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Center of mass using Mach's restatement of Newtonian Mechanics

  • #1

Homework Statement


Use Mach’s restatement of Newtonian mechanics to show that if we define the centre of mass of two particles according to,
⃗ r = (m1 ⃗ r1 + m2 ⃗ r2) / (m1 + m2)

then the center of mass moves according to the equation,
⃗r = [(m1 ⃗u1 + m2 ⃗u2) / (m1 + m2)]t + ⃗r0

where ⃗r0 is a constant vector.

Homework Equations


This is a condensed version of the notes that were given in class but everything is there :

Mach's approach was to use Newton’s third law, which is the only law of the three to actually state a law or prescription of nature, as a starting point.From Newton's third law we can get the expression

a1/a2 = k12 and k12 is a constant.

So then Mach’s reformulation of Newton’s mechanics states that for any two of two or more interacting particles, the ratio of their acceleration will be constant. If we apply a Galilean transformation we obtain,

a1'/a2' = k12 = a1/a2

showing the equations are form invariant.


We relate the constant with the inertial mass .
Mach’s version of Newtonian mechanics is free from definitions and has therefore simplified the theory by making reference only to acceleration a geometric quantity. All other standard notion of Newtonian mechanics can be derived from Mach’s restatement.


The Attempt at a Solution


I fully understand Newton's three laws and I get what Mach's restatement. I just can't figure for the life of me how I can use it to get the desired equation. If anyone has an idea of where to start that would be great!

Cheers!
 

Answers and Replies

  • #2
TSny
Homework Helper
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Look at the 2nd time derivative of
⃗ r = (m1 ⃗ r1 + m2 ⃗ r2) / (m1 + m2)
 
  • #3
2
0
derive R relative to t and then at t=0 R=R0 and m1u1+m2u2=constant
 
  • #4
I did what you said but I get (constant/m1+m2)t + R0 = R

How can a scalar + a vector = a vector ??
 
  • #5
TSny
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Since Mach's statement deals with the accelerations of the two objects, it might be good to somehow relate the formula for the position of the center of mass to the accelerations.

Note ##\frac{d\vec{r_1}}{dt} = \vec{v_1}=## the velocity vector.

What if you take one more derivative? What does ##\frac{d^2\vec{r_1}}{dt^2}## give you?
 
  • #6
2
0
the first derivative to big R relative to t (dR/dt) in left hand side will give u already m1u1+m2u2 in right hand side then then multiply two side by dt and then make integrations two side u will get what u need
 
  • #7
hey, im pretty sure i am in your class cause you quoted the exact pdf the teacher gave us, i really don't understand this question either and don't get any of those answers too..
 
  • #8
@elsayed , it doesn't use mach restatement though
 
  • #9
TSny
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It might help if we knew the mathematical level of this class. Is this a class where you can use calculus?
 

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