Dos particulas de masas m1 y m2 se mueven uniformemente en diferentes circulos de radios R1 y R2

step 1 So once again, welcome to a new problem. Let's just think about the origin and if you're having

Dos particulas de masas m1 y m2 se mueven uniformemente en...

Key Concepts

Trajectory Analysis

Trajectory analysis involves determining the nature of the path traced by a moving object from its position functions. This includes identifying if the path is circular, elliptical, or of another form. In problems involving multiple moving objects, analyzing the resultant trajectory—such as that of the center of mass—requires combining and comparing the individual motions, often revealing whether the combined movement exhibits uniform circular motion or a more complex behavior.

Parametric Representation of Motion

Parametric equations describe the coordinates of moving objects as functions of a common parameter, typically time. In the context of circular motion, they often involve sine and cosine functions to capture the periodic and oscillatory nature of the path. This representation helps in directly analyzing the time-dependent behavior of each coordinate, which is essential for determining trajectories and dynamic properties.

Uniform Circular Motion

This concept refers to the motion of an object moving in a circular path at a constant speed. Even though the speed remains constant, the direction of velocity changes continuously, resulting in a constant centripetal acceleration that points toward the center of the circle. This idea is fundamental in understanding circular trajectories and the kinematics involved in such motions.

Center of Mass

The center of mass of a system is the weighted average position of all the masses in the system. It is calculated by taking each mass’s position multiplied by its magnitude and then dividing by the total mass. Understanding the center of mass is crucial in mechanics, as it often simplifies the analysis of the system’s overall motion, particularly when external forces act uniformly on the system.

Transcript

00:01 So once again, welcome to a new problem.

00:05 Let's just think about the origin and if you're having an xy axis like this.

00:15 So we have an xy axis and obviously we're going to have two particles.

00:20 They have different radiuses.

00:22 We don't want to put the radiuses at the same position, but we're going to hypothesize that the fast one, r1, is at a description.

00:32 Specific radius with components in the x direction being the same as x1 equals to 4 cosine and then y1 being the same as 4 sine of 2t.

00:52 Of course in the z direction we don't have any components.

00:59 We can also hypothesize that the second point, this is the first point.

01:03 Point, this is the second point.

01:09 So our 2 is the same as x2 is the same as 2, cosine of 3 t minus pi over 2.

01:27 And then the y component is the same as 2 sine 3 t minus pi over 2.

01:37 So that's what you're seeing in the problem.

01:42 These are the x and the y coordinates.

01:44 Just to be specific, the y coordinate is on the y axis and the x coordinates on the x axis, that's what you're seeing.

01:54 The first part of the problem is we want to find out what's r1 and what's r2.

02:01 So this is the radius of both particles.

02:08 You know, that's what we want to find out.

02:10 Second thing we want to find out is the x coordinate of the center of mass and the y coordinate of the center of mass.

02:20 That's what we want to do, the x and the y coordinate of the center of mass.

02:25 And then the last thing we want to do is we want to figure out.

02:33 So figure out if the center of mass is circular in motion.

02:49 So by circular in motion we mean that if i had a circle like this and i wanted to figure out the formula for the circle, this is the radius, this is x, this is y, then you can see that x squared plus y squared is going to be the same as r squared.

03:06 So it's something similar to that in terms of finding the radius.

03:11 So we're going to jump in and try and figure out the radius of these two particles.

03:17 We know that if just based off of that, if x squared plus y squared equals to r squared.

03:31 So r squared is the same as x squared plus y squared.

03:35 And then we get the square root of both sides.

03:37 This is plus minus.

03:40 So r is the same as plus minus radical x squared plus y squared.

03:46 Those are the coordinates.

03:49 So in part a we know that x1 of t is the same as we're starting with x1 of t is the same as 4 cosine of 2t and then also we know that y1 of t is the same as 4.

04:12 4 sine of 2 t.

04:17 Those are the pieces of information that were given in the problem.

04:23 So to find the radius, we have to plug in those numbers.

04:26 So we have 1 being the same as radical.

04:31 We start with the x, 4 cosine of 2t.

04:35 We want to square that plus 4 sine of 2t.

04:40 We also want to square that.

04:42 Maybe we use this for it, this is just to make sure we don't have any contradictions.

04:48 Of course we're using this formula right here, pythagoras, so we have r1 is the same as 16 cosine 2x4 2t plus 16 sine squared 2 t that's those are the two numbers we're dealing with and recall, if we have theta right there and this is x this is y this is r you can see the sign of theta is the same as opposite of a hypoteness and cosine of theta is the same as adjacent of hypoteness and since x squared plus y squared equals to r squared uh you can always solve for x by multiplying both sides by r these two cancel out r cosine theta plus uh this this one also multiplies like that.

06:00 So it becomes r sign of data like that.

06:06 Since the x and the y and the r squared, 1 squared that, and then we have used distributed property for the exponent, r squared, cosine squared, data plus r squared, sine squared, data equals to r squared.

06:25 Then we go ahead and divide everything by r squared.

06:29 So, cosine square data plus sine squared data equals to 1.

06:39 So coming back to this problem, we get to see that we can pull out the 16 and have cosine, so sine squared 2t.

06:53 And of course, this whole thing is equivalent to 1.

06:59 That collapses the problem to 16 times 1, which is plus minus 4, but since it's a radius, we can't have negative 4.

07:10 So r1 is the same as positive 4 in terms of the radius.

07:17 So that's what you're seeing in the problem.

07:19 This is connected to particle 1.

07:25 Jumping on to particle 2, we get to see that same process.

07:40 Or r2, you're going to be the same as radical x2 squared plus y2, you know, square that, squared, and this becomes x2, you can recall was the same as 2 cosine, 2 cosine of 3 t, minus 5 of 2 squared, plus 2 sine of 3 t, plus 5 of 2 squared, plus 2, 2, 5 of 2, 2, 2, 2, 2, 2, 2, and 2.

08:20 You want to square that.

08:30 So then this one becomes, you can pull out, distributive property, we end up having 4 cosine squared of 3 t minus 5 o 2, plus we do the same thing for this one, 4, 5 squared 3p minus 5 over 2.

08:55 Then we end up having 4 right there, cosine squared 3 t 5 over 2 plus sine squared 3 t minus 5 2 this the same was 1 so we have r2 the same is radical i want to call it plus minus radical 4 which is then plus minus 2 but we don't really pick the negative because it's radius so we end up with positive 2 being the same as the radius and this is for the second particle.

09:45 That's the particle two.

09:49 And then jumping ahead into part b, our goal was to find the and y coordinate of the center of mass.

09:58 In simple terms, x center of mass is the weighted average of the product of the mass and the distance from the origin in the x direction.

10:15 Of course you can plug in the numbers.

10:20 We don't have to plug in but it's helpful to do that.

10:25 So we have m1 and simplify that.

10:30 We have m1, x1 which is the same as 4 cosine of 2 t.

10:43 M2 x2 is 2 cosine of 3 t minus 5 or 2.

10:52 Then we want to divide everything by hf1 plus l2...