How Do Forces Affect the Center of Mass Displacement in a Two-Object System?

[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, especially how I can use the vector forces? Thank you.

 

[Doc Al]

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.

Looks to me like you got the olive and nut mixed up; it's the olive that is at the origin.

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, especially how I can use the vector forces?

There are two ways to approach this. One way is to treat each "particle" separately: Given the force, find its acceleration, then it's displacement. (Treat each component independently.) Then find the new center of mass at t = 4.6 sec.

Another way, a bit easier, is to treat the nut and olive as a single system. Find the net force on the system (just add the forces). Then, treating the system as a single "particle" (with mass equal to the total mass of both), you can find the acceleration--and then the displacement--of the center of mass directly.

 

[kyle8921]

I would approach this problem by first understanding the concept of center of mass and how it relates to the position of objects in a system. The center of mass is the point at which the mass of an object is evenly distributed, and for a system of objects, it is the point at which the total mass is evenly distributed. This point is also the point at which the net external force acts on the system, causing it to move in a certain direction.

In this problem, we have two objects, the olive and the Brazil nut, with masses of 0.11 kg and 0.82 kg, respectively. The olive is located at the origin (0,0) and the Brazil nut is located at (0.99, 2.1) m. At t=0, two forces are acting on the system, Fo = (4i + 4j) N on the olive and Fn = (-4i - 3j) N on the Brazil nut. We need to find the displacement of the center of mass at t=4.6 seconds with respect to its position at t=0.

To solve this problem, we can use the formula for center of mass, which is given as:

CM = (m1r1 + m2r2 + ... + mnrn) / (m1 + m2 + ... + mn)

Where m is the mass of each object and r is the position vector of each object.

In this case, we have two objects, so the formula becomes:

CM = (m1r1 + m2r2) / (m1 + m2)

Substituting the values, we get:

CM = ((0.11 kg)(0,0) + (0.82 kg)(0.99, 2.1)) / (0.11 kg + 0.82 kg)

= (0.81, 1.72) m

This is the position of the center of mass at t=0.

To find the displacement at t=4.6 seconds, we need to use the concept of acceleration and the formula for displacement, which is given as:

x = x0 + v0t + 1/2at^2

Where x is the displacement, x0 is the initial position, v0 is the initial velocity, t is the time, and a is the acceleration.

In this problem, we have the