Consider the system of particles discussed in Exercise 7.4. Suppose the particles are connected

Consider the system of particles discussed in Exercise 7.4....

Key Concepts

Conservation of Linear Momentum

This concept refers to the principle that in an isolated system, where no external forces act, the total linear momentum remains constant over time. In the context of the given system, even though the particles are connected by a spring which exerts internal forces, these forces cancel each other out due to Newton’s third law, leading to the conservation of momentum in the horizontal (x) and vertical (y) directions.

Center of Mass Motion

The motion of the center of mass of a system is directly linked to the overall linear momentum of the system. When the linear momentum in each direction is conserved, it implies that the center of mass will move at a constant velocity in the absence of external forces. This means that despite any internal interactions like those from the spring forces, the center of mass of the particle system will exhibit uniform motion.

Internal and External Forces

In any mechanical system, it is crucial to distinguish between internal forces, which act between components of the system, and external forces, which act on the system as a whole. Internal forces, such as the spring force connecting the particles, obey Newton's third law and cancel each other out, thereby not affecting the total momentum of the system. This distinction explains why internal forces do not alter the overall motion of the center of mass.

Transcript

00:01 So for our system of three particles, as in the previous problem, we can write down the position vectors for each of the particles.

00:11 For particle a, ra, is simply 3 along the unit vector j.

00:17 The position vector for b, rb, is equal to 1 .2 along the i direction, plus 2 .4 along j, plus 3 along k.

00:33 And the position vector of c relative to the origin is 3 .6i.

00:42 So here we have our position vectors for our three particles.

00:47 For part a of the question, we need to find the mass center of the system.

00:51 And we know that the mass center is defined as follows.

00:57 The masses of the three particles, ma plus mb plus mc, essentially the mass of the whole system, multiplied by the mass center r is equal to the sum of the product of the masses and the position vectors.

01:16 So m -a -r -a plus m -b -r -b plus m -c -r -b plus m -c -r -c.

01:28 So since we know the masses, we get that nine times the mass center r is equal to 3 into 3j.

01:41 Of the mass in the position vectors plus 2, which is the mass of b, into 1 .2i, plus 2 .4j, plus 3k, plus the mass of c, which is 4 kg, into its position vector 3 .6i.

02:06 And from here, we can find that the mass center, r, is equal to 1 .3.

02:17 867 and that's in meters in si units in the i direction plus 1 .533 meters in the j direction plus 0 .667 meters in the k direction so that is our mass center of our system of particles so that's what we need to calculate in part a now in part b we need the linear momentum of the system.

02:55 So we need to first calculate the linear momentum for each particle.

03:04 Once we have the linear momentum for each particle, we can calculate the linear momentum for the system.

03:11 So for each particle, the linear momentum is its mass times its velocity vector.

03:17 And the velocity vectors are given.

03:19 So it's simply again a scalar product.

03:22 So the linear momentum for particle a is m -a -v -a.

03:28 And we get 12 i plus 6j plus 6k the linear momentum for particle b is mb vb again a scalar product gives us 8 i plus 6j for particle c the linear momentum is mc vc and this is minus 8 i plus 16 j plus 8k and hence we have our linear momentum is mc vc and this is minus 8 i plus 16 j plus 8k and hence we have our linear momentum for each particle.

04:12 Now that we have the linear momentum for each particle, we can calculate the linear momentum of the system.

04:19 The linear momentum of the system is the mass of the system m multiplied by the velocity for the system.

04:31 And this is m -a -v -a.

04:35 So the sum of the linear momentum for each particle, plus mb, v -b...

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