Solve 1-D dynamics problem using integration and ?

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Homework Help Overview

The problem involves a 75kg rocket launched upward, with a thrust force decreasing from 5000N to 0N as a quadratic function of time. The rocket is said to reach a speed of 200m/s during the engine's firing interval, raising questions about the integration of acceleration and the implications of changing forces.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss integrating acceleration but encounter difficulties with variables and time. There are attempts to apply impulse-momentum concepts, leading to confusion over negative time results. Questions arise regarding the assumption of constant mass due to fuel burn and the validity of the acceleration expressions.

Discussion Status

Some participants have offered corrections to the acceleration expression and have suggested using the maximum velocity to find the time at which it occurs. There is acknowledgment of the need to clarify what is being solved for, particularly regarding the apex of the rocket's flight.

Contextual Notes

Participants note that the problem is situated within a section of the textbook focused on kinematics and dynamics, implying that conservation laws may not be expected to be applied. The assumption of constant mass is also questioned, which could affect the dynamics of the problem.

Ascendant78
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Solve 1-D dynamics problem using integration and ?

Homework Statement



A 75kg rocket is launched directly upward. The force on the rocket due to its engine decreases from 5000N to 0N as a quadratic function of time. The rocket reaches a speed of 200m/s during the time interval that its engine fires.

Homework Equations



Ft = 5000 - At2
a = -1/75At2 + 66.6 (the force of thrust minus force of gravity divided by mass)

The Attempt at a Solution



I tried integrating acceleration and end up with both a time and a variable A left that I can't seem to get rid of. With no given time or position beyond the start, I'm at a loss as to how to solve this? I even tried impulse-momentum and comparing it to the integrated acceleration (velocity), but I ended up getting a negative time for my answer. Also, this question is in a part of our textbook that is only analyzing kinematics and dynamics, so it seems as if they aren't even expecting you to use conservation laws on it. If anyone can help point me in the right direction, I'd appreciate it.
 
Last edited:
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Ascendant78 said:

Homework Statement



A 75kg rocket is launched directly upward. The force on the rocket due to its engine decreases from 5000N to 0N as a quadratic function of time. The rocket reaches a speed of 200m/s during the time interval that its engine fires.

Homework Equations



Ft = 5000 - At2
a = -1/75At2 + 4265 (the force of thrust minus force of gravity divided by mass)

The Attempt at a Solution



I tried integrating acceleration and end up with both a time and a variable A left that I can't seem to get rid of. With no given time or position beyond the start, I'm at a loss as to how to solve this? I even tried impulse-momentum and comparing it to the integrated acceleration (velocity), but I ended up getting a negative time for my answer. Also, this question is in a part of our textbook that is only analyzing kinematics and dynamics, so it seems as if they aren't even expecting you to use conservation laws on it. If anyone can help point me in the right direction, I'd appreciate it.

You haven't made clear what it is you're supposed to work out.

In any case, there is a big assumption here - that the rocket's mass doesn't change (from fuel burn). Is this justifiable?

Assuming it is, there still seems to be an error in your expression for the acceleration because the 4265 term also needs to be divided by the mass.
 
After you've made Curious3141's correction, I think you are given that the maximum velocity your rocket reaches is 200m/s, right? At that point the acceleration is 0. Use that to solve for the time when that happens in terms of A. Then put that time into your integrated acceleration expression, which must equal 200m/s at that time. That should let you solve for A.
 
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Curious3141 said:
You haven't made clear what it is you're supposed to work out.

In any case, there is a big assumption here - that the rocket's mass doesn't change (from fuel burn). Is this justifiable?

Assuming it is, there still seems to be an error in your expression for the acceleration because the 4265 term also needs to be divided by the mass.

Apologies about that. I corrected the value in the problem and as far as what we are solving for, it is for the time where it reaches the apex.
 
Dick said:
After you've made Curious3141's correction, I think you are given that the maximum velocity your rocket reaches is 200m/s, right? At that point the acceleration is 0. Use that to solve for the time when that happens in terms of A. Then put that time into your integrated acceleration expression, which must equal 200m/s at that time. That should let you solve for A.

Ah, thank you. I was definitely over-complicating this one. Tried it out and got the right value. I really appreciate the help.
 

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