The four masses shown in FIGURE EX12.13 are connected by massless, rigid rods. (b) Find the moment of inertia about a diagonal axis that passes through masses B and D.

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Hey, everyone in this problem, we have a rectangle that houses four masses. And we're show that in the diagram, the masses are linked to using rods of negligible mass. And we're asked to calculate the system's moment of inertia about an axis diagonal through masses one and three. Let's look at our diagram first. And then we're gonna go through the answer traces that we have. So we have in the top right of our rectangle maps one, it has a mass of 0.5 kg in the bottom, right, we have two with a mass of 0.75 kg. The bottom left, we have a mass three with a mass of 0.5 kg. And the top left number four with a mass of 0.75 kg. The width of a rectangle is nine centimeters and the height is 12 centimeter. And we're told that we're calculating the moment of inertia about this axis, this diagonal axis through mass one and three. And so we can draw that on our diagram and we're gonna be rotating about that axis. Now we have four answer choices in this problem all in kilogram meters squared. Option A 0.0078. Option B 0.0052. Option C 0.0044 and option D 0.108. Now let's recall for the moment of inertia eye, this is gonna be equal to the sum of the map multiplied of the distance squared of all of our masses. So in this case, it's gonna be the sum from I equals 1 to 4 of mir I squared. If we write this out, OK. For each term, we can write that I is equal to M one, R one word plus M two, R two squared plus M three, R three squared plus M four air force one, no R one and R three. These are the distances of mass one in three respectively from the axis of rotation. Now mass one and three lie on the axis of rotation. So R one is gonna be equal to R three, which is just gonna be equal to 0 m. OK? So the first term and the third term in our expression are going to go to zero. So now we need to find R two and R four. If we go to our diagram and I'm gonna draw it in blue just so that we can see the difference here on our diagram can see that the distance from mass four or from mass too, we can draw on our diagram and it's gonna go from those masses to a point where we have this perpendicular line to our axis. And that's gonna be the shortest. Just now on the top, we have this width of a rectangle is nine centimeters. Then we have our R four value and our R two value. Now, let's start with R two. We have one side length, this nine centimeter length. We don't have another sideline and we don't have either of the angles other than the right angle. OK. So what we need to do is either figure out one of these sides or figure out an angle in order to calculate R two. And what we're gonna do is calculate the angle theta, OK in the left corner between the diagonal at the bottom of a rectum. And we can calculate that using triangle 123. So we're gonna start by finding data now between triangle 12 and three. OK. This is a right angle triangle. We know the opposite and adjacent sides. So we're gonna use tangent to find data. Now recall that tan the is gonna be equal to the opposite side divided by the adjacent side. Now, the opposite side, that's the length of our triangle 12 centimeters, the adjacent side, that's the width nine centimeters. And so we end up with tangent of data is equal to 12 centimeters divided by nine centimeters. And so beta is going to be equal to the inverse tangent of four and we can simplify 12 divided by nine into four thirds. And if we work this out, we're gonna get a beta value of approximately 53.13 degrees. All right. So now that we have data, let me go back to our diagram. We have the, we found it using triangle 123. OK. This big triangle that takes up half of our rec. Now we can focus on our little triangle where R two is right. So now we're gonna find R two and we're gonna do it in a very similar way. OK. We have an angle, we know the hypotenuse and we're looking for the opposite side. So let's use sine of theta because it's gonna relate those sites that we are interested in. I recall that sign of theta is gonna be equal to the opposite side divided by the hypotenuse. So in this case, we have sign of the angle which we know now 53. degrees is equal to R two, the opposite side divided by the hypotenuse, which is the width of our rectangle 97. We can solve for R two by multiplying both sides by nine centimeters. We get that R two is equal to nine centimeters multiplied by sine of 53.13 degrees. We can work this out and we get that R two is equal to 7.27 m. Now, if we look at our diagram, we can do the exact same thing for R four. But what you'll notice is that this diagram is completely symmetrical about our rotational axis. And so R two is going to be equal to R four. OK. We can find the angle using triangle 143. It's gonna be the exact same which is gonna give us the same sideline. So R two is equal to R four. Now we have those values, we can get back to our moment of inertia calculation. So we have that R four is also gonna be equal to R two, which we found to be 7.2 centimeters. OK? So let's write that out. So we don't forget. And now back to our moment of inertia. OK. We've already said that the 1st and 3rd terms go to zero for a moment of inertia. Let me write it all up. So we don't get confused. M one, R one squared plus M two, R two squared plus M three R three squared plus M four, R four squared R one and R three are zero. So those terms go to zero, we're left with 0.75 kg multiplied by 7.2 centimeters. We're gonna convert to meters by divided by 100. So this is gonna be 0.072 m squared. OK? And we do that so that we can match our answer choices. The answer choices are in kilogram meters squared. So we want to do that conversion Now we're gonna add 0.75 kg multiplied by 0.072 m all squared again. OK. So the mass of two and four was 0.75 kg. And then we have our R two and R four squared respectively. And if we work all of this out, we get that our moment of inertia eye is 0. kilogram meters squared. All right. Mhm. Comparing this to our answer choices in a round date, we can see that the correct answer is going to be option A 0.0078 kg meters squared. Thanks everyone for watching. I hope this video helped see you in the next one.

The four masses shown in FIGURE EX12.13 are connected by massless, rigid rods. (b) Find the moment of inertia about a diagonal axis that passes through masses B and D.

Verified Solution

Key Concepts

Moment of Inertia

Parallel Axis Theorem

Rigid Body Dynamics