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Numerade Educator

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Problem 20 Hard Difficulty

A 5.00-kg particle starts from the origin at time zero. Its velocity as a function of time is given by
$$
\overrightarrow{\mathbf{v}}=\left(6 \mathrm{m} / \mathrm{s}^{3}\right) t^{2} \hat{\mathbf{i}}+\left(2 \mathrm{m} / \mathrm{s}^{2}\right) t \hat{\mathbf{j}}
$$
(a) Find its position as a function of time. (b) Describe its motion qualitatively. (c) Find its acceleration as a function of time. (d) Find the net force exerted on the particle as a function of time. (e) Find the net torque about the origin exerted on the particle as a function of time. (f) Find the angular momentum of the particle as a function of time. (g) Find the kinetic energy of the particle as a function of time. (h) Find the power injected into the particle as a function of time.

Answer

a) $2 t^{3} \hat{\mathbf{i}}+t^{2} \hat{\mathbf{j}}$
b) If $\vec{r}$ and $\vec{v}$ are parallel to each other, then the particles travel in a straight line.
c) $(12 t \hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2}$
d) $60 t \hat{\mathbf{i}}+10 \hat{\mathbf{j}} \mathrm{N}$
e) $\left(-40 t^{3} \mathrm{N} \cdot \mathrm{m}\right) \hat{\mathbf{k}}$
f) $\left|-10 t^{4} \hat{\mathbf{k}}\right|$
g) $90\left(\mathrm{m} / \mathrm{s}^{3}\right)^{2} t^{4}+10\left(\mathrm{m} / \mathrm{s}^{2}\right)^{2} t^{2}$
h) $\left(300 t^{3} \mathrm{W}\right)+20 t \mathrm{W}$

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SR

Sihle R.

October 16, 2020

Video Transcript

Ah, question like this is testing our understanding of the mathematical relationships between different physical properties. Really More than anything else, can we compute a position given a velocity? Can we computer force a torque and angular momentum and so on and so forth? That's ultimately what this question is gonna be testing us. So when I see a question like this, I asked myself two questions. The first. What information have you given us in the second? What information are they asking me to find? Then the bulk of my work in solving a question like this will be on determining how to get from what they've given us toe what they're asking for. So let's start with that first question. What information have they given us? Well, the first thing that they give us is the mass of the particle. The first sentence there tells me that the particle has a mass of five kilograms. The second piece of information they give us is not as explicit as the first, but they tell us that the particle starts at the origin when time T equals zero. So the way I would write this down is that the position are of the particle at a time equals zero is the origin which is zero in the X so zero I hat and zero in the UAE, which would be zero j hat. If you're comfortable, you can just write it as a single zero. But when I'm dealing with vector components that have I had and J had pieces, I like to write them both down just to remind myself that this is in fact a vector quantity and so that I don't forget that it's a vector. The next piece of information that they give is the most important information that we're going to use from the bulk of the problem, which is the velocity as a function of time and again they give us to this in a vector equation. Using the I hat on, the J had notation. They've even thrown some units in there to ensure that we understand which units go with which physical properties. So for now I'm just copying the information down. We've got six meters per second Q T squared I hat and two meters per second squared times t in the Y direction or the J hat direction. So this is all the information that we've been given. The next question I ask myself is, What information do I actually need to solve? And there's a lot. But by breaking this question down into each individual step will find that even though it's asking us a lot of things, each one on its own will lead very nicely into the next or a future one. So the first question they ask us is what is the position is a function of time. The next question they ask is describe its motion qualitatively, so that's not really a calculation. We'll get to that one when we get to it. Next, they ask for acceleration as a function of time, and then they ask for net force as a function of time. You can have the net under there if you want to. It's up to you your personal preference. After Net force, they ask for a net torque, which we symbolise with a little towel, a Greek T. After torque, they ask for a angular momentum again as a function of time. Second last, ask for the kinetic energy as a function of time, which is not a vector, so I'm not gonna drug little vector air over top of it, and same with power. The last thing that they ask for. So let's start at the top with position. How do we find the position given the information we have? So the first question I need to ask myself is, What is the relationship between position and the information I'm given in this case Velocity initial position in mass being velocity, which is velocity is equal to the derivative of position with respect to time. Now, that would be great if we had position and we're trying to calculate velocity. However, because we have the opposite situation, we're just going to do the opposite of this formula. So what's the opposite of a derivative taking an integral. So the position is going to be computed by taking the integral of the velocity with respect. Two time are as a function of time is computed by taking the integral, and I'm going to substitute in the velocity just from the question, I'm gonna drop the units just to make it a little bit easier to right in the integral here, plus to TJ hat all retrospective time. That was a quick look quick, Little refresher in doing inter girls, when I'm doing Integral is of power. I look at the exponents on each variable. I add one to that exponents which in this case would give me three. And then I divide my term by that new exponents. Same thing over here. If there's no exponents written, I've gotten exponents of one. I add one to that to get to. And then I divide my expression by that new exponents. So my next line would look something like this. After taking the integral six t cubed over three in the I had direction, which just goes along for the ride. It's not a variable, so I don't have to worry about it when I integrate j half now because I've done an indefinite integral I need to add in a constant of integration, which could technically be a variable. It could. Sorry, it could technically be a vector. It could have components in the X direction and in the Y direction. So that's the next thing that we're gonna have to do after we simplify this a little bit, we're gonna have to solve for what that constant of integration is But first, I'll just quickly clean up my integration. 6/3 econ simplify to just two t cubed in the I had direction and the twos cancel haute toe. Leave me with just t squared j hat in the jihad direction. So when solving for a constant of integration like this see here we typically will use what are called initial conditions or some other information about the position that has been given to us. In this case, you'll recall the question told us, Hey, at time T equals zero, we know what the position is. It zero I had zero j hat. So buy something t equals zero into my integral here and here for tea and letting r equals zero on this side over here we can actually solve for what might see is gonna be So let's do that. We know that the position AT T equals zero is zero I hat plus zero j hat. So we'll plug that in, and then we also know that that happens AT T equals zero. So that's what we're gonna plug in there. And then we just sold for C looking at each term separately. I've got to time. Zero cute. Well, that's just gonna be zero. Here. I've got zero skirt that's also going to be zero. So on my right hand side, I'm gonna end up with zero. I had plus zero j hat and on my left hand side, I have the same thing. So as it turns out, my C is going to be non existent because I already have zero. I had an zero j hat at Time T equals zero already. So my final answer for our, which I can plug back in to my expression on the right hand side, is just gonna be what I had right here. So there we have our position to t cubed I hand plus T square J head meters. So the next question asks us to describe this motion qualitatively so qualitatively to me means using words and just describing basically, what is the particle doing? I'm gonna look at it in its separate vector components. So first I'm gonna take a look at the X, and then I'm gonna take a look at the y. So let's write thes out as two separate functions. X of T equals to t cubed. And why of T equals T squared. Looking at these two functions, the exponents on the X has a T cubed. Where is the power on the Y function is only T squared. So to me, this means that the function is moving further in the X direction than it is in the wider action. This is further added by the fact there's a coefficient of to here on the X function as well. So I would just describe its qualitative motion as the fact that it's moving in two dimensions, and it's moving further along the X direction than it is in the Y direction as time progresses. If I wanted to sketch very roughly or what that might look like on a graph, it might be doing something like this where it's growing a lot quicker in the X than it is in the UAE. But because it just asked us to graph or describe its motion. Qualitatively, I'm not gonna waste a lot of time plotting an exact trajectory, because that would be more quantitative analysis than is necessary. So that's how I would describe its motion qualitatively, the next question asks us to solve for the acceleration you can ask the same question I did before. What's the relationship between acceleration and the information that I already know? Well, the relationship that comes to mind that, you may recall, is that the relationship between acceleration and velocity is very similar to the relationship between position and velocity, in that it's the derivative of velocity with respect to time. This time the relationship is the white right way around. So all I need to do to solve for the acceleration as a function of time is take the derivative of the velocity. So to do that, I'm again going to just drop my units to keep things simple here and attack it with a derivative. So in case you need a quick little refresher on how the derivative works, when I'm taking the derivative of an expression that has explosions, I look at the exponents. First I multiply the exponents by my coefficient and then I'm gonna subtract one from it. So for my first term, that would give me an acceleration in the I had direction of two times six, which is 12 and t to the power of one, which I don't even have to write. So not going to and that's in the I have direction. If we do the same thing with the with the T here, I have an exponents of one. I take my one down and I multiplied by the 21 times two is just too. And subtracting one from the exponents on the T gives me a T to the power of zero, which is just no t at all or one. So I don't even have to write anything down. And I'm left with just a two. So this would be my acceleration. And if I want to throw my units in there, I can either throw them in at the end like that, or I can put them back in to the individual constants. Whatever you are most comfortable with, place the units there or whatever is expected of you as you're answering this question. So the first, um, constant, you will recall from the question had the units of meters per second cubed. So if you put that in there in the I have direction, um and then the second constant had units of meters per second squared in the jihad direction. So there we go. That is our acceleration. Next, we're gonna see how to solve for force. So let's ask ourselves that question again. How does force relate to one of the physical variables that we've already calculated? The one that comes to mind for me is force equals mass times acceleration. Luckily, we have our mass and we have our acceleration. We just calculated it so we can just shove those two things together to sell for our force. Mass was five kilograms and my acceleration was 12 meters per second. Cubed, um, times t in the I had direction plus two meters per second squared in the jihad direction. So in this case, we're multiplying a scaler by a vector. So mass is the scaler, accelerations, the vector. All I have to do is multiply the scaler by each vector component as I would if I was just distributing this multiplication across two terms in a bracket in order to get my final answer for force. So doing that, we get five times 12 which gives me 60. The units multiply together to make kilogram meters per seconds cubed, or I could just call that Newtons per second to make it a little bit cleaner five times to give me a 10 again. The kilograms multiply to make kilogram meters per second, cubed or again. If I want to simplify the units, we can just rate that as Newton's. And so there's my force. So the next part of the problem asks us to solve for torque the net torque on the particle again, we need to ask ourselves, what is the relationship between torque and another variable that we've already calculated in this case? The formula that we're gonna use is that torque is equal to position our cross product with force. So this one involves taking the cross product of these two vectors. So this question provides us a great opportunity to practice taking the cross product. But otherwise we've already calculated our position, and we've already calculated our force in the last question. So we already have enough information to do this. We just actually have to go through the cross product to calculate the answer to this question. So again, I'm gonna write down our two vectors, just dropping the units to make it a little bit more readable. So our position we had to t cute. I had plus t squared in the J Hat Cross with our force that we just calculated, which is 60 t in the eye hat, plus 10 in the jihad. So when doing the cross product, there are a bunch of different ways to do the cross product. But I'll show you the one that I find most convenient for questions like this where I've got vectors already in the I had J hat notation and the way I like to do this is just by distributing my vectors as they normally would and then resolving the cross product of the individual vector components. So what do I mean by that? Kind of like we did in the previous question, where you multiply the scaler tons of vector. I'm just going to distribute the X component times the X component, the Y component times, the y component, the Y component times X component and the Y component times the y component like I would if I was just multiplying to bracketed expressions by each other. Normally, however, after I do that, I'm gonna have to resolve each of the cross product separately. And to do that, I'm gonna exploit the two relationships that I know of when I think of cross products, the 1st 1 is the cross product of any vector component with itself is always zero. So if I ever have to vector components that cross themselves when they're the same vector So I had cross I had J head crusty had Kwiatkowski head. It's always equal to zero. The other relationship I remember is this site click coal relationship between the I had jihad and K head So I had crossed with Jay. Hat is always gonna turn into key hat and j had cross of k had is always gonna turn into I hat and so on and so forth. Whenever I go this way around the circle, I always get a positive result if I end up going the other way So I had cross K hat, for example. I would still get J hat, but it's going to be negative. Likewise, J Hat Cross I have would give me que hat, but it would just be negative. So now, applying the principles we've just talked about, we can compute the torque component by component multiplying. The first component of the first vector tends the first component 2nd 1 we get to t cubed times 60 t and I like to separate out the vectors afterwards so that we can see exactly how we're going to simplify them. I had cross I had. Next term we get is to t cubed times that 10 and that's in. I had cross J hat second or third term. Rather, we're gonna get is the T squared from the first vector and the 60 t from the second vector. And that's a J had cross I have now I should mention here that order when doing the cross product product matters, as we kind of talked about in this step over here One way we get a positive and one way we get a negative doing our final piece here. We've got t squared times, tense age, a hat cross J hat factor. So applying the principles that we just talked about up here, we can see some of these air just gonna immediately disappear to zero. The 1st 1 because it's an I had crossing I had and the 2nd 1 because it's a J had cross Ajay hat that's also going to disappear to zero the second term here. We haven't. I had cross Ajay hat using our wheel. We can see that this would lead us with a positive K hat. So I'm gonna write that in there as a positive. Can't. On the final term we have to deal with. We've got a J head cross. I had again looking at the wheel that's still going to give us a K hat, but this time, because we went quote unquote backwards, That's gonna be a negative, negative k hat that we're dealing with. So now all we have to do is just clean this up a little bit. To do that, I've just clean up our workspace a little bit, and I'm just going to simplify the two terms that we have left. So the 1st 1 becomes 20 t cubed in the K hat direction, and the 2nd 1 becomes negative. 60 t cubed also in the K head direction. Now, because both of these are T cube terms and they're both in the K hat direction. I can combine them as, like terms and end up with my final torque, which is negative. 40 t cubed in the K hat direction. For now, I'm gonna write that down and then we'll be tackling angular momentum. All right, so we've added our torque answer to our list of answers and you'll notice I've put the units back in Newton meters per second cube, which just came from multiplying the meters by the units that we had in the previous question. Next on the agenda is calculating angular momentum. Two formulas come to mind when we want to calculate angular momentum. Given the information we already have, the first is angular. Momentum is equal to the cross product of position and linear momentum, which is lower case p. You will recall that regular momentum is equal to mass times velocity. So because we know the position and we know the mass and we know the velocity, this is a form that we could use, given the information we have. The only downside here is that it would require taking another cross product. The other relationship that we could exploit is the relationship between torque, an angular momentum which is, like some of our other variables, a time derivative. Since I'm trying to calculate angular momentum given torque, the opposite of this would be that angular momentum is the integral of torque with respect to time. So either of these two formulas we can use to calculate angular momentum. If there's one way that is more comfortable to you use. That way, however, I'm gonna do it, the Inter go away because given the information we have in this problem, it's gonna be more straightforward. And we just did a cross product, so I'd rather not do another one if I don't have to. To calculate this integral, we need to use the torque that we already found and plug it into the integral formula that we just rode above. So that looks something like this again. I'm going to drop the units just to keep things a little bit cleaner here. T cubed K hat with respect to time again taking an integral with the power We have a power of three integral power rule, says I. Add one to that and then divide my answer by my new exponents, which would be dividing by four. And so my final answer in the integral will be, um, negative 10 t to the power of four in the key hat direction. So that's the angular momentum as I said before, If you want to go through and do the cross product, you can the steps would be the same as how we did the cross product to solve for the torque. Um, but you may notice that doing it the integral way was quite a bit quicker. So let's write that down and move on to kinetic energy. So now we're getting into non vector quantities. So kinetic energy and power, that too we have left to solve for neither of which is a vector. But we're still gonna have to use the vector components we already have in order to solve this for these remaining two components. So kinetic energy, same questions before What's a formula that relates kinetic energy and some of the things that we have already calculated? The one that comes to mind is the basic definition of kinetic energy, which is 1/2 m. V squared. So the question that often gets asked when we see a vector squared in a scaler equation like kinetic energy is how do I score of actor and what does that mean? So when we see it like this, what the V squared is actually representing is the dot product of the vector V with itself. Let's do a brief review of the dot product. So let's say I have two vectors A and B and I'm dotting them to each other. Let's assume that they're three dimensional vectors with an I had a J had in the key had component. Or we could also say that they each have an X y and zed component. The dot product basically takes the X components, multiplies them together, and to that adds the Y components, multiplying them together and the zed components, multiplying them together and so on and so forth. If you had any higher dimensional vectors. So what does that look like for our vector here? Well, we've got V squared, which is v dot v. I write that would explicitly, that's again just dropping the units for simplicity. Six. T squared in the eye hat plus two t in the J hat dotted with the exact same thing 60 square in the eye had and to t in the J hat. So the dot product basically says, multiply the like components together and then add them. And that's exactly what I'm gonna dio. So multiplying the two. I had components. I get 60 squared times 60 squared, which is 36 t to the fore and then multiplying. The J have components together. I get to tee times to T, which is four t squared, and that's it. Notice how the I had in the J had both disappear. That's the point or part of the point of the dot product, which is? I should end up with a scaler quantity. The I had and the J had both disappear, and I'm just left with 30 62 the four plus four t squared. So at this point, I'm going to take this and set it aside for now and go back to my kinetic energy equation at the top here. And let's plug in what we know. So, OK, does he go toe 1/2? Well, that's 1/2 we know half my masses five again just dropping the units for now. And my velocity is what we just figured out right here 36 t to the fore plus four t squared, multiplying this all through. I get half times five times 36 is 90 t to the four the half times the five times the four is 10 t squared. So let's actually take the time here to put back in the units for a second and see what happens with all those units. So I've just coached everything up a little bit to give myself a little bit more space. But let's go through and talk about the units here. So the five, which was my mess head units of kilograms, the 36 t to the four plus four t squared. Well, that whole thing has to have units of meters squared per second squared because that was a velocity. But if we go through to the individual terms, let's look at what their units would be. The 36 well, that came from the six meters per second cube. So those units are gonna be meters squared per second to the six. Because that's what a meter per second cube squared would be. Same thing here. This four came from that two meters per second squared, scoring that I get meters squared per second to the power of four. So multiplying all of those units through this 90 is gonna have units of kilograms meters squared per second to the six, and the 10 is gonna have units of kilogram meters squared per second to the fore. Now, those air pretty ugly, so we could simplify them. And there's a number of ways to simplify them. But to me, the way that makes the most sense would be to call them as jewels per second to the fore and jewels per second squared because one jewel is a kilogram meter squared per second squared. So taking out one of those from each of those terms, I'm left with a second of the four on the bottom in the first and a second squared on the bottom in the in the other one. So we're gonna write that one down and move on to the final piece of this puzzle, which is the power? All right. Our final question is calculating power asking the same question we've been asking all along. You know it by heart by now. What's the relationship between power and something we've already solved? If you had answered, power is the derivative of kinetic energy. With respect to time, you are exactly along with where we're going to go. So one final derivative and that will take us to the end of this essay of a question. So power equals D by D. T of 90 key to the fore plus 10 t squared, taking another derivative, remembering our power rule. I take my exponents, multiply it down and subtract one from it. Do that to both terms, and what we end up with is a power that 90 times forgives us 3 60 t cubed plus 10 times two is 20 t for my power. And again we could put the units into the individual terms. Or we could put the units in on the overall expression. And you know what? Since I haven't done that for any of the other ones yet, let's do that for this Final one because we know that power has units of what's so let's go ahead. And just to show you one other way to write the units, we can go ahead and write Watts around the entire thing. So there we have it, those air, all of the different properties that we have calculated