A long, thin rod of mass M and length L is standing straight u...

Question

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 the speed of the tip of the rod?

Accepted Answer

Hey, everyone. Let's go through this practice problem. A rigid uniform bar is attached by its lowest end to a frictionless horizontal hinge. The bar is initially perpendicular to the horizontal plane. The bar of length one m and mass 10.25 kg is released from rest and rotates in a vertical plane. Calculate the speed of the highest end of the bar. When it reaches the horizontal plane, we have four options to choose from. Option A 3.3 m per second. Option B 5.4 m per second, option C 9.4 m per second and option D 29.4 m per second. OK. So this problem can be a little tricky to understand if you can't quite visualize what's going on. So I'm gonna draw a diagram explaining that. So we have an upright bar attached by its lowest end to a frictional horizon, frictionless horizontal hinge. So I'll kind of draw this bar as a vertically positioned thing with its hinge down here and it's oriented along a vertical axis, let's say the Y axis. So when it's released from rest and it starts falling, it's going to fall onto the horizontal axis. So I'm going to draw horizontal axis, just call this the X axis. And when this falls after it falls, it'll be horizontally positioned, kind of like this. So it rotates and then kind of falls around into a circular path until it lands where it should be. It's kind of AAA crude sketch, you should get the idea from this. And they'll say that the mass of the bar is M sub B and the full length of the bar is L sub B. And the problem asks us to find the speed of the highest end of the bar. So the little dot at the top of the bar, we want its tangential speed and we can find the tangential speed by relating it to variables involving the way the bar rotates as it falls. So we want to connect the tangential motion of this point to the rotational motion of the bar. And so the important equation, the key equation to remember with that is that the tangential speed is equal to a radial position to a radial distance from the axis of rotation. In this case, the length of the bar multiplied by the rotational speed. So R omega of course, though, in order for that to work, we first need to find the rotational speed and we can find that using the law of conservation of energy, if we assume that this bar is an isolated system, then the law of conservation of energy applies where the total amount of energy does not change. In other words, as the rod falls, its gravitational potential energy is converted into rotational kinetic energy. But the total amount of mechanical energy stays the same a more detailed way of writing. This would be to say that the initial kinetic energy plus the initial potential energy is equal to the final kinetic energy plus the final potential energy. And, and, and one very useful thing to recognize is that because the bar is rotating and not moving in a translational way, the kinetic energy is entirely consisting of rotational kinetic energy, which means we won't have to bother with splitting up the kinetic energy into two different terms as we sometimes do in other problems that involve rotational energy. And what's even nicer is that there are already some ways we can simplify this energy equation. First consider the fact that the problem tells us the bar is released from rest before it begins rotating. This means that its initial energy is completely potential and it only begins gaining kinetic energy as it falls. So its initial rotational kinetic energy is zero. Whenever it says something is released from rest, we can assume that the initial kinetic energy is zero. Additionally, since the bar falls down to the bottom position, and since potential energy is largely based on how we define the positions, I am going to define the this horizontal axis to be where Y is equal to zero, which means its height, its final height is zero, which by consequence means that its final potential energy is zero. So U sub F also goes to zero. So we've already simplified the energy equation down to the initial potential energy being equal to the final rotational kinetic energy. And lastly one final key piece of information that's going to be important to remember is that the moment of inertia of a bar rotating from one side is equal to one third multiplied by the mass of the bar multiplied by the square of the bar's length. And now we have everything we need to get started on solving this problem. So as we discussed earlier, the initial gravitational potential energy is equal to the final rotational kinetic energy. So let's expand these out into the formulas we're more used to. So gravitational potential energy is as usual equal to the mass of the object multiplied by the gravitational acceleration multiplied by the y position, the vertical position of the object's center of mass. We don't always consider the center of mass and a lot of problems where we're looking at particles and, and and spherical objects or compact objects. But since we have a bar in this case, a long object where the center of mass is quite far away at some distance away from the point we're actually trying to solve for the motion of it's important to make that distinction here. And then for the rotational kinetic energy that's equal to one half multiplied by the moment of inertia multiplied by the square of the rotational velocity. So now let's expand this further by plugging in some things that we have from the problem. So this is M sub B, the mass of the bar multiplied by G. And then the height of the center of mass of the bar is just going to be half of the length of the bar since the center of mass is going to be right in the middle of the bar, which is half of its length. So Y sub C M is just equal to 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 M sub B multiplied by L sub B squared. And this is being multiplied by the square of the angular velocity. And right off the bat, there's a good amount of simplification, we can do both sides of the equation have this M sub B. So they can cancel out both sides of the equation also have a division by two and they also have at least one L. So just one of the L's will cancel out what we're left with is that G is equal to one third L sub B multiplied by the square of the angular velocity. Let's algebraically solve this equation for the angular speed. So that we can put it into the initial equation. We talked about the very first one he mentioned for the V to find out what the tangential speed is. So first off, let's algebraically solve this equation for omega squared by multiplying both sides of the equation by three to get rid of the fraction and dividing both sides of the equation by L sub of B. And so we find that Omega squared is equal to three G divided by L sub B and then taking the square root of both sides to cancel up the square. We find that Omega is equal to the square root of three G divided by L sub B. And this is the final equation for the final angular speed of the bar. So let's put in numbers 9.81 m per second squared or the gravitational acceleration as usual. And then L sub B, well, we're told the length of the bar is one m. So if we put this into a calculator, then we find an angular speed of about 5. radium per second. All right. So now that we've found Omega, let's put that into the equation we talked about at the beginning of the problem that the tangential speed, which is what we want to find is equal to the radial distance to that point multiplied by the angular velocity. Well, the radial distance, as we can see from the graph is just the distance from the axis of rotation in the one side, the lower side of the bar all the way to the point we're looking at which is the very top of the bar. So our R value is just the full length of the bar. So one m multiplied by the angular speed, we just found 5.4 radiance per second, which is of course, just equal to 5.4 m per second. So 5.4 m per second is the final speed of the tip of the bar. And if you look at our multiple choice options, this agrees with option B which similarly states 5.4 m per second. So that's the final answer to this problem. I hope this video helped you out. If you need more practice, please consider checking out some of our other videos which will give you more experience with these types of problems. That's all for now. I hope you all have a lovely day. Bye bye.

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 the speed of the tip of the rod?

Key Concepts

Rotational Motion

Conservation of Energy

Kinematics of Rigid Bodies

Watch next