Determine the moment of inertia about the axis of the object shown in FIGURE P12.52.

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Hey, everyone in this problem, we will ask, what is the moment of inertia of a thin uniform rod with two point masses attached to it rotating about the axis as shown in the figure. OK. So we're shown in this figure, we have our thin rod with a mass of Mr, it's rotating a boat in axis perpendicular to and passing through the center of mass of the rod. OK. We have one point mass with a mass M that is a distance of the length of the rod or OK. And then on the other end, we have this point mass with a mass of three M that is a distance of 1/6 of the length of the rod away from the center of the mass. OK. So we have four answer choices here A through D and all of them are just different expressions. A combining the length of the rod, the mass of the rod, the mass of our point masses into one expression for this moment of inertia. And we're gonna come back to those once we're done working through this problem. So let's get started. Thanks in the moment of inertia. I let's recall what this is gonna be. We have three objects in this system. So we're gonna have the moment of inertia of the rod itself. Then we're gonna have the moment of inertia of point mass one. OK. So we're gonna call that M one and we're gonna have the moment of inertia of point mass two. So M two, we're gonna have to add all three of those together. Alrighty, let's start with the moment of inertia of the rod itself. And we have this thin rod rotating about this axis perpendicular to itself through, through the center of mass. And so the equation that's gonna govern that. Yeah, but the moment of inertia of the rod is going to be 1/ multiplied by the mass of the rod multiplied by the length of the rod squared. OK. And that is an expression and equation that you can look up on a table in your textbook or that your professor has provided. OK. That moment of inertia for this particular object. All right. So if we fill this in with the information we were given in the problem, then the moment of inertia or the rod is gonna be 1 multiplied by an R multiplied by LR squared. All right. So we have that now let's move to point mass one. And we're gonna say that point mass one is the point mass at the top. OK. So this is gonna be M one and the point mass at the bottom is going to be mass and two. OK. So M one is the smaller point mass, all righty. So the moment of inertia for a point mass recall is gonna be the mass M multiplied by the radius squared. And when I say radius, what I really mean here is the distance from that point mass to the axis of rotation. So in this case, the mass is gonna be the mass of this point max, which we're told is capital M and the distance from the point mass to this axis of rotation, we're told in the diagram is one quarter of LR, right. So we have LR divided by four all squared and we can simplify this. This is gonna be M multiplied by LR squared divided by 16. OK. So we've done one point mass, we're onto the second point mass and we're using the exact same equation, OK. It's a point mass. So we are looking at the mass multiplied by the distance squared. So we have M two R two squared. Now our mm two, we're told is three M that's gonna be multiplied by the distance to the axis of rotation, which we can see in our diagram is LR divided by six again squared. Now, if we simplify this one, what we have is three M multiplied by LR squared divided by 36. And we're gonna leave it like that for now. And we're gonna put all of these values together into our one equation for the moment of inertia. OK. So recall the moment of inertia is gonna be the sum of the three moment of inertia from these objects. So we have the moment of inertia. I is going to be one 12 Mr LR squared. What Mlr squared divided by plus three Mlr squared divided by 36. Mhm. So we want to simplify this expression and we want to simplify it in a way that matches one of these answer choices. OK. So we can see we have this common factor of LR squared that we can pull out of our equation. And then we're looking to pull out of a factor of maybe LR divided by six or four or three, something like that. So let's look at our largest factor that we can pull out and we're gonna do a factor of a quarter. OK. So we have the moment of inertia, I is gonna be equal to one quarter, LR squared when we factor one quarter, LR squared. And what we're left with is Mr divided by three plus M divided by four plus three M divided by not now we're getting closer to this final answer, but we can see we have these two M terms, OK? One of them is an Mr, so we can't really simplify that or combine any like terms there, but we have these two terms that both have M in them. So let's try to combine those like terms, OK? And first thing we're gonna do is reduce this second fraction, OK? We have three divided by nine. So we can write that as one third. So we have a quarter, LR squared multiplied by Mr divided by three plus M divided by four plus M divided by three. Now, we want to find a common denominator right now, we have a denominator of four and a denominator of three. So the common denominator is going to be 12. Our equation becomes one quarter. LR squared multiplied by Mr divided by three plus, we have three M less for um all divided by 12. OK. We multiplied the first fraction by three and the second fraction by four to get that denominator. And one final step, we can simplify in the numerator. We get that the moment of inertia I is one quarter multiplied by LR squared multiplied by Mr divided by three plus seven M divided by 12. And that is the final expression for the moment of inertia of this rod with those two point masses. If we compare this to our answer traces, OK? We can see that answer choice A and answer choice B only have LR terms, not LR squared terms. OK? So those are out right off the bat because of that. Then we have option C or D option C is the expression we found we have LR squared divided by four multiplied by Mr divided by three plus seven M divided by 12. And so that is gonna be the correct answer. A option D does not have the correct coefficients on each of those terms. Thanks everyone for watching. I hope this video helped see you in the next one.

Determine the moment of inertia about the axis of the object shown in FIGURE P12.52.

Verified Solution

Key Concepts

Moment of Inertia

Axis of Rotation

Parallel Axis Theorem