https://youtu.be/RLGeK2UZIEE
PHYS 1101: Lecture Twenty, Part Three
Okay. So we’ve got this electric generator turbine that spins at 3600 revolutions per minute. Friction is really small, but it is working to slow the turbine down, and it takes ten minutes for this turbine to coast to a stop. How many revolutions does it make while stopping? Okay. We have seven variables. That’s all we can choose from, because we are going to use these three equations, or some combination, to solve the problem and they only involve our seven variables.
So we must pick one of those variables to solve for, to give us the information that we want. Ultimately, we have to figure out how many revolutions it makes while stopping. Question 23. What’s the variable you’re going to have to assign to that quantity? What are7 we going to go after? Question 24: what variable would you assign to the quantity ten minutes? Question 25: what variable would you assign to 3600 rpm?
Okay. Let me do this problem for you. Here I have just a copy of all of my variables, just a little mental note. The start variables, the end variables, the Alpha in between, and here’s my three equations. The name of the game, do whatever you have to do. Sketch it. Visualize it to be able to decide which of these variables you know, which one you want, and then work with these equations to make it so.
Here’s my sketch. I’m going to start out with this turbine starting out at an angular position of zero and I’m just going to indicate that the thing starts out winding in this counterclockwise direction, so here’s my Omega0, but I know that it’s slowing down. Friction is slowing it down. Eventually it comes to a stop ten minutes later, so I’m going to indicate that my Alpha has to point in the direction opposite to Omega. I have some initial instant.
Let’s call this my start and let’s call this my end. At the start, I have Omega0, I have Theta0, which I have set to zero, and I have time, which I have set to zero. At the end, I’ve got time, I’ve got position, and I have the final angular velocity. My seventh variable is Alpha in between.
Okay. There’s a great sketch for me. Let me now think about the variables that I know and what I want. Okay. Well I’ve already got these zero values written up here. I’m not going to write those again. Let’s just go through each one of them. Here are two I know. Do I know the initial angular velocity? Yes. It’s 3,600 revs per minute. This I know has to be an Omega. Has to be an angular velocity because of the units, and that’s the condition at the start of the problem.
That’s how fast it’s going around initially. So Omega0 is 3600 revs per minute. I’m going to do my conversion right away to get it into radians per second. I know in one minute, I have 60 seconds. There’s my conversion from minutes to seconds. Now I’m going to do my conversion of 2π radians in one revolution. And now I’m left with units of radians per second, and that gives me 376.8 radians per second. That’s this variable. I’ve done all three of these now.
Do I have a value for Alpha? Do I see anything here with units of radians per second squared or degrees per second squared? No, I don’t know it. How about the end, t. Well, I do know the time is ten minutes. Converting that to seconds of course gives me 600 seconds. Okay. There’s my time. The final Theta. Do I obviously know that? No. How about my final angular velocity? Here it says it coasts to a stop. When it coasts to a stop, remember our analog there, if it was at rest, that the velocity’s zero. That’s true for the angular velocity too. Because it’s at a stop at the end of the problem, Omega is zero.
Okay. I’ve thought through all seven of my variables. Which variable do I want? I’ll put that over here. I need to know the revolutions that it makes while stopping. I need to know how many times it’s gone around. I need to know what the angle is that is swept through if I get that total angle, so one time around it’d be 360 degrees and the next time around would be 720, or the first time around is 2π radians, the second time around is 8π
radians. If I know how many times around it’s gone, my final angular position, I can figure out the number of revolutions. So what variable do I want? I want Theta, the final angular position.
Okay. Here’s my list of equations. What should I use to go after it? I have time, I don’t know Alpha, so this has Theta in it, let’s see, number three also has Theta in it, but look, both of this that have Theta have Alpha. I don’t see any way around it. I think I have to solve for Alpha first and then use one of these two equations to get my Theta. I like equation one. I’m going to go with one first. So let me write here, looks like we’ll need Alpha.
Again, both equations that involve Theta have Alpha in it. I don’t see a way around it, so I’m going to use equation one. I have all the values I need to do that. Do a little algebra to rearrange this equation. Alpha is going to be Omega minus Omega0 over time. Plug in values for that, my final Omega was zero, I started out at 376.8 radians per second, and then I have to divide that by 600 seconds.
That gives me an Alpha that is -0.628 radians per second2. This negative, it’s good indication here that things are as should be. Alpha is in the opposite direction to Omega0, so it should have the opposite sign. I sketched this with Omega0 being positive, because it’s counterclockwise, that’s why I made this positive, so Alpha did turn out to be negative.
Okay. Now that we have Alpha, we can use either of these two equations to figure out Theta. My eye’s kind of drawn to equation three. Either one would work. I’m going to try equation three. Equation three is Omega2, so the final angular velocity is the initial one squared plus 2 Alpha times Theta minus Theta0. My initial angle is zero, my final angular velocity was zero, so I have 0 equals Omega02 plus 2Alpha times Theta. I know Alpha now, and Omega0. I’m after Theta. A little algebra rearrangement here and I have that Theta’s minus Omega02 over 2 times Alpha. When I plug in numbers, I end up with 113,000 radians.
Okay. Let me point out something regarding the signs here. Notice that when you’re working with these kinematic equations, like we did in chapter two, you have to be very careful about keeping track of the sign and substituting in the signed quantity, so when I went in and substituted for Alpha, of course I put in the negative quantity, and then likewise for Omega0, that’s inside my parentheses here, that negative quantity then gets squared. A negative times a negative in the squaring cancels with itself. And then this negative that’s out front cancels with the negative of my Alpha, so I do end up with a positive angle position.
This does mean I have wrapped around, in a counterclockwise sense, and my angle has gone from 0 to 113,000 radians. So I’ve gone around a lot of times. How many times? Let’s convert that into revolutions. 1 rev is equal to 2π radians. So I have to take that number and divide it by 2π and I end up with 18,000 revolutions. That’s a lot of times, but that probably makes sense. It said it had low friction. It’s going to go around a lot.
Okay. Let’s do another example. In this case, we’ve got a Ferris wheel that rotates at an angular velocity of 0.24 radians per second. Starting from rest, it reaches its operating speed with an average angular acceleration of 0.030 radians per second2. How long does it take the wheel to come up to the operating speed? Okay. Let’s just look at the numbers and the units and right away work out what they have to represent. The 0.24 has units of angle radians per second.
That has to be an angular velocity. I’m going to highlight that in green. I don’t know if it’s the initial or the final yet. Let’s see, it starts from rest. That’s giving me more angular velocity information. Let me grab my red highlighter, because notice the units on this number. It’s angles per second2. That’s angular acceleration. In fact, it tells us angular acceleration. That’s Alpha.
Okay. We’re asked how long does it take the wheel to come up to operating speed. How long? We’re going to be after t, the final time. I’ve copied here again our list of our seven possible variables. Must work with these seven variables. The first three correspond to the start of the problem, the last three characterize the end instant of time, and of course Alpha has to be constant between these two. And here are our three equations that we can work with.
Let’s draw a quick picture to get started. In fact, I’m going to go grab this picture and just see if I even need to modify it or how do I modify it, because it captures the essence of all of these rotational problems. I have a Ferris wheel, it’s not told me otherwise, so I’m going to assume that the rotational motion is going to be in the positive direction, which is counterclockwise. Here’s our initial angle and our initial time set to zero. I can keep that convention.
The Ferris wheel rotates at an angular velocity of 0.24. That’s telling me about the steady state operation of the Ferris wheel, meaning after the operator hits Go, it’s going to have to wind up. It’s going to have to start from rest, it’s going to accelerate up until it gets up to its operating angular velocity, which is this 0.24 radians per second and we need to figure out how much time it takes to do that.
So, again, I’m going to have an initial and a final. I start from rest. I may go around a few times, but by the time I am at the end instant for this problem, my final scope at that time, which is what I want, I’ll be at a certain angular position. It might be 720 degrees plus 180, who knows? It’s going to be some big angle, but I will be up to my final operating angular velocity. I will be up to 0.24 radians per second and if I start from rest as it describes, then I know, if I started at rest, and I ended up with some bigger positive angular velocity, Alpha has to be in the same direction and so has to have the same sign as Omega. That’s the only way I’m going to wind up starting from rest and get up to this final angular velocity.
So there I have, on my drawing here, my little sketch, I think I have already gone through each of these variables and listed my knowns. Let me summarize it here. My initial time, position, I can set those to zero. Velocity, that is zero, because the scope of my problem is starting from rest, it reaches its operation speed with angular acceleration of this. How long does it take to do that? This is the start and then this is the end where it’s up to speed. The final time is the thing I want. The final angle I don’t know, and it doesn’t ask me for that. What’s the final angular velocity? It’s 0.24 radians per second.
The thing that I want is the variable, t. What equation should I use to get there? Well, equation one and two are the only ones that involve time. Equation two involves the angle, the final angle position, and I don’t care about that information. Let’s if equation one will work. The final angular velocity is equal to the initial plus Alpha times time, and that will work great for me. Use equation one. It’s Omega equals Omega0 plus Alpha times time. I have Alpha, Omega0 and Omega and there’s the variable that I want. So again, I have to be careful of the sign, as always, when I use my three kinematic equations because the sign captures the direction of these quantities and I’ve thought that through up here when I indicated and pictured that the motion was counter clockwise so I could make it positive.
So I have a positive final angular velocity and it speeds up. It winds up so the Alpha is going to be positive. That’s actually a variable, I realize now, I didn’t write down that I do have. Alpha is plus 0.030 radians per second2. Okay. So canceling, or substituting in any zeroes here, just to start me out with a simpler equation, my initial angular velocity is zero, so the equation I have to work with is Omega equals Alpha times time.
That solves to t is equal to Omega over Alpha, so I have 0.24 radians per second divided by plus 0.030 radians per seconds2 and if you check the units here, one of the seconds cancels with one of those and I have a fraction in the denominator, so this denominator comes up top and I will be left with seconds, which is what I want. When I punch these into my calculator, I end up with 8 seconds. And that’s our answer.
Okay. This last example, I have a couple of quiz questions to ask you about it. Now we’re shown a graph and I want to draw an analogy to the graphing exercises that we did in chapter two, because those things we learned also apply. Those are analogous as well. The graph that’s shown is the angular velocity versus time. It says angular velocity, that’s the variable Omega, versus time. We see that at time, t, equals zero, I have a negative radians per second. This would indicate I start out with this object rotating pretty fast in the negative direction. It has a negative Omega of -9 radians per second.
My Omega then is becoming less negative. It actually is slowing down. I have an Alpha that got to be in the opposite direction. My Omega is changing. One second later, I am going slower at -6, but it’s still in the counterclockwise direction. Another second later it’s at -3. Look at that, at three seconds, my angular velocity is zero. That means for an instant, this thing stops rotating. One second later, it’s at plus 3 radians per second. Now it’s starting to rotate in the counterclockwise direction here. At this point now, it’s rotating in the counterclockwise direction. Now Alpha, again, is in this direction because it’s speeding up. A second later, I’m going 6 radians per second clockwise direction.
Okay, The first quiz question for you. This graph, this is a velocity versus time graph. Remember back to chapter two, when you were looking at linear velocity, v versus t, what did the slope of that line tell you? What variable did the slope of that line give you? Look back to chapter two and remind yourself. That’s what you have to answer for question 26. Question 27, then, is draw the analogy to this problem.
You’re looking at angular velocity versus time. What does this slope give you? What would this rise over run be? Well a rise would be units of Omega, radians per second, and the run has units of time. Let me just write the units here. The rise, the units on the vertical axis here is radians per second, and then I have to divide that by a run to figure out a slope. You do the rise over the run. The run would have units of seconds. What quantity is the slope giving you?
Okay. Let’s solve this problem down below. I need to figure out the angular displacement of the wheel from 0 to 8 seconds. Angular displacement, that’s a Theta minus Theta0. I can always start my angle at zero, so I’m after the final angle. That will be my angular displacement. Okay. Here I have copied our graph over again so we can see it as we work the problem and I’ve copied again our main equations to work with and all the variables that we need to consider.
Remember, it asked us for the angular displacement, and that what we need to solve for is this variable Theta and the knowns were the time would be at 8 seconds and then we have the information from this graph. This graph shows us how Omega changes as time goes on. Okay. If we want Theta, there are only two equations that can help us out and we know we have to evaluate it at 8 seconds, so I am drawn to equation two. It’s the only equation that has the information in it that we can use. So let’s write that down here. We’re going to go for and see if this will work for us, the angle we’re after is given by the initial angle, which we start out at, is zero, plus Omega0 t plus one half Alpha t2.
Okay. What things do we know? We know time. We’re supposed to evaluate this equation for 8 seconds, so we have that. Do we know Omega0? This is the angular velocity at time t equals zero. Let’s look at our graph. At time t equals zero, our angular velocity is -9 radians per second. That is then, our angular velocity at t equals zero. -9. I’m going to write that off here to the side for extra note. Let me write it up here. Omega0 is -9 radians per second.
So, we’ve got that one too, Omega0. One-half Alpha. Do we know the angular acceleration? Remember Alpha has units of radians per second every second. That’s radians per second2. It tells us how the angular velocity is changing. Well that is the slope of the line on a graph of Omega versus t. What is the slope of this line?
Well, for any time interval, because it’s a straight line the whole way, I just have to pick any segment here. Let me pick this segment, and I need to evaluate the slope, which is the rise, which is my angular velocity change, over the run, the time. So my slope for that red segment is over this, looks like a two second interval from 3 to 5 seconds, my angular velocity, my rise is from 0 to plus 6. So my rise is plus 6 radians per second and that’s over a run of two seconds. My slope then is 3 radians per second squared. That is Alpha.
I can add to my list now here of knowns that I’ve put below, that Alpha is equal to 3 radians per second2. Now what about the sign? Is the sign correct? Well my slope is positive. Alpha should be positive. I start out with a -Omega0, so these do have opposite sign and it is going to give me that slowing down effect I talked about here, and it will give me a speeding up effect I talked about here.
This wheel starts out rotating counterclockwise, but it’s slowing down. It stops briefly at 3 seconds and then that same Alpha causes it to start winding up in the clockwise direction. Slowing down, comes to a stop, and now it starts speeding up in the counterclockwise. You can think that through more on your own if you want. So I’m set to work with my equation.
In fact, I don’t even have to do any algebra. I can just plug things in. Let me just rewrite it here. I have -9 radians per second for my Omega0 times 8 seconds plus one-half, Alpha is plus 3 radians per second2, times 8 seconds, and I have to square that term. The first term here gives me -72 and that’s in radians. Then I have to add to that number one-half times 3 times 8 times 8. 8 squared is 64. 64 times 3 divided by two and that’s plus 96 radians. And when I combine those, I end up with 24.
So in the first part, I end up with a combination of these two, so my net angular displacement ends up being kind of small, and that’s because I started out rotating this way and that was clocking up a negative angular displacement given by that, but as I slowed down, came to a stop, it then started winding the other way and I ended up having a net positive angular displacement in the counterclockwise direction that was this big. And so from start to finish, the combination leaves me with 24 radians. And that’s the end of lecture 20.