MSB47
Badges: 2
Rep:
?
#1
Report Thread starter 4 years ago
#1
The net external force acting on an object can be evaluated as the rate of change ofmomentum. This turns out to be a more fundamental way of stating the force than the use ofNewton's second law. Using Newton's second law and momentum:






But this limited relationship can be generalized to




I'm not sure how this was mathematically achieved? Can someone explain how this generalization is found?

Thanks for any help in advance
0
reply
astro67
Badges: 10
Rep:
?
#2
Report 4 years ago
#2
(Original post by MSB47)
The net external force acting on an object can be evaluated as the rate of change ofmomentum. This turns out to be a more fundamental way of stating the force than the use ofNewton's second law. Using Newton's second law and momentum:






But this limited relationship can be generalized to




I'm not sure how this was mathematically achieved? Can someone explain how this generalization is found?

Thanks for any help in advance
I suspect the problem is that Newton never wrote F = ma. That is an interpretation that was made of his original statement. There is some debate about his words but he referred to the change of motion of a body being proportional to the force acting on it. This is closer to F = dp/dt than F = ma. That means that nobody generalised from Newton. F = ma is not the starting point and is not actually Newton's second law, although it is often (mis)represented as such. It is a specific formulation of Newton's general law for situations where the mass of a body is constant.
0
reply
MSB47
Badges: 2
Rep:
?
#3
Report Thread starter 4 years ago
#3
(Original post by astro67)
I suspect the problem is that Newton never wrote F = ma. That is an interpretation that was made of his original statement. There is some debate about his words but he referred to the change of motion of a body being proportional to the force acting on it. This is closer to F = dp/dt than F = ma. That means that nobody generalised from Newton. F = ma is not the starting point and is not actually Newton's second law, although it is often (mis)represented as such. It is a specific formulation of Newton's general law for situations where the mass of a body is constant.
so the last equation is not correct? because im wondering how can it be used if both mass and speed are changing in a situation?
0
reply
Lissy14
Badges: 16
Rep:
?
#4
Report 4 years ago
#4
(Original post by MSB47)
The net external force acting on an object can be evaluated as the rate of change ofmomentum. This turns out to be a more fundamental way of stating the force than the use ofNewton's second law. Using Newton's second law and momentum:






But this limited relationship can be generalized to




I'm not sure how this was mathematically achieved? Can someone explain how this generalization is found?

Thanks for any help in advance
------------------------

Force is change of rate of momentum.

Momentum is p=mv
------------------------

To work out the change of rate of momentum (and therefore the force) you can do:

1) ∆p/∆T (change in momentum over change in time) This is the same as change RATE of momentum

or

2) mV/∆T (because p=mv and mass doesn't change)

-------------------------

If mass is changing (like a rocket ejecting fuel and becoming lighter) but velocity stays the same, the force will still change.
So we get the equation:

3) V∆m/∆T (velocity doesn't change)

-------------------------
But this is only an average.
The momentum might increase quickly at the beginning of the time frame (∆T) and slower at the end.
This would mean the force is larger at the start and smaller at the end.
So the force is only an average.

(To make the force more accurate you make the ∆T smaller)
-------------------------
Not sure if this answers you question but I hope it helps
0
reply
MSB47
Badges: 2
Rep:
?
#5
Report Thread starter 4 years ago
#5
(Original post by Lissy14)
------------------------

Force is change of rate of momentum.

Momentum is p=mv
------------------------

To work out the change of rate of momentum (and therefore the force) you can do:

1) ∆p/∆T (change in momentum over change in time) This is the same as change RATE of momentum

or

2) mV/∆T (because p=mv and mass doesn't change)

-------------------------

If mass is changing (like a rocket ejecting fuel and becoming lighter) but velocity stays the same, the force will still change.
So we get the equation:

3) V∆m/∆T (velocity doesn't change)

-------------------------
But this is only an average.
The momentum might increase quickly at the beginning of the time frame (∆T) and slower at the end.
This would mean the force is larger at the start and smaller at the end.
So the force is only an average.

(To make the force more accurate you make the ∆T smaller)
-------------------------
Not sure if this answers you question but I hope it helps
Thanks for your help but what if both mass and velocity are changing at the same time?
0
reply
astro67
Badges: 10
Rep:
?
#6
Report 4 years ago
#6
(Original post by MSB47)
so the last equation is not correct? because im wondering how can it be used if both mass and speed are changing in a situation?
The last equation is correct - F = ma is a specific version of that equation for systems where m is constant, which is the case for discrete bodies in non-relativistic motion. The point is that the last equation is the proper expression of Newton's second law as Newton himself expressed it - it isn't an extension of F = ma that was added after Newton. Statements that say F = ma is Newton's second law of motion are technically wrong although they are common.
0
reply
MSB47
Badges: 2
Rep:
?
#7
Report Thread starter 4 years ago
#7
(Original post by astro67)
The last equation is correct - F = ma is a specific version of that equation for systems where m is constant, which is the case for discrete bodies in non-relativistic motion. The point is that the last equation is the proper expression of Newton's second law as Newton himself expressed it - it isn't an extension of F = ma that was added after Newton. Statements that say F = ma is Newton's second law of motion are technically wrong although they are common.
Ahh ok I see what you mean but say if both speed and mass are changing would you just calculate the change in mass and change in velocity and multiply them together and then divide them by the time period it all occurred?
0
reply
astro67
Badges: 10
Rep:
?
#8
Report 4 years ago
#8
(Original post by MSB47)
Ahh ok I see what you mean but say if both speed and mass are changing would you just calculate the change in mass and change in velocity and multiply them together and then divide them by the time period it all occurred?
How would the mass be changing? If a body is shedding material (e.g. a rocket) or accreting it (e.g. a collapsed star orbiting a giant main sequence companion) then you will need to consider what is happening to this lost or gained mass - what is its momentum before and after it is attached to the body? That will have an impact on the residual body's momentum, in addition to any other force acting. Alternatively, is this a relativistic mass change? If so, you will need to invoke special relativity. The general form of Newton's second law will still hold but p = \gamma*mv, where m is the (constant) rest mass so you need to evaluate the rate of change of that quantity.
0
reply
MSB47
Badges: 2
Rep:
?
#9
Report Thread starter 4 years ago
#9
(Original post by astro67)
How would the mass be changing? If a body is shedding material (e.g. a rocket) or accreting it (e.g. a collapsed star orbiting a giant main sequence companion) then you will need to consider what is happening to this lost or gained mass - what is its momentum before and after it is attached to the body? That will have an impact on the residual body's momentum, in addition to any other force acting. Alternatively, is this a relativistic mass change? If so, you will need to invoke special relativity. The general form of Newton's second law will still hold but p = \gamma*mv, where m is the (constant) rest mass so you need to evaluate the rate of change of that quantity.
well when mass is lost in a rocket it is accelerating the rocket upwards which increases it's velocity too right? So both mass and velocity are changing simultaneously?
0
reply
Lissy14
Badges: 16
Rep:
?
#10
Report 4 years ago
#10
(Original post by MSB47)
Thanks for your help but what if both mass and velocity are changing at the same time?
F=∆P/∆T
Change in momentum can be calculated by doing several things…

  1. M x ∆V (if v is changing and m is constant)
  2. V x ∆M (if m is changing and v is constant)
  3. Or P2 – P1


Method 3 is used when both mass and velocity are changing.
It calculates the initial moment and takes it away from the final momentum. This gives the change in momentum

_______________________________

So ∆P=P2-P1
and,
F=(P2-P1)/∆T
1
reply
MSB47
Badges: 2
Rep:
?
#11
Report Thread starter 4 years ago
#11
(Original post by Lissy14)
F=∆P/∆T
Change in momentum can be calculated by doing several things…

  1. M x ∆V (if v is changing and m is constant)
  2. V x ∆M (if m is changing and v is constant)
  3. Or P2 – P1


Method 3 is used when both mass and velocity are changing.
It calculates the initial moment and takes it away from the final momentum. This gives the change in momentum

_______________________________

So ∆P=P2-P1
and,
F=(P2-P1)/∆T
ahh ok thanks for your help
0
reply
Lissy14
Badges: 16
Rep:
?
#12
Report 4 years ago
#12
(Original post by MSB47)
ahh ok thanks for your help
Your welcome ^-^
0
reply
morgan8002
Badges: 20
Rep:
?
#13
Report 4 years ago
#13
(Original post by MSB47)
The net external force acting on an object can be evaluated as the rate of change ofmomentum. This turns out to be a more fundamental way of stating the force than the use ofNewton's second law. Using Newton's second law and momentum:






But this limited relationship can be generalized to




I'm not sure how this was mathematically achieved? Can someone explain how this generalization is found?

Thanks for any help in advance
This is not always correct.
Force is defined as the rate of change of momentum.

F = \dfrac{dP}{dt}

F = \dfrac{d(mv)}{dt}

F = m\dfrac{dv}{dt} + v\dfrac{dm}{dt}(product rule)

F = ma + v\dfrac{dm}{dt}
0
reply
MSB47
Badges: 2
Rep:
?
#14
Report Thread starter 4 years ago
#14
(Original post by morgan8002)
This is not always correct.
Force is defined as the rate of change of momentum.

F = \dfrac{dP}{dt}

F = \dfrac{d(mv)}{dt}

F = m\dfrac{dv}{dt} + v\dfrac{dm}{dt}(product rule)

F = ma + v\dfrac{dm}{dt}
Thankyou very much that makes a lot of sense i understand it now
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Bristol
    Undergraduate Open Afternoon Undergraduate
    Wed, 23 Oct '19
  • University of Exeter
    Undergraduate Open Day - Penryn Campus Undergraduate
    Wed, 23 Oct '19
  • University of Nottingham
    Mini Open Day Undergraduate
    Wed, 23 Oct '19

Have you made up your mind on your five uni choices?

Yes I know where I'm applying (116)
65.91%
No I haven't decided yet (36)
20.45%
Yes but I might change my mind (24)
13.64%

Watched Threads

View All