# What does it mean for a particle to be free?

1. Dec 7, 2008

### catstevens

1. The problem statement, all variables and given/known data
From an inertial reference frame S, the vector position of a particle of mass
m1 = 1kg is given by r1(t)=(tx-hat - t^2y-hat)m.
The vector position of a particle m2=2m1 is given by r2=(t)=(tx-hat +t^3y-hat)m

Is particle #1 free along the y-direction? Explain

2. Relevant equations
If the particle is free along the y-direction then the y's would equal to 0.

3. The attempt at a solution

r1= -t^2y
r2=+t^3y
=not free?
(I am actually unsure if my definition of free is true)

2. Dec 7, 2008

### Fredrik

Staff Emeritus
"Free" means "unaffected by forces", or equivalently that the potential is constant. In this case, it means you have to check if it's accelerating in the y-direction or not.

3. Dec 7, 2008

### catstevens

Particle #1's motion along the y-direction is -t^2.
Would this mean that it is not free?

4. Dec 8, 2008

### Fredrik

Staff Emeritus
That's its position at time t. What's its acceleration at time t?