Classical Dynamics of Particles and Systems
5th Edition
ISBN: 9780534408961
Author: Stephen T. Thornton, Jerry B. Marion
Publisher: Cengage Learning
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Chapter 5, Problem 5.16P
To determine
The force of attraction between the sphere and the sheet.
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Chapter 5 Solutions
Classical Dynamics of Particles and Systems
Ch. 5 - Prob. 5.3PCh. 5 - Compute directly the gravitational force on a unit...Ch. 5 - Prob. 5.7PCh. 5 - Prob. 5.8PCh. 5 - A planet of density 1 (spherical core, radius R1)...Ch. 5 - Prob. 5.16PCh. 5 - A thin disk of mass M and radius R lies in the (x,...Ch. 5 - A point mass m is located a distance D from the...
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- A merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.3 meters, and a mass M = 251 kg. A small boy of mass m = 42 kg runs tangentially to the merry-go-round at a speed of v = 1.2 m/s, and jumps on. Randomized Variables R = 1.3 metersM = 251 kgm = 42 kgv = 1.2 m/s a)Calculate the moment of inertia of the merry-go-round, in kg ⋅ m2. b) Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round. c) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy.arrow_forwardA merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.3 meters, and a mass M = 251 kg. A small boy of mass m = 42 kg runs tangentially to the merry-go-round at a speed of v = 1.2 m/s, and jumps on. Randomized Variables R = 1.3 metersM = 251 kgm = 42 kgv = 1.2 m/s a) The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry-go-round when the boy is half way between the edge and the center of the merry go round? b) The boy then crawls to the center of the merry-go-round. What is the angular speed in radians/second of the merry-go-round when the boy is at the center of the merry go round? c)Finally, the boy decides that he has had enough fun. He decides to crawl to the outer edge of the merry-go-round and jump off. Somehow, he manages to jump in such a way that he hits the…arrow_forwardA merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.3 meters, and a mass M = 291 kg. A small boy of mass m = 42 kg runs tangentially to the merry-go-round at a speed of v = 1.8 m/s, and jumps on. Randomized VariablesR = 1.3 metersM = 291 kgm = 42 kgv = 1.8 m/s Part A- Calculate the moment of inertia of the merry-go-round, in kg ⋅ m2. Part B- Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round. Part C- Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy. Part D- The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry-go-round when the boy is half way between the edge and the center of the merry…arrow_forward
- A merry-go-round is a playground ride that consists of a large disk mounted to that it can freely rotate in a horizontal plane. The merry-go-round shown is initially at rest, has a radius R = 1.5 meters, and a mass M = 251 kg. A small boy of mass m = 41 kg runs tangentially to the merry-go-round at a speed of v = 1.8 m/s, and jumps on.Randomized VariablesR = 1.5 metersM = 251 kgm = 41 kgv = 1.8 m/s (a) Calculate the moment of inertia of the merry-go-round, in kg ⋅ m2. Part (b) Immediately before the boy jumps on the merry go round, calculate his angular speed (in radians/second) about the central axis of the merry-go-round.(c) Immediately after the boy jumps on the merry go round, calculate the angular speed in radians/second of the merry-go-round and boy.(d) The boy then crawls towards the center of the merry-go-round along a radius. What is the angular speed in radians/second of the merry-go-round when the boy is half way between the edge and the center of the merry go round?(e) The…arrow_forwardConsider a satellite in elliptical orbit around a planet of mass M, and suppose that physical units are so chosen that GM D 1 (where G is the gravitational constant). If the planet is located at the origin in the xy-plane, then Explain the equations of motion of the satellite? Let T denote the period of revolution of the satellite. Kepler’s third law says that the square of T is proportional to the cube of the major semiaxis a of its elliptical orbit. In particular, if GM D 1, then?arrow_forwardConsider a satellite in elliptical orbit around a planet of mass M, and suppose that physical units are so chosen that GM D 1 (where G is the gravitational constant). If the planet is located at the origin in the xy-plane, then Explain the equations of motion of the satellite?arrow_forward
- We are presented with a particle of mass (m = 4.56 kg) which is in uniform circular motion going around a fixed point O. It had a radius of orbit (r = 3.63 m) and it experiences a net force of magnitude F = Br toward the fixed point O, where B = 2.75 Nm. What is the period of the particles orbit?arrow_forwardCalculate the effective gravitational field vector g at Earths surface at the poles and the equator. Take account of the difference in the equatorial (6378 km) and polar (6357 km) radius as well as the centrifugal force. How well does the result agree with the difference calculated with the result g = 9.780356[1 + 0.0052885 sin 2 0.0000059 sin2(2)]m/s2 where is the latitude?arrow_forwardA particle P of mass 2m can slide along a smooth rigid straight wire. The wire has one ofits points fixed at the origin ’O’, and is made to rotate in a plane through ’O’ with constantangular speed Ω. Show that r , the distance of P from O, satisfies the equation r ̈ − (Ω^2)*r = 0 Initially, P is at rest (relative to the wire) at a distance a from O. Find r as a function of t in the subsequent motion.arrow_forward
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