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Problem 7 In the figure 1, two particles each with mass m = 0.85 kg, are fastened to each other; and to a rotation axis at 0, by two thin rods, each with length d-5...

Question

Problem 7 In the figure 1, two particles each with mass m = 0.85 kg, are fastened to each other; and to a rotation axis at 0, by two thin rods, each with length d-5.6 cm and mass M =1.2 kg: The combination rotates around the rotation axis with the angular speed 0-0.30 rad/s. Measured about 0, what are the combination'$ (a) rotational inertia and (b) kinetic energy?

Problem 7 In the figure 1, two particles each with mass m = 0.85 kg, are fastened to each other; and to a rotation axis at 0, by two thin rods, each with length d-5.6 cm and mass M =1.2 kg: The combination rotates around the rotation axis with the angular speed 0-0.30 rad/s. Measured about 0, what are the combination'$ (a) rotational inertia and (b) kinetic energy?



Answers

In Fig. $10-37$ , two particles,
each with mass $m=0.85 \mathrm{kg},$ are fastened to each other, and to a rotation
axis at $O,$ by two thin rods, each with
length $d=5.6 \mathrm{cm}$ and mass $M=$
1.2 kg. The combination rotates
around the rotation axis with the angular speed $\omega=0.30 \mathrm{rad} / \mathrm{s}$ . Measured
about $O,$ what are the combination's
(a) rotational inertia and (b) kinetic energy?

Okay in this question we have to particles attached to two rods. So we have situation looks like this. Each rod is 5.6 centimeters long. Or 0.56 m. Each Rod has a mass of 1.2 kg and each Point Mass has a point mass of .85 kg. So our moment of inertia is the first thing that we need to figure out. And the way that we're gonna do that is we're going to consider this to be one long rod with two point masses. So the moment of inertia for two point masses is M R squared plus M M R squared. I just use prime to donate uh you know, the two different ones plus the moment of inertia of a rod about the end of the rod, which is what we have in this situation. A moment of inertia is one third M L squared. Okay, so now let's plug in. So our inner point mess As a massive .85 kg and radius from the pivot point .056 m squared. Plus Our 2nd mess has pointed five kg and it's at a double of that radius or .112 m Squared plus one third. Our total mass of the rod is 1.2 plus 1.2 is 2.4 kg times The length, which is .112 meters squared. Right? The total length we're considering it just to be one longer big rod. I'm gonna plug this in. We get I equals point 0 to 3 kg meters squared. Now we want to calculate the kinetic energy kinetic energy just like linear B one half mv squared for rotational is one half I omega squared. And they give us Omega in the questions we have one half times are moment of inertia. That would just solve. 4.0-3. I'm leaving on your chest for room In our angular velocity is .30 radiance per second. So we get a kinetic energy .0010 jewels. So

This problem covers the concept of the battle access to them. And from the paradoxes tour um the rotational inertia about the 0.0. Is a Cuban rental. The rotational inertia about the center of mass plus the mass of the object into the distance of the center of mass from the excess reputation. So from this we can write for part a the rotational inertia of the system. About always a cuba random the rotational inertia of the road about its center of mass. That is Capital M into the square upon 12 plus The mass of the road into the spire of the distance from zero. of the center of mass. That is t upon to square this for the object. M. Small. And we can write uh traditional. Let's hear about point toys. Mm The square Plus for the 2nd road the moment. Traditional we'll see about the center of mass is mps were upon 12 plus ah mm into three. Dearborn. Two squares. Okay. Plus for the mar small um m into jodi square. So the rotational inertia about zero is uh 8.3 capital M. D. Square. Yes, five small in the square. Or we can write the rotation in us. Yes, the square upon three into eight M plus 15 small in. So this is the rotational inertia about the ex exportation and the solution of party. Now, for part B, the rotational kinetic analogy is half of the rotation and I'll see about. Oh, into omega square are the rotational kinetic energy is half of the generational inertia is The square upon three into eight M plus 15 into omega square african. Right? The rotational kind of technology as they square up all six into eight M plus 15. And omega square

In this solution, there is a figure given. So first I will draw this figure and explain the data. So you can see that in the figure there are three particles is given, whose mouth is, that is forage particle masses given that is 23 g. And these three particles are fashioned to three wards of length. That is given this equal to well centimeter. So here I'm doing this figure that is given. So this is the first particle, this is the second particle and this is the third particle and I can say that this is the first door whose land this status city, then this is the second door whose land is D. And this is the third door whose land is that is D. And yeah at this point masses for what for I considered these three particles that is Cuban. There is M is equal to 23 g. So and this is the point named SQL and at this point, oh so for the information is given, that is they're totally assembly protests around Point B. With anglo Israel is Cuban. That is omega is equal to 0.85 radiant per second. And so now you can see that in the kitchen three parts is given. So I will answering the all parts one by one. So let's start answering the execution. So in the part A. It is asking that we have to find the rotational inertia of assembly. That means we have to find the rotational inertia about point or of system. So it'll readiness. That is. We know that the rotational inertia for point about a vertical axis. There is if I consider that this is a vertical axis and this is the point at which the particle is situated with masses M. And the distances. Studies from the vertical axis, perpendicular distances. That is our then for particle, the rotational inertia about this vertical axis we reorganize there is M. R. Is square. So use this concept then I will say that the rotational inertia of the assembly will be written as there is I considered I one plus I to plus I three. So for I one the rotation inertia will be that is mass of particles for all for all three particles masses him. So that means given the Dcm and now our is changed according to particle. So for first particle that are well written eyes with respect to point. Oh, I considered I will wait at this day. That means the square. No, for then I do will be it nice. That is the rotational inertia for second point about point or will be I considered mass multiplied by distance with the square two point. Oh, that is I consider D plus D. That means today. So that means M. R. Square. That means emptiness of to the square plus. Now, similarly I three will return eyes. That is rotational inertia of third particle. So I considered mm times of total decisions with respect to point or will be that is D plus D. Plus D admits three D. It's choir. So mhm. No. Mhm. Simplify it. Then I will get the answer for party. That is answer is Mhm. After reading I will get that is 14 MD square. So now put the value of M. And these human there is M is 20 g and 23 g and these 12 sentimental. Then I can see that in as a unit. The M will return is that is 23 multiplied by 10 to the power of monastery kg. And they will be recognized in the A. C. Unit against the 12 times of 10 to the power of minus two m. So for this data, that is human in this step then I will get the rotation in this year of assembly about point or will be equal to at least 14 times of and will return as that is 23. Multiplied by 10 to the power of minus three kg. Multiplied by the square. That means 12 times of 10 to the power of minus two Holy square. So after swallowing final value is that is equal to 1.1, sorry, 4.6 times of 10 to the power of minus three kg meter square. So this is the answer for part A. Now come to the next part. That is part B. In which solution is asking that we have to find the magnitude of angular momentum of middle particle. That means mhm. For this vertical we have to find the magnitude of angular momentum. So we know that the angular moment of general formula is studied. L. Is equal to that is I Oh my God! So yeah, I will be for middle particle. I can say that that will be I two. And omega is given that is by which the assembly is rotating about 20.0. That is given omega angular speed. That is equal to 0.85 18 per second. So use this data and I considered it will be recognized that is angular momentum of the middle particle about point or will be organized. That is it will be that is I considered yes that is mm times of our square. I will be there is two days where that means I consider it mass multiplied by okay to the square then omega. So omega is given that is 0.85 Ready in per second. No for the value of mass M. Of article and that is the that is given length of war. So I will get the angular momentum mhm of mhm. Middle particle about point full. Yeah that is equal to that means L. Is equal to 23 times of 10 to the power of minus three kg. Multiplied by two, multiplied by 12 times of 10 to the power of minus two. Holy square multiplied by 0.85 So simplify it. Then answer is studies one point one time. So I'll stand to the part of monastery kg meters square once again. So this is a powerful part. B now come to the next part that is part C. In which it is asking that we have to find the magnitude of angular momentum of the assembly. So I considered the my needed of angular momentum of assembly will be written as that is equal to that is L assembly will be equal to I system times of omega. So I say to me which I had already. God in party value will be that is used this value from party and omega is given in the question that is 0.85 radian per second. So the data then I will say that the magnitude of angular momentum of assembly about white or will be recognized. That is I system is that is 4.6 times of mhm 10 to the power of monastery kg meter square multiplied by omega is that is 0.85 radian per second. So no simplify eight. Then I regret the answer that is equal to 3.9 multiplied by 10 to the part of minus three kg metal square per second. So this is as well for part C. That means magnitude of angular momentum of the assembly about point or will be written as that is 3.9 times of 10 to the power of minus three kg meter square parts again. So I had saw the question completely. Thank you.

For this problem. On the topic of angular momentum, we have three particles each of mass 23 g. Fast and fast into three rods of length 12 cm and negligible mass. The legit assembly rotates about a 0.0 with an angular speed of 0.85 ratings per second. And we want to find the rotational inertia of this assembly. The magnitude of the angular momentum of the middle particle as well as as well as the magnitude of the angular momentum of this assembly. About zero. Now a particle contributes M. R squared to the rotational inertia where r is the distance from the original to the particle. So the total donation rotational inertia i is equal to um times three d. All squared plus M Times two d. All squared plus mm time's D squid which is 14 M. D. Squared. Yeah. And we know M as well as D. So this is 14 times 2.3 Times 10 to the -2 Kg times 0.12 m Squared. Which gives educational national 4.6 Times 10 to the -3 kg. Me too squared. And for part B the angular momentum of the middle particle was given by L. M. And L. M. Is its rotational inertia. I am times omega. And so this is for M. D squared for the middle of particle times omega Which is four times a mass of 2.3 times 10 to the -2 kg Time 0.12 m Squared times the angular speed of 0.85 radiance per second. So the middle particle has angular momentum 1.1 Times 10 to the -3 kg meter squared the second. Mhm. And lastly for part C, the total angular momentum. His eye times omega which is 14 and D squared, which is a total rotational inertia times omega. Again, if we put our values in, we get this to be 14 times two point three Times 10 to the -2 kg Time 0.12 m 2ared Time 0.85 radiance per second. So calculating we get the total angular momentum to be 3.9 times 10 To the -3 kg meter squared per second.


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