How Do You Calculate the Distance a Burning Body Descends Under Gravity?

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In summary, the conversation discusses a problem involving a body with equal masses of inflammable and non-inflammable material descending freely under gravity while burning at a constant rate. Using momentum considerations, the conversation shows that the body descends a distance before all the inflammable material is burnt. One individual was having trouble with the second part and after some discussion and corrections, they were able to successfully solve the problem.
  • #1
devious_
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I'm having trouble with the following question. Can anyone please give me a push in the right direction?

A body consists of equal masses [itex]M[/itex] of inflammable and non-inflammable material. The body descends freely under gravity from rest. The combustible part burns at a constant rate of [itex]kM[/itex] per second, where [itex]k[/itex] is a constant. The burning material is ejected vertically upwards with constant speed [itex]u[/itex] relative to the body, and air resistnace may be neglected. Show, using momentum considerations, that

[tex]\frac{d}{dt}[(2-kt)v] = k(u-v) + g(2-kt)[/tex]

where [itex]v[/itex] is the speed of the body at time [itex]t[/itex]. Hence show that the body descends a distance

[tex]\frac{g}{2k^2} + \frac{u}{k} (1 - \ln 2)[/tex]

before all the inflammable material is burnt.


I managed to do the first part, but I have no idea how to approach the second.
 
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  • #2
devious_ said:
I'm having trouble with the following question. Can anyone please give me a push in the right direction?

A body consists of equal masses [itex]M[/itex] of inflammable and non-inflammable material. The body descends freely under gravity from rest. The combustible part burns at a constant rate of [itex]kM[/itex] per second, where [itex]k[/itex] is a constant. The burning material is ejected vertically upwards with constant speed [itex]u[/itex] relative to the body, and air resistnace may be neglected. Show, using momentum considerations, that

[tex]\frac{d}{dt}[(2-kt)v] = k(u-v) + g(2-kt)[/tex]

where [itex]v[/itex] is the speed of the body at time [itex]t[/itex]. Hence show that the body descends a distance

[tex]\frac{g}{2k^2} + \frac{u}{k} (1 - \ln 2)[/tex]

before all the inflammable material is burnt.


I managed to do the first part, but I have no idea how to approach the second.

I'd like to see how you got the first part. I keep getting another term in that derivative.
 
  • #3
(For simplicity, ~d = [itex]\delta[/itex].)

(m + ~dm)(v + ~dv) - ~dm(v - u) - mv = (m + ~dm)g . ~dt

Multiplying everything out and letting ~dt, ~dm and ~dv -> 0:
m(dv/dt) + u(dm/dt) = mg

dm/dt = -kM
m = -kMt + C

When t=0, m=2M, so C =2M. Hence:
m = 2M - kMt = M(2-kt)

Substituting this in the ODE:
M(2-kt)(dv/dt) + u(-Mk) = M(2-kt)g
(2-kt)(dv/dt) - uk = (2-kt)g

Substract -vk from both sides:
(2-kt)(dv/dt) - vk = uk - vk + (2-kt)g
d[(2-kt)v]/dt = k(u-v) + g(2-kt)

I hope that's readable. :tongue2:
 
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  • #4
devious_ said:
(For simplicity, ~d = [itex]\delta[/itex].)

(m + ~dm)(v + ~dv) - ~dm(v - u) - mv = (m + ~dm)g . ~dt

Multiplying everything out and letting ~dt, ~dm and ~dv -> 0:
m(dv/dt) + u(dm/dt) = mg

dm/dt = -kM
m = -kMt + C

When t=0, m=2M, so C =2M. Hence:
m = 2M - kMt = M(2-kt)

Substituting this in the ODE:
M(2-kt)(dv/dt) + u(-Mk) = M(2-kt)g
(2-kt)(dv/dt) - uk = (2-kt)g

Substract -vk from both sides:
(2-kt)(dv/dt) - vk = uk - vk + (2-kt)g
d[(2-kt)v]/dt = k(u-v) + g(2-kt)

I hope that's readable. :tongue2:

OK. That looks good. I had totally messed up the split of the mass into 2M with the rate of conversion being a CONSTANT kM.

I've almost got the result, but I'm off by a sign. I'm confident I'm on the right path, so see what you can do with this. Rewind your derivation to

(2-kt)(dv/dt) - uk = (2-kt)g

add the uk to both sides and divide by the (2-kt) factor. Then you can separate to get

dv = f(t)dt

Integrate and impose v(t=0)=0 to evaluate the constant. After you get the velocity, integrate from t = 0 to the time of burnout, which from your mass equation is t = 1/k, to find the distance. I get the first term fine. There is an integral of lnxdx that I think I am messing up at the moment because I have a minus in front of the 1 in the [1-ln2]term. I'll see if I can fix it, but you can probably get it.
 
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  • #5
OlderDan:
Possibly, your sign flaw comes from the fact that:
1) u is given as the fuel's SPEED relative to the body, rather than the relative velocity
2) Thus, with positive direction parallell with gravity, the relative velocity of the fuel is -u

At any rate, I did that mixup at first.. :blushing:
 
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  • #6
arildno said:
OlderDan:
Possibly, your sign flaw comes from the fact that:
1) u is given as the fuel's SPEED relative to the body, rather than the relative velocity
2) Thus, with positive direction parallell with gravity, the relative velocity of the fuel is -u

At any rate, I did that mixup at first.. :blushing:

Thanks for the suggestion, but that's not it. That would have been a more pervasive problem. It was something very basic though. As one of my British professors used to say "The trouble with you guys (American students) is you don't know how to integrate." This was simply a matter of not correctly changing limits when I changed the integration variable. When I do it right, I get the answer given. At least I can plead "time of day" factors; that time stamp on my post is my local time :rolleyes:
 
  • #7
I appreciate the help. Thank you!

Apparently my problem was finding the time of burnout. I was substituting m=0 instead of m=M, so I always got t=2/k which made things very messy.

I think the reason you got a minus was because of the integral of ln(2-kt), as you said. Maybe you forgot the minus next to kt?

Anyway... Thanks again. :smile:
 

What is impulse?

Impulse is the change in momentum of an object. It is equal to the force applied to the object multiplied by the time interval over which the force is applied.

What is momentum?

Momentum is a measure of an object's motion and is equal to the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

How is impulse related to momentum?

Impulse is directly related to momentum. The impulse applied to an object is equal to the change in momentum of that object. In other words, the greater the impulse, the greater the change in momentum.

What is the equation for impulse?

The equation for impulse is: I = F * Δt, where I is impulse, F is force, and Δt is the time interval over which the force is applied.

How is the law of conservation of momentum related to impulse?

The law of conservation of momentum states that the total momentum of a system remains constant unless acted upon by external forces. Impulse is a way to quantify the change in momentum, and according to Newton's Third Law, every action has an equal and opposite reaction. Therefore, the impulse applied to one object will result in an equal and opposite impulse applied to another object, keeping the total momentum of the system constant.

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