Calculating Rail Gun Distance for Escape Velocity

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Homework Help Overview

The problem involves a conducting bar sliding over horizontal rails, with a focus on calculating the distance required for the bar to reach escape velocity using the principles of electromagnetism and kinematics. The context includes a magnetic force acting on the bar due to a constant current and magnetic field.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using kinematic equations to relate acceleration, distance, and velocity, with some suggesting different methods to approach the problem. There is also a focus on correcting initial formula usage and exploring the implications of time in the calculations.

Discussion Status

The discussion is active, with participants offering different perspectives on the approach to the problem. Some have identified potential errors in their calculations and are reconsidering their methods, while others are encouraging exploration of multiple approaches.

Contextual Notes

Participants are navigating the complexities of the problem, including the assumptions made about forces and the simplifications involved in the calculations. There is an acknowledgment of the role of gravity in a real-world scenario, which is not fully addressed in the current setup.

lylos
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Homework Statement



A conducting bar with mass=25.0 kg and length L=51.0 cm slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I= 2400A in the rails and bar, and a constant, uniform, vertical magnetic field B = 0.480 T fills the region between the rails.
See figure:
railgun.jpg


It has been suggested that rail guns based on this principle could accelerate payloads into Earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the Earth (11.2 km/s).
For simplicity, assume the net force on the object is equal to the magnetic force, as in parts A and B, even though gravity plays an important role in an actual launch into space.


Homework Equations



I know the magnetic force is 588N.
F=ma
V = at
X = at^2


The Attempt at a Solution



588/25 = 23.52 m/s^2
11200 m/s = 23.52 m/s^2 * t
t = 476.190

x = 23.52 m/s^2 * 476.19^2
5333322.666672 m

Am I missing something, it says that is incorrect...
 
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I think the best approach here might be with the kinematics eqn that relates velocity to distance along which body is uniformly accelerated, since Vi=0 (body at rest): V^2=2a*x
 
ah, my formulas were wrong... it's x=1/2at^2 and v=at... I should've caught that...
 
you miss my point maybe. there is no need to mess with time, but certainly you can get the right answer using that approach. Try it both ways!
 

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