040717_Gold_Exercises

# 040717_Gold_Exercises - Goldstein Chapter 1 Exercises...

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Unformatted text preview: Goldstein Chapter 1 Exercises Michael Good July 17, 2004 1 Exercises 11. Consider a uniform thin disk that rolls without slipping on a horizontal plane. A horizontal force is applied to the center of the disk and in a direction parallel to the plane of the disk. • Derive Lagrange’s equations and find the generalized force. • Discuss the motion if the force is not applied parallel to the plane of the disk. Answer: To find Lagrangian’s equations, we need to first find the Lagrangian. L = T- V T = 1 2 mv 2 = 1 2 m ( rω ) 2 V = 0 Therefore L = 1 2 m ( rω ) 2 Plug into the Lagrange equations: d dt ∂L ∂ ˙ x- ∂L ∂x = Q d dt ∂ 1 2 mr 2 ω 2 ∂ ( rω )- ∂ 1 2 mr 2 ω 2 ∂x = Q d dt m ( rω ) = Q m ( r ¨ ω ) = Q 1 If the motion is not applied parallel to the plane of the disk, then there might be some slipping, or another generalized coordinate would have to be introduced, such as θ to describe the y-axis motion. The velocity of the disk would not just be in the x-direction as it is here. 12. The escape velocity of a particle on Earth is the minimum velocity re- quired at Earth’s surface in order that that particle can escape from Earth’s gravitational field. Neglecting the resistance of the atmosphere, the system is conservative. From the conservation theorme for potential plus kinetic energy show that the escape veolcity for Earth, ingnoring the presence of the Moon, is 11 . 2 km/s. Answer: GMm r = 1 2 mv 2 GM r = 1 2 v 2 Lets plug in the numbers to this simple problem: (6 . 67 × 10- 11 ) · (6 × 10 24 ) (6 × 10 6 ) = 1 2 v 2 This gives v = 1 . 118 × 10 4 m/s which is 11.2 km/s. 13. Rockets are propelled by the momentum reaction of the exhaust gases expelled from the tail. Since these gases arise from the raction of the fuels carried in the rocket, the mass of the rocket is not constant, but decreases as the fuel is expended. Show that the equation of motion for a rocket projected vertically upward in a uniform gravitational field, neglecting atmospheric friction, is: m dv dt =- v dm dt- mg where m is the mass of the rocket and v’ is the velocity of the escaping gases relative to the rocket. Integrate this equation to obtain v as a function of m, assuming a constant time rate of loss of mass. Show, for a rocket starting initally from rest, with v’ equal to 2.1 km/s and a mass loss per second equal to 1/60th of the intial mass, that in order to reach the escape velocity the ratio of the wight of the fuel to the weight of the empty rocket must be almost 300! Answer: This problem can be tricky if you’re not very careful with the notation. But here is the best way to do it. Defining m e equal to the empty rocket mass, m f is the total fuel mass, m is the intitial rocket mass, that is, m e + m f , and dm dt =- m 60 as the loss rate of mass, and finally the goal is to find the ratio of 2 m f /m e to be about 300....
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040717_Gold_Exercises - Goldstein Chapter 1 Exercises...

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