What is the angular momentum vector of the 500 g rotating bar in FIGURE EX12.44? Give your answer using unit vectors.

Video transcript

Hi everyone. In this practice problem, we will have a 300 g rectangular slap rotating about its center. As shown in the figure we're being asked to calculate the angular momentum factor of the slap where the slap is 30 centimeter long and 12 centimeter wide. And we're being asked to also express the final answer using the appropriate unit factors. We have a figure given in the problem statement, you're depicting the system of the slap that we have with its dimension. The options given for the angular momentum are a 0. times 10 to the power of negative three J kilograms meter meter squared per seconds. B 3.39 times 10 to the power of negative three I kilograms meters squared per seconds. C 3.93 times 10 to the power of negative three J kilograms meter squared per seconds and D 0.36 times 10 to the power of negative three K kilograms meters squared per second. So the rectangular uh slap is going to be a rotating rigid body. So the angular velocity or omega in this case will be equal, which is given in the problem statement are essentially listed in the diagram or the figure given which is going to be nine pr pm, right. So, so Omega is going to be nine pr pm and we wanna convert that into radium per second. So we wanna take nine P and multiply that with two pi divided by 60 radium per seconds and calculating that, that will essentially give us the angular velocity Omega to then be equals to three pi radium per seconds. Next, we want to move on to calculate the moment of inertia of a rectangular slap about its center which is going to be equal to one divided by multiplied by M multiplied by L squared. So the L in this case is going to be the total length of the slap. So that will essentially be 12 centimeter just like so which we have to convert into meter which is 0.12 m. So in this case, we can substitute all our values. So I will then be equal to one divided by 12 multiplied by the mass which is going to be 300 g or in si 0.3 kg multiply that with the L squared or the length toner characteristic line which is 0.12 m squared. And then calculating that we will get the moment of inertia of the rectangular slap about its center to be equal to 0.36 times 10 to the power of negative 3 kg meter squared. Finally, we can calculate the angular momentum. And if we recall the uh formula for the angular momentum L will be equal to I multiplied by Omega. So I is 0.36 times 10 to the power of negative kg meter squared multiply that with Omega. So Omega is three pi radiance per seconds and then calculating that we will then get the angular momentum to then be equal to 3.39 times 10 to the power of negative 3 kg meter squared per seconds. The uh problem statement also ask, ask us to express the final answer using the appropriate unit factors. So all you need to do is to use the flamings right hand rule, right hand thumb rule to determine the direction of the angular momentum. So the direction of L will then be in the positive X direction. So essentially L will then be equal to 3.39 times standard power of negative three I kilograms meter squared per seconds. So the correct appropriate answer is going to be 3. times 10 to the power of negative three I kilograms meter squared per seconds, which will correspond to option B in our answer choices. So option B will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this video and thank you so much for watching. Bye bye.