Download meet me plus you equals

download meet me plus you equals

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download meet me plus you equals

M one is gonna contribute m one times its distance from the axis squared would be a, so we do a squared, and let's say b is just the length of this string, so b just represents that length, and similarly c represents that length, and we're gonna assume the radii of these masses are small. I had to draw 'em big so we could see 'em, but it's easiest if you consider them to be small, 'cause then we don't have to take into account their actual radius. So we'd add to this, that's this m one a squared is just the contribution to the moment of inertia that's being contributed by just m one, so we have to figure out the contributions from each of these other masses, so we'll have m two times its distance from the axis.

It isn't gonna be b, it's gonna be all the way, so that's gonna be a plus b squared, and then if you wanted to find the contribution from m three so that you'd get the total, you'd have m three times, well, it'd be a plus b plus c squared, this would be the total moment of inertia for the entire system, which says it's gonna be more difficult, right? The more mass you add into the system, the more sluggish it is to acceleration, the more difficult it is to rotate.

More on moment of inertia

So how could we make this three mass system easier to rotate? Let's say you were tired of requiring so much torque to move this thing, you wanna make it easier to rotate. One thing you can always do is just take your masses and move them toward the axis, i. If you do that notice all of these rs are gonna get smaller, if you reduce the r you're gonna get less moment of inertia, and that object's gonna be easier to rotate, easier to angularly accelerate, you can whip this thing around easier if the mass is more toward the axis.

So this makes sense, think about a baseball bat. If you had a baseball bat, so if you got this baseball bat, this is not the best drawing of a baseball bat, but you've got a baseball bat. If you swing it from this end where this is the axis, it's hard to rotate, right? You've got all this heavy mass over here at the end, but if you swing it instead where this is the axis, if you just turn it around and swing it from this end, where this is the axis, now you've made it so most of the mass is near the axis, and if you do that, the radius of that mass is gonna be smaller, and if the radius is smaller it's gonna contribute less to the moment of inertia, less to the rotational inertia, it's gonna be easier to swing.

So you can swing a baseball bat really easy if you hold it by the fat end, compared to the actual end you're supposed to hold, you can swing this faster. It's probably not a good idea, you've probably not gonna hit the ball very far, but you'll be able to swing it much faster 'cause that moment of inertia's gonna be smaller.

And then the other thing we could do, we could always just reduce the masses. If you can make the mass less you reduce the moment of inertia, and if you can move those masses toward the axis, you reduce the r, you reduce the moment of inertia or the rotational inertia. But what if you don't have point masses at all?

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I mean, we don't always have situations where the thing that's rotating are a bunch of point masses, what if you had something more like this, where it was like a rod that had its mass evenly distributed throughout the entire rod, and it rotated in a circle. I mean, we couldn't use this formula now because this assumes that all the mass is rotating at some radius, r, but for this rod, only the mass at the end of the rod is rotating at the full length of the rod.

The mass that's closer to the axis is gonna have a smaller radius, it'll only be rotating at part of the length. This would only have a radius of L over two, and this part right here would only have a radius of maybe, L over eight.

So how do we figure this out? We can't just say the total mass of this rod, if this rod has a total mass m, and a total length L, we cannot say that the moment of inertia of this rod about its end is gonna be mL squared, that's just a lie. This total mass is not rotating all at a radius of length L, only the little piece at the end is rotating with a radius of length L. The rest of this mass is having its contribution to the rotational inertia diminished by the fact that these masses are getting closer and closer to the axis, so what do we do?

Well we can't use this, let's get rid of this. The truth is you have to use calculus to derive the formula for these continuous objects, and it's fun. You can do integrals and you can solve for these moments of inertia, that's one of my favorite calculations to do, it's kinda like a puzzle.

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You can solve for the moments of inertia, but if you don't know calculus, that would just look like witchcraft to you, so I suggest you learn calculus and try it, 'cause it's really fun, but I'm just gonna give you the result.

It turns out the moment of inertia for this rod is gonna be, and without knowing the exact answer, we should be able to say, is it gonna be bigger than, less than or equal to mL squared. We should be able to say, it's gotta be less than mL squared, it's not going to be mL squared, it's gonna be less than this because mL squared would be if all of the mass were at the full length of the rod for their radius.

Then you would put mL squared. If you could melt this rod down into just a ball, and put that ball at the very far end, you'd be maximizing its rotational inertia, 'cause you'd put all of the mass with the same largest radius r, but some of this mass is in here.

Some of this mass is only at L over two, or L over four, or at L over eight. So those little pieces of mass are having their rotational inertia contribution diminished, so we're gonna have less than Ml squared. So this is for a rod with the axis at the end of the rod. So that's the moment of inertia for a rod rotating about an axis that's at one of the ends of the rod, but what if we move this axis to the center?

What if we move the axis here so that this whole rod rotates around a point in its center.

Le Chatelier's principle (video) | Khan Academy

Do you think the moment of inertia of this rod that's the same mass and length that it was, we're just rotating it about the center, do you think this moment of inertia is gonna be bigger than, smaller than or equal to what the moment of inertia was for a rod rotated about the end.

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