A long, thin rod of mass M and length L is standing straight up on a table. Its lower end rotates on a frictionless pivot. A very slight push causes the rod to fall over. As it hits the table, what are (a) the angular velocity

Video transcript

Hey, everyone. Let's go through this practice problem. A paper cutter is made of a uniform bar attached by its lower end to a horizontal base by a frictionless axle. The bar of mass M sub B and length, L sub B is initially at rest and perpendicular to the base. When the bar is barely moved from its initial position, it rotates around the axle, determine the expression of the angular velocity at the instant. The bar strikes the base. We have four multiple choice options to choose from. Option A, the square root of three G divided by L sub B. Option B, the square root of three M sub B G divided by LSA B, option C three G divided by L sub B and option D three M sub B G divided by L sub B. All right. So this problem can be kind of tricky if you're not sure what's going on. So the first step is to draw a diagram. And so, and it's in the, in the cutter's initial case, it's, it resembles a vertically positioned bar with a axle that it's rotating around right at the bottom and in its final position it's resting on the perpendicular, it's resting on the base that is perpendicular to its initial position to its final position would look like this where it's now horizontal, but around the same axis. And so when you close it, it, it rotates around that axis to end up in the closing position. The problem tells us that the mass of the bar is M sub B and the length of the bar is L sub B. Now, in order to find the expression for the angular velocity, we're going to have to find a relationship between the angular velocity and the other variables that are relevant such as the length of the bar and its mass. And one equation we have that can relate these variables is the law of conservation of energy. If we assume that this situation is an isolated system, then that means the total amount of its mechanical energy remains constant or rather the change in mechanical energy is zero. Or alternatively, if we were to write it out in a more expansive way, the initial kinetic energy plus the initial potential energy is equal to the final kinetic energy plus the final potential energy. However, one thing that's very useful for us to know that makes this pretty simple is the fact that since the bar is moving only rotationally and not translation, that means that the kinetic energy for the bar always only exists in a rotational form. So this means that we won't have to break up the kinetic energy term into tangential kinetic energy and rotational kinetic energy like we often have to do in other problems that involve rotational kinetic energy. What's even nicer about this problem is that there are already some ways we can simplify this energy equation. So first off, the problem tells us that when the bar is barely moved from its initial position, it rotates. So what we essentially have here is a case where the bar's initial kinetic energy before it starts falling is zero. It initially has no rotational kinetic energy and just gravitational potential energy before that energy is converted into kinetic energy as it begins falling. So we can cross out the case of I rotational term right off the bat because it goes to zero. Another thing to note is that since the bar is falling down to some bottom position, remember that when we're dealing with potential energy, it's up to us to define the positions that are relevant for that potential energy. So to make things simple, I am going to define the position where the cutter has reached its bottom position to be where Y is equal to zero, which means that the final potential energy can also be seen as zero. So really our energy formula can mo most simply be written as the initial potential energy is equal to the final rotational kinetic energy. Now another thing to note before we start putting in equations and doing math is that we're gonna want to have an expression for the moment of inertia of the bar. So recall that the moment of inertia of a bar as it's being rotated around an axis positioned on one end of the bar is equal to one third multiplied by the mass of the bar multiplied by the square of its length. So now that we are aware of that, we have all the tools we need to begin billing out our energy equation. So as we mentioned earlier, the initial potential energy is equal to the final rotational kinetic energy. Let's expand this out into variables in terms that we're more familiar with. So the gravitational potential energy is of course equal to the mass of the bar multiplied by G the gravitational acceleration multiplied by the vertical position of the center of mass of the object. We don't always include the center of mass term in energy problems because a lot of the times we're dealing with uh particles or compact objects where the position of the center of mass is essentially the same as anywhere else on the object. But since in this problem, we're dealing with a long bar where the center of mass is not at the same place as everything else and will follow a different path. It's important to keep in mind where the center of mass of the bar is, and this is equal to the rotational kinetic energy. The final rotational kinetic energy which is equal to one half multiplied by the moment of inertia multiplied by the square of the angular speed of the bar. All right. So now let's make some more substitutions. So M G Y sub C M, so the mass of the bar, as we discussed is M sub B, this is being multiplied by G and then multiplied by the vertical position of the center of mass of the bar. And so if you look at our diagram, the center of mass of the bar is obviously going to be right at the center of the bar, which if we consider the way the height is laid out, that's going to be at half of the bar's length. So for Y sub C M, we could just plug in the length of the bar. L sub B divided by two and this is equal to one half multiplied by the moment of inertia of the bar, which as we discussed earlier is one third multiplied by the mass of the bar multiplied by the square of the length of the bar. And this is being multiplied by the square of the bar's final angular speed. Let's do some more simplifications. Both sides of the equation include the mass of the bar and sub B. So those can be canceled out. There's also a division by two on both sides of the equation. And both sides of the equation also have at least one L sub B on the right hand side of the equation, there's an L sub B squared. So only one of those L sub BS will be canceled out. And the equation we're left with says that G is equal to one third multiplied by L sub B multiplied by the square of the angular speed. And all that's left for us to do is solve this equation algebraically or omega squared. So divide, dividing both sides of the equation by one third L sub B, we find that Omega squared is equal to three G divided by L sub B. Lastly, if we take the square root of both sides of the equation to get rid of the square, then we find that Omega on its own is equal to the square root of three G divided by L sub B. And this is our answer to the problem. If you look at our multiple choice options, this agrees with option B stating that the angular velocity is equal to the square root of three G divided by L sub B. So option A is the correct answer to this problem. And that's it for this problem. I hope this video helped you out. If you think you need more practice, please check out some of our other videos which will give you more practice and more experience with these types of problems. But that's all for now. I hope you all have a lovely day. Bye bye.