A 2.00-kg rock has a horizontal velocity of magnitude 12.0 ms when it is at point P in Fig. E10.35.
(a) At this instant, what are the magnitude and direction of its angular momentum relative to point O?

Question

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35.

Diagram showing a rock's motion with velocity and distances labeled for angular momentum calculation.

(a) At this instant, what are the magnitude and direction of its angular momentum relative to point O?

Accepted Answer

Hey everyone. Welcome back in this problem. We have a 1.5 kg stone launched horizontally with a velocity of 14 m per second at the top of a 10 m high cliff. At point A, A recording camera is located on the ground below at a 0.15 m from the base of the cliff at Point B and were asked to determine the magnitude and direction of the stones angular momentum relative to the camera at the moment it is launched. So we have our diagram here on the left. We have this cliff at the top of the cliff on the edge, we have 0.8, we have the stone being launched 14 m per second horizontally to the left. This cliff is 10 m tall And then on the ground 15 m away from the cliff, we have the camera at Point B. So let's just draw this out in terms of a position. So let's say we put the camera point B at the origin, then point A be located in the first quadrant. I had an x value of 15 m. I know why value of 10 m. Andy Stone is being launched from point a to the left. Alright, let's take up and to the right to be positive, then the velocity of the stone V is going to be negative 14 m per second because it's acting in the opposite direction as a positive direction. Alright, now we're asked to find the angular momentum relative to the camera. So we put our camera at the origin and we want to find the angular momentum recalled that the angular momentum L can be given as our cross with P where R is the position vector of the stone with respect to the camera and P is a linear momentum. Okay. Now, the linear momentum we know is equal to MV. And so we can write the angular momentum L as R crossed with um V. Now M is just a constant and so we can just multiply it in this cross product. Alright. So we need to start with our okay. What is the position vector of the stone with respect to the camera? Well, we've drawn it out like this with the camera at the origin that allows us to write that position vector R easily. And the position vector R it's going to be this vector that joins B to A And it is going to be the vector with x component 15 m And y component 10 m both positive because this is pointing to the right and upwards or positive direction. What the velocity vector B? Well in the x component, it's pointing to the left which we said is a negative direction. So the x component of the velocity is going to be negative 14 m/s. And this is a horizontal launch, okay. It's launched perfectly horizontal, which means all of that speed, all of that velocity is in the X direction, none of it in the Y direction. And so the Y component of the velocity is going to be zero m per second. So we have our, our vector, we have RV vector and we know that the mass M of the stone is 1.5 kg. So we have everything we need to calculate this vector L okay the angular momentum vector. So angular momentum vector L is going to be equal to R which is 15 m common 10 m to us. The mass oops not plus sorry cross product with the mass 1.5 kg times the velocity vector V negative 14 m per second, comma zero m per second. We can distribute this 1.5 kg to each of these terms because it's just a constant being multiplied out front of a vector. So we have L is equal to 15 m, comma 10 m cross product with The vector negative 21 kilogram meter per second, comma zero kilogram meter per second. Now we do the cross product, okay. People do this in different ways, whichever way you prefer to do the cross product or whichever way you prefer your or your professor prefers that you do. The cross product is fine. We're gonna go ahead and do it like this. We're gonna take the determinant of the matrix three by three matrix made up by the following. The first row is gonna be I hat J hat and K hat indicating the three directions the unit vectors. The second row is going to be made up of the r vector. Okay. So the I component is 15, The J component is 10, the K component is zero and the last row is going to be made up of the velocity vector. So the I component is negative 21 the J component is zero and the K component is also zero. Now we can take the determinant using cramer's rule and we're going to expand about this last column because we have these two zeros here. Um So that's going to make it easier, that's going to simplify our calculations. And so we have the first element in the last row which is sorry, the last column, which is K hat. And we're gonna multiply that by the two by two matrix determinant made up of the values if we remove the row and column that Kay had is in okay. So we take out the first row, we take out the last column and we're left with the matrix with the first row 15, 12th row negative 21 0. And we're gonna take the determinant of that two by two matrix. Then we're gonna add, we're gonna move down in our column. So we're sticking with column three. And now we're on the second element which is zero multiplied by the determinant of the matrix made by removing the row and column of that element. So we have zero times I had J hat in the first row negative 21 0 in the second row. And then we are going to do and that should be a negative, not positive. And then we're gonna do the same for that last column. We have zero. And we're gonna multiply by the matrix determinant if we remove that Roman column. So we have I hat J hat 15 10. Now we've moved through these last two pieces fairly quickly. The reason for that is because they both have these terms zero in front. Okay. So both of these are just going to go to zero. We don't need to worry about the determinants. They're the only one we have to worry about is the first one. We have K hat times the determinant of the matrix with first row 15, 10 and second round negative 21 0. Now to take the determinant of a two by two matrix, we're going to multiply the diagonal starting with the first element In the first column, first row. So we have 15, we multiply it the diagonal which is zero. And then we're going to subtract the multiplication of the opposite diagonal -21 times 10. This is going to give us a value of the first term goes to zero, negative 21 times 10 is negative 210 minus negative 210 gives us a value of positive times K hat. And so our momentum vector L angular momentum vector L is going to be 210. And our unit kilogram meter squared per second. How did we get that unit? We have meters and then we have multiplying by kilogram meters per second. So this is our momentum vector L This tells us that the magnitude is going to be 210. We have nothing in the I component, nothing in the J component, 210 in the K component. So 210 is the magnitude. And what about the direction? Okay. This is positive this K hat component. So we imagine that it's going to be out of the page, but we can use our right hand rule to check, let's go back up to our diagram if we point our fingers in the direction of our and curl them in the direction of V R thumb points out of the page. Okay, just as we expected. And so if we look at our answer choices, we found that the magnitude of the angular momentum was 210 kg meters squared per second, the direction is out of the page. And so we have answer choice f thanks everyone for watching. I hope this video helped see you in the next one.

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35.
(a) At this instant, what are the magnitude and direction of its angular momentum relative to point O?

Verified Solution

Key Concepts

Angular Momentum

Linear Momentum

Position Vector