Rocket + Connected Body Dynamics problem

[Mohammad Ishmas]

OP is confusing the velocity of the rocket with the constant relative velocity of the expelled gas.

sorry but what does "OP" mean ? , And yeah thanks for pointing that error out I'll change that

 

[jbriggs444]

sorry but what does "OP" mean ? , And yeah thanks for pointing that error out I'll change that

OP is internet slang for "original poster". That is you in this case.

It is also used for "original post". Context usually makes it clear which is meant.

 

[Mohammad Ishmas]

OP is internet slang for "original poster". That is you in this case.

It is also used for "original post". Context usually makes it clear which is meant.

ok thanks

 

[kuruman]

sorry but what does "OP" mean ? , And yeah thanks for pointing that error out I'll change that

Good. Let's see your revised problem description.

 

[Mohammad Ishmas]

OP is confusing the velocity of the rocket with the constant relative velocity of the expelled gas.

Let us just reconsider the problem as a whole by removing the rocket. There is a capsule of 625 kg attached to blocks of 200kg,100kg,50kg and 25 kg respectively it is losing its mass from at the rate of 12.5 kg/s from the 25 kg block and then to 50 kg then 100 kg ...

If the acceleration after 5 seconds is 56.666.... m/s^2
then find tensions between 50 kg block and 100 kg block , 100 kg block and 200 kg block and also find contact force between 200 kg block and 625 kg capsule

 

[Mohammad Ishmas]

Let us just reconsider the problem as a whole by removing the rocket. There is a capsule of 625 kg attached to blocks of 200kg,100kg,50kg and 25 kg respectively it is losing its mass from at the rate of 12.5 kg/s from the 25 kg block and then to 50 kg then 100 kg ...

If the acceleration after 5 seconds is 56.666.... m/s^2
then find tensions between 50 kg block and 100 kg block , 100 kg block and 200 kg block and also find contact force between 200 kg block and 625 kg capsule

Also in the reconsidered problem everything is being on the x axis so the mg doesn't mean anything

 

[jbriggs444]

Let us just reconsider the problem as a whole by removing the rocket. There is a capsule of 625 kg attached to blocks of 200kg,100kg,50kg and 25 kg respectively it is losing its mass from at the rate of 12.5 kg/s from the 25 kg block and then to 50 kg then 100 kg ...

If the acceleration after 5 seconds is 56.666.... m/s^2
then find tensions between 50 kg block and 100 kg block , 100 kg block and 200 kg block and also find contact force between 200 kg block and 625 kg capsule

Can you please state the complete problem as you would pose it to your students?

As it stands, you have not stated that an external force applied to the 200 kg block is responsible for the observed 56.666... m/s² acceleration.

Stepping back a bit...

A variable mass problem is sometimes posed in terms of a railroad car (a hopper car) which is pouring out its contents into a container below as it rolls past. Or one can invert it and pour material onto a conveyer belt.

A multiple mass problem is normally posed in terms of stacked blocks being pushed sideways. Perhaps on a frictionless surface.

A realistic rocket problem would have the large stage at the bottom pushing on the smaller stages and payload above. The mass flow rate would not be constant from one stage to the next. Instead, the mass flow rate for the initial stage would be large compared to the mass flow rate for the final stage.

Although Robert Goddard's initial design had the rocket nozzle near the top of the assembly with the rocket frame dangling behind, this is not at all standard. I am not entirely sure how much of the classic picture below is rocket and how much is launch stand. Goddard's later designs moved the nozzle to the bottom.

 

[kuruman]

Let us just reconsider the problem as a whole by removing the rocket. There is a capsule of 625 kg attached to blocks of 200kg,100kg,50kg and 25 kg respectively it is losing its mass from at the rate of 12.5 kg/s from the 25 kg block and then to 50 kg then 100 kg ...

If the acceleration after 5 seconds is 56.666.... m/s^2
then find tensions between 50 kg block and 100 kg block , 100 kg block and 200 kg block and also find contact force between 200 kg block and 625 kg capsule

Your reconsidered problem ignores the issue with your design that I pointed out right from the start.
You cannot push with a string.

I will refrain from spending more time to post here unless I see that you take our advice and suggestions under consideration.

 

[Mohammad Ishmas]

Your reconsidered problem ignores the issue with your design that I pointed out right from the start.
You cannot push with a string.

I will refrain from spending more time to post here unless I see that you take our advice and suggestions under consideration.

well let us say that this system is being pulled from 625 kg block with the data given below

 

[jbriggs444]

well let us say that this system is being pulled from 625 kg block with the data given below

I notice that you have not given any data below. So I will harvest the data from up-thread.

We begin with blocks of 625, 200, 100, 50 and 25 kg arranged in a row from left to right on a frictionless surface, all connected with ideal massless cords.

We apply a leftward force of 62,500 N on the 625 kg block. We are invited to imagine that this is from a rocket
nozzle on that block. The nozzle is supplied with 12.5 kg/s of liquid propellant which is expelled with a relative velocity of 5000 m/s rightward.

The propellant is supplied in turn from the smaller blocks, starting at the right and working left. The blocks are pure propellant. Any tank is negligibly massive. Once the propellant in a block is exhausted, the block itself is discarded.

Consider the situation after 5 seconds has elapsed.

a) What is the acceleration of the assembly at this point?
b) What is the tension in the cord connecting the 625 kg block to the 200 kg block behind it?


I would analyze the problem by first calculating the system state at 5 seconds into the scenario. The trailing 25 kg tank would have been exhausted in the first two seconds. It will have been discarded and is now irrelevant. The 50 kg tank will have lost 37.5 kg of its initial mass. 12.5 kg remains.

So we have 625 + 200 + 100 + 12.5 kg = 937.5 kg total assembly mass at this time.

This assembly is subject to a net external force of 62,500 N.

Assuming that the assembly moves rigidly (as it obviously should) this means that it is accelerating at a rate of $a = \frac{f}{m_\text{tot}} = \frac{62500}{937.5} \approx 66.7$ m/sec²

[Since gravity is not a factor, we will not be adjusting the acceleration to 56.7 m/sec²]

Ignoring the 625 kg block in front, the mass of the remainder of the assembly is currently 200 + 100 + 50 + 12.5 = 312.5 kg. This assembly is accelerating leftward at approximately 66.7 m/sec². So it must be subject to a leftward net force of $f_\text{net} = m_\text{tail} a \approx 312.5 \times 66.7 = 20833$ N.

Naively, this is the tension in the cord between the 625 kg and 200 kg blocks: about 20,833 N.

If I've not muffed the calculation. [Edit: One muff corrected. I counted the 50 kg block when I should not have].


However, there is a momentum flow that we have not accounted for.

We are piping liquid propellant forward from the 12.5 kg (remaining) tank to the 625 kg block where the rocket nozzle is attached.

If the pipe is narrow, the velocity of propellant in the pipe times the 12.5 kg/sec flow rate may amount to a significant force. This would be a retarding force on the trailing block and a propulsive force on the leading block. The tension in the cord[s] would need to increase to reflect this.

If the pipe is wide and long, the mass of propellant in the pipe may be significant. This messes up our accounting for the mass of the various assembly bits at any point in time -- where do we count the mass of the propellant in the pipe?

So let us assume that the whole assembly is short and the pipe is wide so that the mass and momentum transfer from the unburned propellant in the pipeline may both be safely neglected.

 

[Mohammad Ishmas]

Guys I sorted out how to find tension in the initial problem consisting of rocket, but there is still one issue that If I take net acceleration that is subtracting value of g from acceleration in upward direction, and then do not account for weight force for FBD of each block as I have already taken net acceleration then the values of tensions are uniform.

But let us say that we take acceleration only in upwards direction and do not ignore weight force then the value of tension force is not force

NOTE that I've also attached a string between 625 kg and 200 kg block which was not in initial problem

 

[jbriggs444]

Guys I sorted out how to find tension in the initial problem consisting of rocket, but there is still one issue that If I take net acceleration that is subtracting value of g from acceleration in upward direction, and then do not account for weight force for FBD of each block as I have already taken net acceleration then the values of tensions are uniform.

Please show your work.

Are you saying that the calculated tensions are identical regardless of the acceleration of gravity? Yes, I agree. So what?

 

[Mohammad Ishmas]

Please show your work.

Are you saying that the calculated tensions are identical regardless of the acceleration of gravity? Yes, I agree. So what?

I'll post the work ,
the calculated tensions are coming identical if we take net acceleration and then as we don't have to subtract mg. in my pdf , I have taken net acceleration but have also subtracted mg which is wrong. BUT STILL what I am saying is that if we take acceleration in upward direction only and then subtract mg and calculate tensions the tensions coming are not uniform means they are not matching with the previous method of taking net acceleration