(II) Two lightweight rods 24 cm in length are mounted perpendi...

Question

(II) Two lightweight rods 24 cm in length are mounted perpendicular to an axle and at 180° to each other (Fig. 11–35). At the end of each rod is a 480-g mass. The rods are spaced 42 cm apart along the axle. The axle rotates at 4.5 rad/s .

(a) What is the component of the total angular momentum along the axle?

(b) What angle does the vector angular momentum make with the axle? [Hint: Remember that the vector angular momentum must be calculated about the same point for both masses, which could be the cm.]

Accepted Answer

Welcome back everyone in this problem. In an engineering lab, a balancing wheel has 2 40 centimeter rods each with a 450 g mass at its end mounted 180 degrees apart on a central axle, the rods are spaced 60 m apart along the axle and the set of spins at four radians per second. First, we want to calculate the axial component of the total angular momentum. And second, determine the angle between the angular momentum vector and the axle with a hint that we can compute the angular momentum vectors for both masses from a consistent reference point, possibly the center of mass. For our answer choices. A says that the axial component is 0.3 kg per square meter per second. And the angle between the angular momentum vector and the axle is 25 degrees B set at 0.48 kg meters squared per 2nd and 31 degrees C 0.58 kg meters squared and 37 degrees and D 0.76 kg meters squared per 2nd and 90 degrees. Now, let's first make sense of our problem here by playing around with our diagram for a bit. OK. Now we can determine the entire angular momentum vector relative to the system's center of mass at the moment depicted in this diagram here that I've highlighted in blue halfway between their distance of 60 centimeters or 30 centimeters away from each mass. In this configuration, we can also put X and Y axis based on our axle because now we can put the Y axis through our axle, OK. A horizontal X axis through our center of mass. OK. And now in this configuration, the positive X axis extends to the right, the positive Y axis vertically along the axle and the positive Z axis emerges perpendicular to the plane of the diagram facing the observer. In this problem. As we said, in the first part, we want to calculate the axial component of the total angular momentum. In other words, if we can find the total angular momentum, then the axial component would be the component along the Y axis or the J component of that vector. Now let's designate our mass at the top as mass A and the mass at the bottom as mass B. OK. And assuming the system undergoes a counterclockwise rotation when viewed from above the rod, then the velocity of mass A points in the positive Z direction. In other words, VA is positive while the velocity of mass B points in the negative Z direction. In other words, VB is negative, I know with that idea. In mind, I should uh yes, let me, I should isolate this here. OK. And with that in mind, now, we can figure out our total angular momentum because recall, recall that our total angular momentum will be equal to the angular momentum of A plus the angular momentum of B where angular momentum is the cross product of the position vector and our linear momentum. In other words, we can write this then OK, as R A crossed with P A plus RB crossed with PB and we can incorporate our velocity here even further to say, since our linear momentum is equal to our mass multiplied via velocity, then we can rewrite this as the mass multiplied by the cross product of R and P plus the cross product of RB and P, sorry, not P A va my apologies. OK. R A and VA plus the cross product of RB and VB. So now we're getting somewhere. But what do we know about that speed? V? Well, we also know that V is equal to the angular momentum omega multiplied by R. And we know that our angular momentum is four radiance per second. So when we multiply that by 0.4 m, then we get our velocity to be equal to 1.6 m per second. So now we can find a cross product for A and B for our angular momentum and then eventually find the J component. Let's go ahead and do that. So applying what we have here, OK. Let me bring it down here. That means our total angular momentum is going to be equal to a mass multiplied by the cross product. OK? Of A, OK. And A R A and VA so that's 0.4 m. OK. 0.3 m in the Y direction or in the J direction and 04 K. OK. And for our velocity, remember earlier, we said the velocity of mass A points in the positive Z direction. So in other words, it's going to be 00 V. Then we're adding that cross product to the cross product of B OK. Where the cross product of B notice here be from our point of reference is 0.4 m to the right and 0.3 m downward. So again, we're gonna have IJK for 0.4 m negative 0.3 m zero. And now the velocity of mass B points in the negative Z direction. So that's 00, negative V. Now we can find a cross product for both of our terms and simplify. Now, remember uh here this is gonna be positive. Our result here is going to be positive for I negative for J. And uh let me fix that here positive for K same thing for our other cross product and not to find our cross product. I'm just going to do one in the interest of time. But remember we block out those that we have marked here and multiply what we have left to figure out the cross product for I. Ok. So that's going to be 0.3 m multiplied by V minus zero, multiplied by zero, which would just give us 0.3 V. Ok. So now when we start to multiply, we're gonna have 0.30 m multiplied by V I and now we're doing it for a minus. OK? Let me take again, let me take these out the cross product here. We're gonna have negative 0.4 m multiplied by V. Let's minus negative 0.4 m V. And that's for J and then for K, OK, they were gonna block all these parts here, the row and column for K. So it's gonna be negative 0.4 multiplied by zero plus 0.3 multiplied by zero with a positive result. And that would just be equal to zero. So this is the cross product of a likewise, if we do the same thing for B, I'll give you a minute to try to do that one on your own. But when you do it, you'll find that you get negative 0.3 m multiplied by negative V minus 0.4 m multiplied by negative V. And this component will be J and our first one would have been I, OK. Now we can go ahead and uh simplify, OK? Because here when you go ahead and simplify our components. Com combine the I components and the J components. You should find that you get two MV multiplied by 0.30 m I plus 0.40 m J let's substitute because, because we know our velocity and we know our mass, so that's going to be two multiplied by 0.45 kg, multiplied by 1.6 m per second, multiplied by the vector 0.3 M plus 0.4 MJ. Now when we multiply through, we finally find our total angular momentum to be 0.432 I OK plus 0.576 J OK, kilogram meters squared per second. So now we have our total angular momentum. Remember earlier, we said we want the axial component of our total total angular momentum and that axial component means the part that's along the y axis because that's where our axle is on. So that means we want the J component and thus the axial component. OK. The axial component, the J component is going to be to two significant figures, 0.58 kg square meters per second. OK. So that's our axial component. We've solved the first part of the question. Next, we need to figure out we need to figure out the angle between the angular momentum vector and the axle. Let's take a step back to visualize exactly what that means. So now we have our total angular momentum. OK. And if we look at it relative to our axle from our uh system that we had set up, OK? For our Y, let me put that in black here for our Y and X axis. OK. Then no, it's 0.432 I plus 0.576 J. It would look something like this. This is our angular momentum, our total angular momentum L and we want the angle between the angular momentum vector and the axle which would be our angle here. Our angle theta. OK. No, for this angle remember that for L it has a vertical component and a horizontal component which rather let me write it here, a vertical component and a horizontal component where our vertical component is adjacent to theta or horizontal component is opposite to theta. And thus we can use the tangent function to help us figure out theta because by that idea, this means the tangent of theta is going to be equal to the X component divided by the Y component. And thus theater is going to be equal to the inverse tangent of that ratio. We know our X component is 0.432 mm. And we know our Y component is 0.576. So now when we put this into a tangent function and solve, yeah, then we get our anger theta to be equal to 36.86 degrees or to two significant figures 37 degrees. Thus, the angle between the angular momentum vector and the axle is 37 degrees and the axial component of the total angular momentum is 0.58 kg square meters per second. If we look back at our answer choices that tells us, then that c is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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