Solve 1-D dynamics problem using integration and ?

AI Thread Summary
The discussion revolves around solving a 1-D dynamics problem involving a 75kg rocket with a thrust force that decreases quadratically over time. The main challenge is integrating the acceleration while accounting for the changing force and ensuring the mass remains constant. Participants highlight the need to correct the acceleration equation and emphasize that the maximum velocity of 200m/s occurs when acceleration is zero. By substituting the time at maximum velocity into the integrated acceleration expression, the variable A can be solved. The original poster successfully resolves their confusion with the help of clarifications from others.
Ascendant78
Messages
327
Reaction score
0
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:
Physics news on Phys.org
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.
 
Last edited:
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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top