A rocket sled of 2.5 Mg (tare) burns 90 kg of fuel a second and the uniform exit velocity of th

A rocket sled of 2.5 Mg (tare) burns 90 kg of fuel a second...

Key Concepts

Rocket Propulsion and Momentum Conservation

This concept is based on Newton's third law, where the rocket ejects mass in one direction and gains momentum in the opposite direction. The conservation of momentum principle explains how the force (thrust) is developed through the expulsion of fuel, and it is fundamental in deriving the equations governing a rocket's motion.

Variable Mass Dynamics

In rocketry, the mass of the vehicle changes over time as fuel is burned. This requires a modified approach to Newton’s second law, often leading to the application of the rocket equation, which relates the velocity of the rocket to the changing mass and the exhaust velocity of the fuel. Understanding variable mass dynamics is crucial for analyzing the motion and performance of a rocket.

Thrust Generation from Fuel Burn Rate and Exhaust Velocity

Thrust is generated when mass is expelled from the rocket at high speed. The magnitude of the thrust force is directly proportional to the fuel burn rate and the exhaust velocity. This relationship is key to determining how much fuel is necessary to achieve a desired speed, as higher burn rates or exhaust velocities provide greater thrust.

Drag Force Proportional to Velocity

In this context, the resistance or drag force acting against the motion of the rocket sled is modeled as being directly proportional to its velocity. This linear drag model simplifies the analysis of energy losses during acceleration, and it must be accounted for in the equations of motion since it affects the net force and the resultant acceleration of the sled.

Integration of Equations of Motion with Drag and Variable Mass

Calculating the fuel requirement involves setting up and solving differential equations that incorporate both the variable mass of the rocket and the drag force (which is a function of velocity). Integrating these equations allows one to determine how the velocity evolves over time, and ultimately, how much fuel must be burned to overcome both inertial and resistive forces to reach the target speed.

Transcript

00:01 In this problem, we have a rocket sled, and the goal of the problem is to find how much food will it take to have the maximum speed be 150 meters per second.

00:13 We are told the mass flow rate, we are told the velocity relative to the sled of the ejected fuel, we are also told that there is a resistor force, some constant times the velocity v and we're given the k we're told that the the fuel is ejected at mswick pressure so we don't have to worry about that that pressure contribution to the thrust and the mass that said that slide they give us as 2 .5 times 10 to 6 grams which is 2 .5 times 10 to 3 kilograms that's the mass of the sled and this is a make it explicit there we go all right so let's look at the let's look at the rocket equation this is the this is dealing with the thrust on the left -hand side remember this is horizontal so we don't have to worry about gravity m dot u r this is from the third law this is what is going on down well if you assume the sled is going to the right then this is what is being, this force here is what is on the fuel being ejected to the left.

01:33 But from the third law, that means it's going to be on the, to the right on the sled.

01:39 Minus kv is equal to capital m, dv, d, t.

01:48 But to be more precise, if we're going to think of this as what we're going to be doing calculus, m is a function of time.

01:58 Let me just write this down.

01:59 Some more total mass that means sled fuel at time t so really it should be written like this and how that mass is lost is it is it a simple loss is it complicated the engines do different things different times whatever this there's a functional we can't do we can't do any type of integration if we don't know this is as a function of time and i'm going to write that out here now m has a function of time it's massive sled plus the mass of the fuel minus the mass flow rate times the time so that's going to be taking away and now let me write out this is mass fluid this is our goal to get that quantity is our goal so let's look at this capital m function let's look at it because in the integration we're going to need to worry about limits.

03:18 So t equals zero, we just have sled and the fuel.

03:27 So m is equal to ms, and i'm just going to call this m0.

03:42 At t equals t1, this is where i'm going to say this is the cutoff.

03:47 You know, this is where you're at the maximum, and we want to be at 150 meters for second at this point.

03:54 M is equal to ms.

03:56 We've used up all the fuel.

03:59 So this gives me then that t1 is going to be mass of the fuel divided by m dot.

04:09 So we know what t1's value is.

04:12 So we now have all limits of integration we're going to need.

04:18 So if we knew this, we can get the number of seconds it takes to get to that point.

04:25 But we have to exhaust.

04:27 And let me put that in here.

04:30 Oh, fuel exhausted.

04:39 That defines maximum, right? if you still was more fuel, isn't it going to be going faster? you know, in the next second? certainly.

04:47 Okay, so let's look at that equation again.

04:52 M .u .r.

04:54 Minus kv, is equal to m -0, minus m .t t, tv, dt.

05:05 That's who we have.

05:09 Bring everything that has a v in it let's cross multiply notice there's a t in this and there's a v in this so d t over m0 minus m dot t so all we have here all we have here is constants and t and we'll do the same on this side dv m dot ur minus kv so all we have here are constants and v.

05:47 So we want.

05:50 Okay.

05:50 Now let's set up the integration.

05:53 Zero.

05:55 We're going to assume it's, as i did before, we're starting at zero time.

06:00 T1d -t, m -0 minus m dot t is equal zero v -max, div, m dot u .r minus kv.

06:20 So remember, this is constant.

06:22 This is constant.

06:22 This is this is constant.

06:27 So how do we want to proceed here? well, let's introduce some new variables to make it explicit what we're going to get out of these integrations.

06:35 Let me introduce alpha.

06:38 Alpha is going to be m0 minus m .t, and that gives me the d t is going to be minus d .m.

06:58 So d alpha, this goes away minus dt.

07:05 So that's what we get there.

07:07 And we can do the same for the v.

07:09 I mean to do as beta.

07:10 Beta is equal to m.

07:13 U .r minus kv.

07:16 This gives me then the dv is minus d beta over k.

07:23 Very similar structure.

07:26 So now we can use those to be right...