### Video Transcript

The object shown in the diagram has a uniform density. Which of the points š“, šµ, š¶, and š· is closest to the center of mass of the object?

So weāre trying to find the center of mass of the object. Letās start by underlining the important bits of the question so that we donāt miss anything. Now, weāve been told that the object in the diagram has a uniform density. This is quite important, and weāll come back to that in a second. What we are asked to do is to say which one of the points š“, šµ, š¶, and š· is closest to the center of mass.

Letās start by quickly discussing the uniform density part. This statement serves to make our life much easier when weāre trying to answer this question. Thatās because if the object has a uniform density, then the mass of the object is spread evenly across it. For example, letās say we have this rectangle. It also has a uniform density. That means that mass is spread evenly across this rectangle. And therefore its center of mass is just the geometrical center of the object. And that happens to be roughly here.

Now letās say we have another block, another rectangle, that looks the same, but the green half of the rectangle is more dense than the pink half. Well this means that thereās more mass in the green half of the rectangle. Therefore, the center of mass is no longer here. But it shifts very slightly to the left, to roughly about here. How much it shifts depends on the difference in density between the two halves of the block?

This is because more mass is distributed in the green half. And therefore the center of mass must shift to that side. Whatās the reasoning behind this? Well the center of mass is the point at which an objectās weight acts. In other words, if we were to balance the object on a fulcrum, then the most stable place to balance it is directly underneath the center of mass.

If we go back to our first rectangle, then this is the best place to balance it. However, if we look at our second rectangle, itās no longer in the center but rather here. This has to do with moments because more of the mass of the object is distributed on the left-hand side. Therefore, there will be a larger force on the left-hand side, again because of the larger mass, than there will be on the right-hand side.

So in order to balance this out, we need to shift the fulcrum to the left. However, we donāt need to worry about that in this question because weāve been given a shape with uniform density. Therefore, the center of mass of the object is basically the geometrical center. So what we can do is to find the geometrical center of the object in both this direction and this direction. The point at which they intersect will be our center of mass.

Now letās start with the geometrical center in the left-to-right direction. Well thatās quite simple because the shape is symmetrical about a central axis. In other words, the right-hand side of the object is a mirror image of the left-hand side. And so the center of mass must lie on this orange dotted line. Now this is expected because all of the possible solutions š“, šµ, š¶, and š· do lie on this line already.

So letās swiftly move on to finding the vertical center of geometry. We can start by imagining that this half block here is not part of the object. If it didnāt exist, then the geometrical center of the object is pretty simple to find. The center of the object is here. Now we can think about adding that half block back in. Weāre adding some extra mass on top here. So we can think about what happens to the center of mass when we add this half block.

Well, letās not imagine that weāve got a square and then a half block on top of it in the middle of the object. Instead letās imagine this to be one object in itself. We can ignore the three blocks to the left and three blocks to the right because theyāre symmetrical to each other and they do nothing to move the center of mass of the object up or down. Therefore, letās just consider the pink block.

Whatās the center of mass of the pink block assuming itās a uniform density block, which it is? Well itās just the geometrical center of the object, which is about here. Therefore the center of mass has shifted upwards to about here. And in our diagram, the nearest point to the center of mass is šµ. Thatās this point here. Therefore, our final answer is that point šµ is nearest to the center of mass of the object.