Physics Ninja looks at the Rocket equations for Thrust and Speed. Both equations are derived using conservation of linear momentum and are used to solve two example problems.
Пікірлер: 98
@deviarchana79403 жыл бұрын
Random person on KZbin. You saved me.
@dan81824 Жыл бұрын
Thank you so much… you’ve solved so many problems to me, you something like a physics god for me from now ❤
@ricardolagoperez99983 жыл бұрын
thk so much!! Clearly and fully explained, perfect for teaching!! when´s the next step...? Jet engines!!
@wernerhartl20693 жыл бұрын
Great explanation. Thanks.
@claudiorebelo2 жыл бұрын
Awesome explanation!
@vanshyadav45333 жыл бұрын
You sir are a life-saver!
@ganeshr5 Жыл бұрын
at the end of 90 secs, the rate of change of mass falls by 16.67% compared to at the end of 1 sec; therefore the average acceleration falls by the same amount. If you want to calculate the final velocity using instantaneous acceleration we have to numerically integrate at intermediate points, say every sec, to get an answer closer to 3466 m/s. You will get a velocity much lower if the final acceleration of 20.83 m/sec^2 is used in v = u + at.
@mef93273 жыл бұрын
Fascinating. Great video. Is it possible to calculate change in velocity if you don't know the exhaust velocity but do know the engine's thrust and change in mass? For example, an engine consumes fuel at a constant rate and produces a constant 100kN for 60 seconds. The initial mass is 50 tons and its final mass is 30 tons. It seems F=ma (a=F/m) could be used with some careful math. F=ma itself, though, yields two different results depending on what is averaged mass or acceleration (acceleration for average mass a=100/40=2.5 but average acceleration t(0) & t(60) = average(2.00 & 3.33) = 2.67.
@hadisardari15645 ай бұрын
great and the best teacher ever.
@andrewvelasco75573 жыл бұрын
This is great
@mircea_ene Жыл бұрын
How would this differ if the rocket takes off (i.e. accounting for gravity). Is the total thrust force the thrust minus G? How does it influence the v(t)
@pinkibag51233 жыл бұрын
Please give the definition of exhaust speed again in details.
@koungmeng4 жыл бұрын
10:45 did you forget to divide by m to get acceleration? correct me if I'm wrong.
@PhysicsNinja4 жыл бұрын
Oops, thanks for pointing that out. Yes there should be an m in denominator.
@papalegba67593 жыл бұрын
@@PhysicsNinja there should be a letter 'h' in the word 'thrust' too. you're not very smart, are you?
@Adeloye10003 жыл бұрын
@@papalegba6759 You are just a sad little man. Does that comment you feel smarter?
@papalegba67593 жыл бұрын
@@Adeloye1000
@Adeloye10003 жыл бұрын
@@papalegba6759
@arturo3511 Жыл бұрын
at 5.06 if dm is negative shouldnt we write +Vfuel*dm. If we write this way Vfuel is positive,dm is negative therefore -Vfdm is positive. Is the momentum of the gas negative or positive in the ground frame ?
@cmac7777 Жыл бұрын
thank you sir
@arundas2267 Жыл бұрын
THANKS.
@joeymosafi11212 жыл бұрын
i think that at 10:45 you should hvae made it 1/m so when you intergrated you could have got the natural log of M due to log laws. also what i love to do is take the inverse function of log which is e and subtract mf/m0 to the other side and bring Vf/V0 to the other side. otherwise great video!!
@ganeshr5 Жыл бұрын
10:47 you forgot to divide by m on the right hand side for the acceleration on the left
@maksymilian5071 Жыл бұрын
The derivation is mathematically correct, and the result is right but the justification for writing (-Vf) rather than (+Vf) at (4:44) is wrong. The velocity of the fuel mixture with respect to the external observer can easily have the same direction as the velocity of the rocket itself which is why it should be written with a plus sign (+Vfuel) and the term (V-Vex) determines the direction of this velocity. This mistake would have easily been avoided if the vector notation had been used. The reason why there in fact SHOULD be a minus there sign has to do with dm. Actually, the dm that appears in the term (m+dm) is NOT the same as the one in the term Vfuel*dm. While the first one denotes the mass DECREASE of the rocket, the second one is the mass INCREASE of the exhaust cloud. Their absolute value is the same but the signs must be different. So, this equation should have the following form: mV = (m+dm)(V+dV)+Vfuel*(-dm) This is of course mathematically equivalent but has a coherent physical justification behind it. I encourage the Physics Ninja to respond to this comment and justify his reasoning :).
@anthonystark7011Ай бұрын
Pls how was the formula for vfuel derived
@enyioma83 жыл бұрын
Quick question. I thought I knew what "relative to the rocket means". Is it the speed the gas moves because the rocket is moving, or is it just the speed used to push the rocket? Lastly, it's hard for me to understand how that the speed relative to the earth is different from the speed relative to the rocket. I'm wondering what makes them different 🤔
@irwanahmed0012 жыл бұрын
the speed of the ejected gas/mass relative to the rocket means the speed of the mass if observed by someone present in the rocket. imagine you are in the rocket, traveling at the same speed as the rocket, and are viewing the gas/mass being ejected. whatever speed you view the gas/mass being ejected at is the speed of the gas/mass relative to the rocket/you. the speed of the ejected gas/mass relative to the earth is different from the speed of the gas/mass relative to the rocket as both the mass and rocket are moving at two different velocities when observed from the earth's point of reference. however, if viewed from the rocket's point of reference, the speed of the gas/mass is viewed as the speed relative to the rocket, as whoever in the rocket is observing the gas/mass being ejected would be moving at the same velocity as the rocket, technically making them stationary, when viewed relative to the ejected mass.
@beegyoshi28282 жыл бұрын
Totally not learning this for fun
@hwanheebyun41424 ай бұрын
more so use leibniz derivative notation for clarity?
@loizostseriotis36792 жыл бұрын
Can I ask something ?Why the velocity of the rocket changes from v to v+dv instead of remaining v after the fuel burned ?
@gauravGupta-bk2sw11 ай бұрын
The thrust force, which is due to ejection of the gases...gives the rocket an acceleration, and therefore, an increment in its speed!
@jacksonchan68568 ай бұрын
very clear
@PhysicsNinja8 ай бұрын
Glad to hear that
@kareemsalessi4 жыл бұрын
Based on your calculations, what is the minimum required acceleration for a vertical rocket-launch??? Also, what is the minimum TWR for a similar rocket??? Thanks. K.S.
@harshitjajoria96893 жыл бұрын
It's based upon your rocket weight To launch a rocket, Total thrust generated should be 30% more to take your rocket in air
@kareemsalessi3 жыл бұрын
@@harshitjajoria9689 30% more than what???
@harshitjajoria96893 жыл бұрын
@@kareemsalessi 30% more weight thrust Like if your rocket's total weight is 70 kg so you need a thrust of 100 kg thrust for take off 👍
@kareemsalessi3 жыл бұрын
@@harshitjajoria9689 According to WHAT ???
@user-jg3ub6ni2w2 ай бұрын
i am doing this same math for missile considering the missile is cruising and neglecting the thrust used in lift,But problem is no one is using drag component here,can some one tell where I can learn what I am asking plzz thanks in advance!!
@vasilisalexopoulos81712 жыл бұрын
could you please explain how you turn the expression to positive in 12:53 ?thanks
@jdkdjd3126 Жыл бұрын
he flipped the fraction within the ln() from m/m0 to m0/m
@bobbyw19953 жыл бұрын
hi, please could you explain at 3.55 why Vfuel = V-Vex, instead of Vfuel = V+dv-Vex ?
@PhysicsNinja3 жыл бұрын
even if you did include the extra dv term in the end it multiplies with a dm term and this is a higher order term because it involves 2 infinitesimal quantities. We neglect all of those higher order contributions.
@bobbyw19953 жыл бұрын
@@PhysicsNinja great, thank you!
@sudarshann71943 жыл бұрын
How can dm
@PhysicsNinja3 жыл бұрын
dm is a small difference in mass (m_final-m_intial) and will be a negative number because the m_final is smaller then m_initial
@skepticsapiens41492 жыл бұрын
Nice
@sigma7208 Жыл бұрын
@10:43 There is a slight mistake. you forgot to divide by the mass in the acceleration equation. Thanks for helping alsways!
@PhysicsNinja Жыл бұрын
Yes, sorry about that
@gauravGupta-bk2sw11 ай бұрын
The idea of positives and negatives is scarier than pennywise
@victoreduardoramirezhuerta70143 жыл бұрын
Hi Ninja of physics its an honor to virtualy meet you can you answer my question please , what if I just add momentums but with signs included ... i mean I take dm as positive and velocity to the right as positive ... then i would have .... MV = ( M - dm)(V+dv) + (V - Vex)dm and the result is Vex*dm = M*dv , the only thing that changed from the original is the sign but in this case dm is positive , but when I integrate I feel awkward because dm is an increase from 0 to the entire mass that is ejected ...... is my observation right ? can i solve problems doing it that way ?
@papalegba67593 жыл бұрын
this guy doesn't even understand that the rocket & exhaust are part of the same system, so he's essentially defining COM as m1*v1=m1*v1. then when he wants the rocket to accelerate past its exhaust velocity (which is a provable physical impossibility), he redefines COM as m1*v1=(m1*v1+m1*v1). a simple free body diagram proves he is spamming pseudoscience.
@theglobalsuccess17653 жыл бұрын
What i see in ur observation everything is right except the case of adding momentum of rocket and fuel's while they are moving in opposite directions. Also the case vex is relative velocity in my case the velocity of fuel becomes vf=vex-v.
@stuartgray58773 жыл бұрын
@@theglobalsuccess1765 dont pay any attention to papa legba. He is literally an imbecile, and could not pass a high school physics exam if his life depended upon it For EXAMPLE, he has said that C4 would NOT even explode in a vacuum. AND he does not know that combusting hydrogen and oxygen makes water.
@koungmeng4 жыл бұрын
yay rocket science
@AryanSharma-xt7cl Жыл бұрын
Great 💗💙💬
@frankdimeglio82162 жыл бұрын
CLEARLY, gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites (ON BALANCE); as the stars AND PLANETS are POINTS in the night sky. Consider TIME (AND time dilation) ON BALANCE. By Frank DiMeglio
@joel.1433 ай бұрын
Where’s the mass at the equation 1 (acceleration equation)
@PhysicsNinja3 ай бұрын
Oops, forgot it
@elephantwalkersmith15333 жыл бұрын
wow! You discovered an equation for Trust! Great!
@adikisela4093 жыл бұрын
thereis gravity in space,its miss conception that is weightless sensation mean as no gravity in object because the escape velocity of it and pull force gravity of earth in balance.
@govindasharman4253 жыл бұрын
It'd have been more American if you shooted this video itself Wait...