Use energy conservation to determine the forces

[imy786]

Homework Statement

Consider an isolated system comprising two particles of masses m1 and m2, whose position vectors, in an inertial frame, are x1 and x2 and whose velocity vectors are v1 and v2. The interaction of the particles may be described by an energy function

E = 1/2 m1v1^2 + 1/2 m2v2 ^2 + U(x1, x2).

(a) Suppose that U = −k/r2, where k is a positive constant. Use energy conservation to determine the forces acting between the particles.

Homework Equations

E = 1/2 m1v1^2 + 1/2 m2v2 ^2 + U(x1, x2). (1)

U = −k/r2, (2)

The Attempt at a Solution

substituting U = −k/r2 into (1) gives

E = 1/2 m1v1^2 + 1/2 m2v2 ^2 −k/r2

then don't know what to do to determine the forces ,
i would think that gravtional forces are acting between the particles.

 

[Dick]

What does r2 mean? r usually denotes distance from x1 to x2. So r2=r^2 or just r? In either case, the force is related to the derivative of the potential energy function U with respect to distance. Why is this post titled 'relativity'?

 

[imy786]

r2 means rsquared which should be r^2
relativity, momentum forces...
well the question was longer , and other parts involved relativty, if u want me to post the rest of the question let me know.

 

[Dick]

If the potential is k/r^2 then it's not gravity. And you're using nonrelativistic expressions for kinetic energy. Maybe you'd better post the full question.

 

[imy786]

(i)Which aspect of Newtonian relativity requires
U to depend only on the separation vector x1 − x2?

(ii) Which further aspect of Newtonian relativity requires U to depend only on the magnitude r = |x1 − x2| of the separation vector?

and part (iii) is given as question (a)

 

[Dick]

Ahhhh! NEWTONIAN relativity. Did you figure out (i) and (ii)? For (iii) I'll reiterate, the force is the negative derivative of the potential. What is the direction of the force?

 

[imy786]

no couldn't do (i) and (ii).

force will act towards the particles

 

[Dick]

What ARE the principles of Newtonian relativity?

 

[imy786]

Newtonian relativity-The special principle of relativity states that physical laws should be the same in all inertial reference frames, but that they may vary across non-inertial ones

 

[Dick]

So in particular you should be able to translate and rotate positions without affecting the physics (such as forces or potentials). What does this tell you about (i) and (ii)?

 

[imy786]

U= -Gm1m2/ x1-x2

(i)Which aspect of Newtonian relativity requires
U to depend only on the separation vector x1 − x2?

mechanics.

 

[Dick]

If you translate the system in space, x1 and x2 change but their difference remains constant, right? Is that important?

 

[imy786]

yes it is important..well is it gravtional (netwonain relatiity that is reuqired)

 

[Dick]

yes it is important..well is it gravtional (netwonain relatiity that is reuqired)

Is that a question? Can you answer what principle for (i)?

 

[imy786]

(i)the principle of conservation of energy

 

[Dick]

The principle is translational invariance - a special case of frame invariance. Likewise, what would rotational invariance have to do with (ii)?

 

[imy786]

(ii) Which further aspect of Newtonian relativity requires U to depend only on the magnitude r = |x1 − x2| of the separation vector?


rotational invariance-

a vector quantity is rotationally invariant if its value remains the same under a rotation of its input vectors. Both the dot product and the cross product are rotationally invariant, while vector addition and scalar multiplication, in general, are not.

 

[Dick]

Basically, yes. x1-x2 is not rotationally invariant but |x1-x2| is. Now, what about your force. The latest version of the potential you sent looks more like -k/r than -k/r^2. Which is it and which force law does that lead to?

 

[imy786]

U = −k/r^2 this is the potenetial that is given in the question.

Newtons force law of gravition:

F= Gm1m2/ r^2

 

[Dick]

Fine. But if the potential is -k/r^2 then what is the force? (Don't repeat 'gravitation').

 

[imy786]

gravtional potentional

 

[Dick]

Not a sentence.

 

[imy786]

i think-

the force between the objects will be gravitonal potential.

 

[Dick]

Potential is not a force. It's a potential. Gravitational potential has the form -k/r. Your potential has the form -k/r^2, you say. Hence your potential is different and is not gravity. QED. What would the force law coming from your potential be?

 

[imy786]

well FORCE= mass* accelration,

acceleration being gravtional

what doee QED mean?.

What would the force law coming from your potential be?

F= KM1M2/ |x1 − x2|

 

[Dick]

QED just means 'thus it is proved'. You are giving me random answers for what the force is. You said you were given that the potential is -k/r^2. Now forget everything else except that. There is a general relationship between potential and force. What is it?

 

[imy786]

FOrce= qd

where q is the potential and d is the distance

 

[imy786]

f = m g,

U = m g y

Fx = dU / dx,

Fy = dU / dy

w = m g,

E = K + U

 

[Dick]

Most of these are either wrong or not applicable to the problem.
F=-dU/dr is the one you want.

 

[imy786]

- dU/ dx= F (x)

U (x) = - integral between x0 to x : F (x) dx + U (xo)

 

[imy786]

- dU/ dr= F (x)

U (r) = - integral between r0 to r : F (r) dr + U (ro)

 

[Dick]

Why are you talking about INTEGRATING!?? Just differentiate U!

 

[imy786]

i got the formula from here...were it says integral

http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html

 

[Dick]

Why didn't you use the part where it says 'derivative'? You HAVE U, you want to find F.

 

[imy786]

is this correct?

 

[Dick]

Yes. So it's a r-cubed force instead of an r-squared force like gravity.

 

[imy786]

is this correct

a (i) (ii)

 

[Dick]

You are correct.

 

[imy786]

r cubed force...is that a correct name for a force??

 

[Dick]

r cubed force...is that a correct name for a force??

No, it's not official. I just said it to point out that the force is related to a cube not a square (like gravity).

 

[Dick]

(iv) provide brief explanation in each case.

(a) Is the force law in (a-=r cubed force) attractive or repulsive?

repulsive as it has positive sign

(b)Does it satisfy Newton’s third law?

yes, as the object is two particles are opposing forces on each other
(c)Is it an inverse square law of force?

yes- it is inversely porptional
(d)Does it conserve linear momentum?

yes-momentum is alwayz converved in a colliosn

(e)Does it conserve angular momentum?

as momentum is alwayz conserved, so does angular momentum

[are these correct for these quesitons}

a) It's attractive. The potential was given as -k/r^2. Not k/r^2. Reverse your sign.

b) Give a clearer statement of Newtons third law.

c) WHY oh WHY would you say it is inverse square when you have just computed it to be inverse cube?

d) Linear momentum conservation is the same as Newtons third law.

e) Does your force exert any torques in the system?

 

[Dick]

so what is the r-cubed force called?

----------------------------------------
n3rd law- "For every action, there is an equal and opposite reaction."

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(c)Is it an inverse square law of force
no its an inverse cube

(d)Does it conserve linear momentum?

yes-momentum is alwayz converved in a collioion and is same as Newtons 3rd law

(e)Does it conserve angular momentum?

no, as there is no torque not applied to the system.

(is this correct)

Inverse r cubed laws don't have any particular name. ARE YOU SURE THAT IN YOUR STATEMENT YOU DIDN'T MEAN TO SAY U=-k/r (then it would be gravity). To show Newton's third law you want to show that the force m1 exerts on m2 is equal and opposite to the force m2 exerts on m1. Then linear momentum is conserved. If there is no torque applied by the force then angular momentum must be conserved. How would you show there is no torque??

 

[imy786]

moment= force (torque * perpendicular distance)

so if there is no moment on the foice there would be no toruq

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quesitons does say

U= -k/ r^2

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can this be a mistake in the quesiton

 

[Dick]

moment= force (torque * perpendicular distance)

so if there is no moment on the foice there would be no toruq

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quesitons does say

U= -k/ r^2

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can this be a mistake in the quesiton

If that's what the question says, then the answer is our inverse cubed force. The forces are directed straight from one particle to the other - so there is no perpendicular component to cause a torque. Is that what you mean?

 

[imy786]

yeah ,

well are all the quesitons..all ok..now and correct..and any parts still needed more clarifcaiton??

 

[Dick]

If you are comfortable with the answers, then I am.

 

[Dick]

It's certainly true that if something is always conserved - then it is conserved. But that's not a very interesting answer. On the other hand you can SHOW N3LAW works. Why not cite that as evidence?