The three masses shown in FIGURE EX12.15 are connected by massless, rigid rods. (b) Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page.

Video transcript

Hi, everyone. In this practice problem, we are being asked to determine the moment of inertia of a system of masses. We will have a triangle housing three different spheres or three different point masses as shown in the figure below. The spheres are actually linked using bars of negligible mass. And we're being asked to determine the moment of inertia of the system about an axis perpendicular to the page and going through mass too. The triangle given or the figure given is showing mass one and mass two as 0. kg. Mass three as 0.52 kg. Uh mass one and two is separated by 0.3 m. Mass one N three is separated by 0.25 m and mass two N three is separated by 0.25 m as well. The options given for the moment of inertia of the system of masses is a 0.86 kg meters squared B 0.3 23 kg meter squared C 0.2 35 kg meter squared and D 0.64 kg meters squared. Awesome. So we will consider the three spheres or the three masses connected by bars of negligible masses as a rigid body. So therefore, the moment of inertia is actually going to be given by I equals sigma of M I multiplied by R I squared where the sigma is representing all the masses in our system combined. And M I is the mass of our particular system. And are I as the distance for for the mass to the axis of rotation itself? So we are told to determine the moment of inertia about an axis perpendicular to the page and going through mass too, since the axis passes through mass two. So therefore, mass two does not contribute to the total moment of inertia or to the total. I, so I'm just gonna block out mass two right here so that we are not paying our attention to it because this is going to be our axis of rotation just like so awesome. So now I'm gonna represent the moment of inertia as I two because it is going through an axis going through mass two. So I two is going to be, I'm gonna first look at mass one. So I two is going to be mass one multiplied by R. The R is the distance between mass two and mass one. So I'm gonna represent that with R I R squared and we wanna plus that with M three. And we want to multiply that by our 23 or the distance between mass two and mass three squared just like. So, so I two equals M one multiplied by R 12 squared plus M three, multiplied by R +23 squared. So we know all of this information from the figure given. So now we want to just substitute off those information into this expression that we have right here for our I two. So M one is 0.35 kg multiplied that by R 12, which is 0.3 m squared. And then plus M three, which is 0.52 kg multiplied that by R 23 squared, which is 0.25 m squared. Just like so and doing this calculation, I two will then equals two point 064 kilograms meter squared just like. So so the moment of inertia for the system of masses shown in the figure in the triangle configuration shown about an axis perpendicular to the page and going through mass two is going to equals to 0. kg meter squared, which will correspond to option D in our answer choices. So answer D with the moment of inertia for our system of masses being 0.64 kg meter squared is going to be the answer to this particular practice problem. So if you guys still have any sort of confusion on this one, please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you.