Car acceleration if resistance forces don't exist

[russ_watters]

You choose Earth ground as ref. frame when calculate kinetic energy. I am in the car engine,if I travel 0km/h ,100km/h or 2000km/s for engine doesn't make difference, he don't feel nothing .engine just feel acceleration when fuel consumption rise...

No, the car and ground interact in a very real/non-arbitrary way to make the car move. You can't transform that away by changing reference frames.

 

[Lnewqban]

Kinetic energy raise with speed for rocket as well,but rocket(constant thrust) will have same acceleration and fuel burn from 0-100km/h and from 100-200km/h, so know that Ke raise with speed don't tell me nothing.
If we neglect fuel loss over time.

I hope you have changed your take on the above statement.

https://en.wikipedia.org/wiki/Work_(physics)#Work–energy_principle

 

[Jurgen M]

Car wheel don't act at Earth c.g. so car has tendency to push Earth back and rotate,translation+ rotation ?
If car is very very big he will push Earth just like little stone when comes at tyer?

 

[A.T.]

Let imagine that car with constant 500HP accelerate but resistance forces don't exist (aero drag,internal friction in engine and transmision,tyer rolling resistance etc etc..)
neglect fuel loss over time..

From 0...

This is already unphysical. With no loses, any non-zero power input would produce an infinite propulsive force at velocity = 0. In reality the propulsive efficiency tends towards zero when the velocity relative to the propulsive medium (here ground) does.

 

[A.T.]

Car wheel don't act at Earth c.g. so car has tendency to push Earth back and rotate,translation+ rotation ?
If car is very very big he will push Earth just like little stone when comes at tyer?

View attachment 291839

The common CoM of planet and car will stay fixed. But the planet's CoM will oscillate as the car drives around it.

 

[Jurgen M]

This is already unphysical. With no loses, any non-zero power input would produce an infinite propulsive force at velocity = 0. In reality the propulsive efficiency tends towards zero when the velocity relative to the propulsive medium (here ground) does.

I started this topic to find out what is the main reason/contribution why car accelerate slower at higher speeds,because I always doubt that air can slow down acceleration so much.It must be something else but is hard to understand with pure logic/intuition because delta V is the same.

 

[PeroK]

Car wheel don't act at Earth c.g. so car has tendency to push Earth back and rotate,translation+ rotation ?
If car is very very big he will push Earth just like little stone when comes at tyer?

The calculations are more complicated if we consider the rotation of the Earth, so let's simpify things by imagining a car of mass $m$ pushing against a large object of mass $M$. The car has an engine which releases energy and creates relative motion between the mass $m$ and $M$. The engineering details are not relevant to show where your mistake is.

First, we look at the initial rest frame. The car moves in one direction at $v$ and the large object in the other direction at speed $V$. By conservation of momentum we have:$$mv = MV$$ Now, we calculate how much energy is needed:$$E = \frac 1 2 mv^2 + \frac 1 2 MV^2 = \frac 1 2 mv^2 + \frac 1 2 M(\frac{mv}{M})^2 = \frac 1 2 mv^2(1 + \frac m M)$$And if $M$ is much larger than $m$, we see that $\frac m M \approx 0$ and we have: $$E \approx \frac 1 2 mv^2$$ We also see that the amount of energy needed increases as $v^2$. E.g. if we want to car to travel at $2v$ relative to the large object, then we need $4$ times the energy.

Now, we try your trick of achieving a speed of $v$ and then moving to the rest frame of the car. What you say is that the car now has zero velocity again. But, what you forget is that in this frame of reference the large object now has speed $v$. The KE of the system in the car's rest frame is: $$E_1 = \frac 1 2 M v^2$$ And the initial momentum is $p_1 = -Mv$ (if we take the direction of the car to be positive). I just want to stress that you cannot simply ignore this, as you have been doing in all your posts.

In any case, we now want to increase the relative speed of the car and object to $2v$, which means approximately getting the car moving at speed $v$. Again, by conservation of momentum we have $$mv - M(v + V) = -Mv$$ and again we have $$mv = MV$$ Now the total KE of the system is:
$$E = \frac 1 2 mv^2 + \frac 1 2 M(v+V)^2 = \frac 1 2 mv^2 + \frac 1 2 Mv^2(1 + \frac m M)^2$$
$$E = \frac 1 2 mv^2 + \frac 1 2 Mv^2(1 + 2\frac m M + \frac{m^2}{M^2}) = \frac 1 2 mv^2 + \frac 1 2 Mv^2 + mv^2 + \frac 1 2 mv^2(\frac m M)$$ Again, the last term is approx zero and we have the change in energy: $$E - E_1 = \frac 1 2 mv^2 + mv^2 = \frac 3 2 mv^2$$ And that is the same change in energy as we found in the original rest frame, where $$E' - E'_1 = \frac 1 2 m(2v)^2 - \frac 1 2 mv^2 = \frac 3 2 mv^2$$
It doesn't matter whether you change reference frames of not, the change in kinetic energy is always the same. As it must be, as the energy came from the fuel consumption, which is independent of reference frame.

 

[PeroK]

PS there is also a very serious point that you can never understand physics if you don't want to get your hands dirty by doing calculations.

 

[Jurgen M]

PS there is also a very serious point that you can never understand physics if you don't want to get your hands dirty by doing calculations.

I agree but I can't do calculation if my logic is completely wrong..I didnt know how to start...
In car frame I forgot for Earth,I behave like Earth don't exist

 

[phystro]

I think it was confusing because of this detail (I now see that you must use u not du ---> P=Fu, as I mentioned in my first post):
A) The graph of Kinetic energy vs velocity wouldn't be a linear graph (it has already been posted). You would need 4 times more kinetic energy to achieve two times higher velocity, with the same acceleration.
B) The graph of Power vs velocity would be linear (P=Fu - for constant acceleration). This means that you would need exactly 2 times more power to achieve two times higher velocity.
C) This is explained because power is the time derivative of energy. i.e: $P= dE/dt$

 

[A.T.]

is hard to understand

The key relationship is quite simple:
power_engine * efficiency = power_propulsion = force * velocity

 

[meekerdb]

But why would object will be harder to accelerate when object travel at higher speeds if speed is relative term?
I don't understand this concept at all..

So you don't agree with comments from member @m4r35n357 in site below?
https://physics.stackexchange.com/q...ct-at-higher-speeds-if-speed-is-relative-term

It's because you have posed the problem in terms of constant power, where power is a relative value, because kinetic energy is a relative value. If the car had a rocket engine it could have constant thrust and acceleration which are absolute values.

 

[meekerdb]

The easy solution is to equate energies: 1/2 mv^2 = Pt so v = sqrt[2Pt/m].

 

[Jurgen M]

It's because you have posed the problem in terms of constant power, where power is a relative value, because kinetic energy is a relative value. If the car had a rocket engine it could have constant thrust and acceleration which are absolute values.

So car with rocket engine will gain benefits because of this,in drag racing?
Wait the minute, if rocket engine has constant thurst that mean increase in power with speed?

Where this increase in power come from,how explain this in reference frame of rocket without violate physics laws?

 

[A.T.]

how explain this in reference frame of rocket without violate physics laws?

Do the calculation in both frames and compare. But don't forget to include the kinetic energy of the expelled exhaust gases.

 

[jack action]

Where this increase in power come from

Increase in fuel consumption.

They may have a greater capacity to create power, but they don't do magic.

 

[jbriggs444]

Increase in fuel consumption.

No. This is not correct.

The context here is the question of how a rocket can succeed in adding kinetic energy to the payload at a higher rate when going from 100 mph to 200 mph than it can when going from 0 mph to 100 mph.

The fuel consumption is identical for both situations.

The missing bit in the energy accounting is looking at the kinetic energy in the unburnt fuel and in the exhaust stream. For a fast-moving rocket, you have higher kinetic energy in the unburnt fuel and lower kinetic energy in the exhaust stream compared to a slow moving rocket. That's where your extra power comes from. Less (and eventually negative!) power goes to the expended fuel while more power goes to the payload.

If you do the energy accounting properly, you will see that the chemical energy in the expended fuel matches the total increase in kinetic energy of exhaust stream plus rocket. You can further see that this holds regardless of what reference frame you choose to use for the calculation.

 

[PeroK]

No. This is not correct.

The context here is the question of how a rocket can succeed in adding kinetic energy to the payload at a higher rate when going from 100 mph to 200 mph than it can when going from 0 mph to 100 mph.

The fuel consumption is identical for both situations.

... to do the calculations. We have a rocket of mass $M+m$, which expels a small mass $m$ at speed $v$ relative to the rocket. By conservation of momentum, the rocket will accelerate by approximately $\frac m M v$. And the total gain in KE is $$\Delta KE = \frac 1 2 mv^2(1 + \frac m M) \approx \frac 1 2 mv^2$$

If the rocket expels a second mass $m$ with the same relative speed, then we have the same increase in speed and the same gain in KE. In this case, we can see this by changing to the new rest frame of the rocket and observing that we have an identical scenario to the first expulsion. To be precise the acceleration will be slightly greater, as the mass of the rocket has reduced to $M$. As the rocket reduces in mass, these continual mass expulsions will become ever more efficient.

The difference from the previous analysis where the Earth was involved is that we do not have the external large mass to take into account when we change frames.

We can check by doing the analysis in a frame of reference where the rocket has any speed $u$.

Initial momentum $p_0 = (M+m) u$ and initial KE $E_0 = \frac 1 2 (M+m) u^2$.

After the expulsion, we have momentum $p = M(u + \Delta u) - m(v-u)$, hence by conservation of momentum we have (as before) $$\Delta u = \frac m M v$$ And the change in KE is:
$$\Delta E = \frac 1 2 M (u + \Delta u)^2 + \frac 1 2 m(v-u)^2 - \frac 1 2 (M+m) u^2$$
Which after some simplification gives, as before:
$$\Delta E = \frac 1 2 mv^2(1 + \frac m M)$$
Again we see that changing the reference frame does not change the increase in KE.

 

[meekerdb]

So car with rocket engine will gain benefits because of this,in drag racing?
Wait the minute, if rocket engine has constant thurst that mean increase in power with speed?

Where this increase in power come from,how explain this in reference frame of rocket without violate physics laws?

Same answer. Power increases because you're calculating it relative to an increasing velocity. Total energy and momentum are conserved, including the exhaust and chemical energy.

 

[Jurgen M]

Rocket speed is 10km/s,exhaust speed relative to rocket is 10km/s and it is always constant if burn rate is constant.

From initial frame at rest,rocket speed is 10km/s and exhaust speed is zero.
Ke of exhaust is zero.There is no thrust in this frame? confused..

 

[PeroK]

Rocket speed is 10km/s,exhaust speed relative to rocket is 10km/s and it is always constant if burn rate is constant.

From initial frame at rest,rocket speed is 10km/s and exhaust speed is zero.
Ke of exhaust is zero.There is no thrust in this frame? confused..

Thrust is related to momentum conservation.

In your example, the exhaust started at $+10km/s$ and ended at $0 km/s$. In that frame, the expellant lost KE. And the remaining rocket gained KE.

 

[A.T.]

Rocket speed is 10km/s,exhaust speed relative to rocket is 10km/s and it is always constant if burn rate is constant.

From initial frame at rest,rocket speed is 10km/s and exhaust speed is zero.
Ke of exhaust is zero.There is no thrust in this frame? confused..

Thrust is equal but opposite to the rate of change of momentum of the expelled mass, which is non-zero:
https://en.wikipedia.org/wiki/Momentum#Relation_to_force

 

[Jurgen M]

Thrust is related to momentum conservation.

In your example, the exhaust started at $+10km/s$ and ended at $0 km/s$. In that frame, the expellant lost KE. And the remaining rocket gained KE.

Can rocket travel faster then exhaust speed(measured in relation to rocket)?

 

[A.T.]

Can rocket travel faster then exhaust speed(measured in relation to rocket)?

Yes.

 

[PeroK]

Can rocket travel faster then exhaust speed(measured in relation to rocket)?

Speed is frame dependent. In Newtonian mechanics, the speed of the rocket can be any number and the speed of the exhaust be that speed plus or minus the relative speed.

In relativistic mechanics, we have the limitation of the speed of light, but the same applies. The speed of the exhaust may be greater or less than the rocket depending entirely on your chosen frame of reference.

 

[PeroK]

PS if the rocket is less massive than the expellant, then in the initial rest frame of the two the rocket gains more speed.

 

[Jurgen M]

Kinetic energy is frame dependent, just as velocity is.

Momentum is proportional to velocity and is frame dependent too, just as velocity is.

Is this true?

 

[PeroK]

Kinetic energy is frame dependent, just as velocity is.

Momentum is proportional to velocity and is frame dependent too, just as velocity is.

Is this true?

Absolutely. These dynamic quantities must be frame dependent. Acceleration, however, is not frame dependent. That's why we can have Newton's second law (for any inertial frame): $$F = ma$$
In fact, this establishes the important idea that we need to define mass and force in such a way that they are not dependent on the choice of inertial reference frame.

 

[A.T.]

Kinetic energy is frame dependent, just as velocity is.

Momentum is proportional to velocity and is frame dependent too, just as velocity is.

Key is that kinetic energy is non-linearly dependent on velocity, while momentum is linearly dependent on velocity. So changes in kinetic energy are frame dependent, while changes in momentum are frame invariant (for inertial frames).

 

[Jurgen M]

In rocket reference frame there is no increase in power?

 

[PeroK]

In rocket reference frame there is no increase in power?

If we are talking about the power in terms of the energy released by the engine, then that must be invariant.

This thread seems to be going round in circles. You seem to be asking almost random questions without showing much evidence of having digested the content of what is now 66 posts.

 

[Jurgen M]

If we are talking about the power in terms of the energy released by the engine, then that must be invariant.

I need this confirmation.
Generally I see increase in power at rocket as "mathematical manipulation" , I don't see anything useful in real physical sense in it.

 

[A.T.]

I don't see anything useful in real physical sense in it.

You can't see what is useful in real physics without doing real physics.

 

[Jurgen M]

You can't see what is useful in real physics without doing real physics.

If I am understood correctly ,in inertial frame(initial frame at rest) rocket increase power ,in non inertial frame(rocket) power don't increase.

Look to me it is just metter of perspective,I don't know how to call it...

 

[A.T.]

Look to me it is just metter of perspective,I don't know how to call it...

Frame dependent.