Center of Mass of a System Problem

[catstevens]

Homework Statement

From an inertial reference frame S, the vector position of a particle of mass
m1 = 1kg is given by r1(t)=(txhat - t^2yhat)m.
The vector position of a particle m2=2m1 is given by r2=(t)=(txhat +t^3yhat)m

Assume t=1second
Find the position, the velocity and the acceleration of the center of mass of the composite system: xCM vCM a CM

Homework Equations

m1 x1(1) + m2 x2(1) / m1 + m2

The Attempt at a Solution

3xkg/ 3kg = 3x-direction

velocity = x
acceleration = 0

I do not think this is right, could some one help?

 

[Saladsamurai]

There is an x and y component to the position vector. You need to take both into consideration.

 

[thomate1]

Your attempt at a solution is not entirely correct. The formula you have used to calculate the center of mass is correct, but your values for the positions of the particles are not. The correct positions for the two particles at t=1 second are:
r1(1) = (1xhat - 1yhat)m
r2(1) = (1xhat + 1yhat)m

Using the formula for the center of mass, we get:
xCM = (m1x1 + m2x2) / (m1+m2)
xCM = [(1kg)(1xhat-1yhat)m + (2kg)(1xhat+1yhat)m] / (1kg+2kg)
xCM = (1xhat+m) / (3kg)
xCM = (1/3)xhat + (1/3)yhat

So the position of the center of mass at t=1 second is (1/3, 1/3) in the x-y plane. This means that the center of mass is located at a distance of 1/3 units from the origin, in the direction of the vector (1,1).

To calculate the velocity and acceleration of the center of mass, we need to take the derivatives of the position formula with respect to time:
vCM = (1/3)xhat + (1/3)yhat
aCM = 0

So the velocity of the center of mass is constant and equal to (1/3, 1/3) and the acceleration is zero. This makes sense, since the particles are moving in opposite directions and their velocities cancel out, resulting in a stationary center of mass.

In summary, the position, velocity, and acceleration of the center of mass of the composite system at t=1 second are:
Position: (1/3, 1/3)
Velocity: (1/3, 1/3)
Acceleration: (0, 0)