A small 10.0-g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50.0 g and is 100 cm in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 cm/s relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps?

Video transcript

Hi everyone. Let's take a look at this practice problem dealing with angular momentum in this problem. One end of a stationary rod is connected to a pivot on a smooth horizontal surface. The rod can rotate freely without friction. On the other end, a 15 g squirrel stands still given that the rod has a mass of 75 g and it's 100 and 20 centimeters long. Calculate the angular speed of the rod just after a squirrel jumps off in the horizontal direction perpendicular to rod with a speed of 25 centimeters per second relative to the surface. Assume that the rod has a uniform mass distribution. We're given four possible choices as our answers. Choice A is 0.125 radians per second. Choice B is 0.250 radiance per second point C is 1.50 radiance per second. And choice D is none of these. Since we're asked to calculate the angular speed, we're going to want to look at the uh conservation of angular momentum here recall that for angular momentum to be conserved, that means that my initial angular momentum L I has to be equal to my final angular momentum. LF Now here initially, I have no angular momentum since both the rod and the squirrel is stationary. So since nothing's moving, I have no angular momentum, so my initial I is going to be equal to zero and that's gonna be equal to my final angular momentum. Well, I have two contributions to that. I have the angular momentum of the squirrel which will label as LS and then we'll have the angular momentum of the rod which will label as LR. However, I do need to be careful because angular momentum is a vector quantity. And so if the Squire jumps off in one direction, it's gonna push back on the rod. And so it's going to move in the other direction. That means that these two angular momentum are going to be moving in the opposite direction. So one's gonna be positive and one's gonna be negative. So we'll just for convenience sake that the um angular momentum of the rod to be negative, that means that my total final angular momentum is going to be LS minus LR. Now, we need to recall our formulas for our angular momentum. And we actually have two of these, we have one for a point object, we'll have L as equal to MVR where M here is the mass of that object. V is the um speed of that object. And R is the distance from the axis of rotation to that object the other form that we're going to want to use is one dealing with our moment of inertia and the angular speed. So we'll have L is equal to I omega. So here I is the moment of inertia and omega is our angular speed. So we need to calculate the um angular momentum for both the squirrel and the rot. Let's start with a squirrel. And since it's going to be treated as a point particle, we're going to want to use that form of the angular momentum, we will have LS is equal to, will have the mass of the squirrel, which is MS multiplied by its speed, which is vs and multiplied by its distance from the axis of rotation. Well, since it's sitting at the end of the rod distance from the axis of rotation, in this case, the pivot point is just the length of the rod which will indicate as a lower case L. Now I need to calculate the angular momentum for the rod. So we'll have LR is going to be equal to. In this case, we're gonna use the one with the moment of inertia and angular speed will have IR multiplied by omega R. Now, I was told that this was a rod and it's rotating about an axis about its end point. And so I can calculate the moment of inertia for that. So I have LR is going to be equal to and recall the moment of inertia formula for a thin rod that is retained about an axis about its end point. And that is one third multiplied by its mass, which will label as Mr multiplied by its length squared, which is the lowercase L squared. And that quantity still is multiplied by omega R. So today we have our two angular momentum so we can plug that back into our angular momentum or conservation of angular momentum formula where I um plug in those values, I am going to um simplify this expression. I'm gonna move the LR to the other side of the equation. So I have LR is going to be equal to LS. And now I can plug in the expressions I had for each one of those. So for LR, I have the one third multiplied by Mr elsewhere Omega R and this has the equal R formula for the um LS which is the MS multiplied by vs multiplied by L. Now I'm wanting to calculate the angular speed. So that's gonna be my Omega R in this formula. Let's solve this equation for that. So I have the mega R can be equal to um I can cancel out when the factors of about and then I just need to divide over by the one third MRL. So doing that gives me free M SBS divided by Mr lowercase L. So I have values for everything on that right hand side, I can plug those in. We'll have a mega R is gonna be equal to three multiplied by the mass of the squirrel, which we're told is 15 g. And we really should convert that into kilograms by dividing by 1000. That will be 0.015 kg. Just so we have everything in si units. Then we're going to multiply that by the speed of the squirrel, which is the 25 centimeters per second. And so convert that into meters per second and you need to divide by 100 that'll be 0.25 meters per second. And that product gets divided by the quantity of by mass of the rod, which is the 75 g which we need to convert into kilograms again. So it's gonna be divided by 1000. That would be 0.075 kg and that gets multiplied by its length which is 100 and 20 centimeters. So I need to convert that into meters by divided by 100 that gives me 1.2 m. So I can plug these values into my calculator and I get omega R is equal to 0.125 radiance per second if you're wondering where the units of radiance comes from. So I divided the kilograms out and I divided the um meters out as well. That left me just one seconds. However, um I'm dealing with an angular quantity here. So it has this hidden units of radiance that's where the radiance actually comes from anytime you're dealing with an angular quantity, the radiance is sort of hidden in that. So now I have my final numerical answer, I can compare that to the answers that I was given for my choices. And this corresponds to answer a. So just to recap here, we started off by looking at the conservation of angular momentum. And we actually had to use two different forms for the angular momentum, one for a point object, and one for a continuous object. And once we used those two formulas and set them uh the angular momenta equal to each other, we were able to solve for the angular speed. So I hope that this has been useful and I'll see you in the next video.