What Determines the Center of Mass Movement in a Two-Object System Under Force?

[sophzilla]

A big olive (m = 0.11 kg) lies at the origin of an xy coordinate system, and a big Brazil nut (M = 0.82 kg) lies at the point (0.99, 2.1) m. At t = 0, a force Fo = (4i + 4j) N begins to act on the olive, and a force Fn = (-4i -3j) N begins to act on the nut. What is the (a)x and (b)y displacement of the center of mass of the olive-nut system at t = 4.6 s, with respect to its position at t = 0?

I first started approaching the problem by doing E(sigma)mixi/Emi, and the same for the y-direction. So, for x-direction, it would be:

(.99molive + 0mnut)/(.82kg + .11kg)

for the y-direction, it would be:

(2.1molive + 0mnut)/(.82kg + .11kg)

I don't even know if I did those correctly.

For the rest, they give you the force in both directions and the duration time (4.6 sec). I have to find the displaceent, which means I first have to find the center of mass for 0 seconds and then for 4.6 seconds.

Can someone help me with how to approach this problem? Thank you.

 

[Hootenanny]

I think you have got the nut and olive vectors mixed up. Here's the steps I would now take;

1. Convert the intial positions into unit vectors (ai + bj)
2. Convert the forces into accelerations
3. Use kinematic equations to find the position vector of the nut and the olive
4. Find the centre of mass

Alternatively, as Doc Al mentioned on the other thread, you could treat it a single system.

~H

 

[kyle8921]

I would approach this problem by first understanding the concept of center of mass and how it relates to the displacement of an object. The center of mass is the point at which the mass of an object is equally distributed in all directions. In this case, the olive and the Brazil nut are two objects with different masses and positions, and the center of mass of the system will be affected by the forces acting on them.

To solve this problem, we can use the formula for calculating the center of mass, which is given by:
xcm = (m1x1 + m2x2)/ (m1 + m2)
ycm = (m1y1 + m2y2)/ (m1 + m2)

Where xcm and ycm are the x and y coordinates of the center of mass, m1 and m2 are the masses of the olive and the Brazil nut respectively, and x1, y1 and x2, y2 are their respective positions in the xy coordinate system.

Now, let's calculate the center of mass at t=0 seconds:
xcm(0) = (0.11kg x 0m + 0.82kg x 0.99m)/ (0.11kg + 0.82kg) = 0.13m
ycm(0) = (0.11kg x 0m + 0.82kg x 2.1m)/ (0.11kg + 0.82kg) = 1.85m

This means that at t=0 seconds, the center of mass of the system is located at (0.13m, 1.85m).

Now, let's calculate the center of mass at t=4.6 seconds:
xcm(4.6) = (0.11kg x 4.6s x 4m/s + 0.82kg x 4.6s x (-4m/s))/ (0.11kg + 0.82kg) = -0.18m
ycm(4.6) = (0.11kg x 4.6s x 3m/s + 0.82kg x 4.6s x (-3m/s))/ (0.11kg + 0.82kg) = -1.1m

This means that at t=4.6 seconds,