with passengers in gondolas at the top and bottom. Obtain equations that would allow one to determine the contact forces at both locations. At the top, downward is the +r direction. Cr + wr = m(v 2 /r) -C + mg = m(v 2 /r) C = mg - mv 2 /r = m(g - v 2 /r) At the bottom, upward is the +r direction. Cr + wr = mv 2 /r C - mg = mv 2 /r C = mg + mv 2 /r = m(g + v 2 /r) 4

5 Satellites in Circular Orbits Example A: (a) At what speed, standing on Earth, would you have to throw a rock, parallel to the ground, in order that it travel around the globe, assumed to be a perfect sphere, eventually striking you in your back, ignoring air resistance and obstacles such as trees, buildings and hills? G = 6.67 x N-m 2 /kg 2 M = 5.98 x kg R = 6.38 x 10 6 m (about 4,000 miles) mar = Fr mv 2 /R = GMm/R 2 Get v = (GM/R) 1/2 = 7906 m/s =17,683 miles per hour (b) How long would it take the rock to strike you? t = Circumference/Speed = 2π (6.38 x 10 6 )/7906 = 5070 s = minutes = 1.41 hours Derive the relationship between the radius of a planetary circular orbit and its orbital speed. Example C: Two planets have circular orbits about a certain star. Planet 1 has an orbital speed of 30 km/s. Planet 2 is four times farther from the star than Planet 1. What is its orbital speed? Recall, the product rv 2 is the same for all planets orbiting a particular star in a circular path: r1v1 2 = r2v2 2 GMm/r 2 = mv 2 /r rv 2 = GM = constant v2 = (r1/r2) 1/2 v1 = (1/4) 1/2 30 = 15 km/s 5

6 Suppose an alternate universe exists in which the gravitational force law is F = Gm1m2/r 4, but all other laws of physics are the same as in this universe. How does the orbital speed of planets in circular orbits depend on the orbital radius in the alternate universe? F = ma GMm/r 4 = mv 2 /r r 3 v 2 = GM = constant A planet orbiting a star in the universe described at the left has a speed of 20 km/s. What is the orbital speed of a planet that is six times farther away from the star? r2 = 6 r1 k = r1 3 v1 2 v2 = (r1/r2) 3/2 v1 = (r1/6r1) 3/2 v1 = (1/6) 3/2 20 = 1.36 km/s 6