Your engineering team has been assigned the task of measuring the properties of a new jet-engine turbine. You've previously determined that the turbine's moment of inertia is 2.6 kg m^2. The next job is to measure the frictional torque of the bearings. Your plan is to run the turbine up to a predetermined rotation speed, cut the power, and time how long it takes the turbine to reduce its rotation speed by 50%. Your data are given in the table. Draw an appropriate graph of the data and, from the slope of the best-fit line, determine the frictional torque.

Question

Your engineering team has been assigned the task of measuring the properties of a new jet-engine turbine. You've previously determined that the turbine's moment of inertia is 2.6 kg m^2. The next job is to measure the frictional torque of the bearings. Your plan is to run the turbine up to a predetermined rotation speed, cut the power, and time how long it takes the turbine to reduce its rotation speed by 50%. Your data are given in the table. Draw an appropriate graph of the data and, from the slope of the best-fit line, determine the frictional torque. <TABLE>

Accepted Answer

Hey, everyone in this problem, we're told that in the wind industry, several research projects are underway to investigate the effect of the friction torque of the wind turbine pitch bearings on wind power efficiency. To estimate the friction torque loss is a researcher measures the angular speed variation of a wind turbine inside a laboratory. The moment of inertia of the turbine is 3 kg meters squared. The researcher gives different initial angular suits to the turbine and measures the time required for the turbine to reach 25% of its initial angular speed. And the collected data is shown in the table below. And we're asked to calculate the magnitude of the friction torque. We are told to note that once the turbine reaches the desired initial angular speed, it will be subject only to frictional forces. OK. Now a hint we're given here is to plot gra using the data from the table. So let's get into this one. We have the data in the table that we're given. OK. So we have initial angular speed omega knot. Hey, this is given an R PM 404 67 5 34 601 and 668 and then we have the corresponding time and seconds. OK. So 1214, 1618 and 20 respectively, we're given this blank graph. So we have time and seconds on the X axis, the initial angular speed in radiance per second on the Y axis. And then we have four answer choices for that friction torque all in Newton meters. Option A 7.9 option B 8.4 option C 9.3 and option D 10. So let's start with the graph. Hey, this was our hint. We want a graph. So let's go ahead and do that. And if we're thinking about our graph and we want the initial angular speed omega knot in radiance per second instead of an RP MS. OK. So if we think about that first value 400 R PM, OK, which is revolutions per minute in order to convert this to radiance per second. And we can recall that we have two pi radiant per revolution and one revolution makes a full circle, two pi radiant in one full circle. And then we can multiply again by one minute per 60 seconds. When we do this, the unit of revolution will divide out, the unit of minutes will divide out and left with radiance per second. OK? And in this case, it's gonna be around 41 0.888 radiance per second. OK. So we're gonna add to our table here and we're gonna add Omega knott in radiance per second and we're gonna repeat the same process for each of these values. So for the first one, we know it's 41.888. If we do the same with 467 that's gonna be 48.904. Next 1 55.92 just taking approximate values rounding a little bit here, but trying to keep some significant digits 62.937 for 601. And finally for 668 it's gonna be 69.95. Ok. So we have all of the data now in the right units that we need for a graph. So let's go ahead and graph. So starting with 12 seconds. So delta T is gonna be 12 seconds and we have Omega knot is 41.88. So just about here, then we're going on to 14 seconds and it's gonna be 48.904 somewhere around here, 16 seconds, 55 0.92 18 seconds, 62.93 and finally 20 seconds and 69.95. And what we can see is that when we grant these, we get fairly close to a nice straight line. And if we were to graph these on really, really exact kind of grid, we would probably get a much nicer line than my kind of rough graphing here. But we get a straight line. OK. So we can represent the relationship between the initial angular speed and the time as a linear relationship. OK. So that's the graph we have the question now is how do we relate that to the tour? And so recall that the torque to can be written as the moment of inertia eye multiplied by the angular acceleration. Alpha. Hm. And what do we know about alpha? OK. Well, we know that alpha, the angular acceleration can be written as the change in angular speed or change in angular velocity delta omega divided by the change in time. Well, this is just the final angular velocity omega F minus the initial angular velocity omega not divided by delta T. In this case, we're told that Omega F OK. So that final angular speed is gonna be 25% of the initial speed. OK. That's the time it takes to reach 25%. OK. So Omega F is actually gonna be 0.25 Omega N 25% of the initial. So alpha is 0.25 Omega knot divided ome uh minus Omega knot sorry, divided by delta T. And we can simplify this as negative 0.75 Omega knott divided by delta T. And that's our alpha value. Now, when we go back to our torque equation, we're gonna substitute in this alpha value. But what we're gonna do is we're gonna ignore the negative and we're asked for the magnitude of this torque. So we're gonna take the absolute value that's just gonna give a 0.75 omega knot divided by delta T. Maybe we can drop that negative for this case when we're just looking for the magnitude. So artur Tau can then be written as the moment of inertia. I multiplied by 0.75 omega knot divided by delta T. So we're starting to get somewhere. OK. If we take a look at the equation, we have Omega knot and delta T and those are both variables that are represented in our graph. So in our graph, we're graphing omega not in terms of delta T. OK. So the initial angular speed in terms of the time, so let's try to rearrange our torque equation so that we are writing Omega knot as a function of time, right, that will then match with our graph. So if we do that, we can write that Omega N is equal to the tort multiplied by delta T divided by the moment of inertia eye multiplied by 0.7. OK. Let's go ahead and rearrange this or simplify a little bit dividing by 0.75 or dividing by three quarters can be written as four thirds. And so we have four multiplied by the torque tail divided by three, multiplied by the moment of inertia. I all multiplied by delta T. Yeah. Now the moment of inertia, I, we know. So let's go ahead and substitute that in. Now we get four tau divided by the moment of inertia, I is three M um kilogram meters squared. We're given that in the problem. So three multiplied by three gives us nine kilogram meters squared. And all of this is still multiplied by delta T. So now we have an equation Omega N is equal to four tau divided by 9 kg meters squared multiplied by delta T. We're writing Omega not as a function of delta T. We know that that graph is linear. So if that graph is linear, then the term multiplying the T and so this entire four Tau divided by 9 kg meters squared must be the slope of our graph. OK. So now we know that four Tau divided by 9 kg meters squared must be equal to the slope of the graph. We can calculate the slope of the graph. We have the values to do that. OK. So what we can say is that for ta divided by 9 kg meters squared must be equal to the slope, calculating the slope. What that means is that Tau is gonna be the only unknown in our equation and we'll be able to solve war, which is the value we want, we want that friction toward Tau. Now this slope we can take as rise divided by run and we can choose any two points along that line. OK. What we're gonna do is take the final point and the initial point. OK. So the furthest to the left and the furthest to the right, if we take the furthest to the right, OK. We know that Omega not in that case is 69.953. OK. Radiance per second. And we're gonna subtract the initial an M angular speed. OK. That's the change in the Y value or the rise of our graph. So 41.888 radians per second divide by the change in time. OK. The final time we had was 20 seconds, the initial time was 12 seconds. So in the denominator 20 seconds minus 12 seconds. If we simplify the right hand side, we get four tow divided by 9 kg meters squared is equal to about 3.508125 ratings per second squared. We can multiply both sides by 9 kg meters squared. Divide by four. A tow will be 9 kg meters squared divided by four, multiplied by 3.508125 radians per second squared. The units we're left with is gonna be kilogram meter squared per second squared. They recall that a Newton is a kilogram meter per second squared. So this is gonna be Newton meters exactly those units we want. And when we work this out, we get that Tau is approximately equal to 7.89 33 Newton meters. So this was a really interesting problem. We were able to graph the initial angular speed versus the time. And you use that to calculate the torque by recognizing that this gives us a linear relationship. So that term we had in our equation must be equal to the, if we round to two significant digits, we can see that the correct answer is going to be option a 7.9 Newton meters. Thanks everyone for watching. I hope this video helped see you in the next one.

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Chapter 12, Problem 12

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