Hello I'm Dr Brian Tang, I'm a college lecturer in Engineering Science at Balliol College. I'm James Cohen, I'm an associate professor in Engineering Science and also a tutorial fellow at Balliol College and we'll be telling you about the manner in which Engineering interviews are conducted here at Oxford. Typically you'll have an interview with two colleagues from a college and we'll spend between 30 minutes and an hour typically asking you a little bit about yourself, why you want to study Engineering, what motivates you, where your interests lie, and then we'll move into the more technical matters where we'll discuss some maths and physics problems. The maths and physics problems that we cover will not extend beyond the things you've dealt with at school, we're not expecting you to have done your own independent research, but they will challenge you, where we want to see how you think so don't worry if you don't know the answer to the questions straight away, that's kind of the idea, and don't believe that we're looking to trick you because that's simply not true. What we're interested in is how you think and how you approach problems, and what we're really looking for is someone who can take the guidance, think for themselves and apply it themselves and then keep going, so it's not a simple matter of how you're getting to the answer but it turns into a discussion. And the reason for this is the interviews are intended to mimic the tutorial environment, so we're looking to see if you will thrive here at Oxford and not just whether or not you went to a really good school for instance, so hopefully you'll see with the demonstration interview that it shouldn't be a scary experience and a lot of our candidates find that it's actually quite enjoyable!
Welcome to the interview. Before we begin I just want to confirm do you prefer Sam or Samuel? I prefer Sam. Alright excellent, nice to meet you Sam, I'm James Kwan and this is my colleague. I'm Brian Tang, yeah, lovely to meet you, thank you for giving- I'm sorry, lovely to meet you both. Nice to meet you, and we will be giving you the interview here at, I guess in this case Balliol College, and of course the structure of this interview is basically we're gonna ask a few quick questions about your interests in Engineering and then we'll quickly jump to maths and physics problems and really just, you know, talk your way through everything and relax a bit. And I'm going to quickly draw your attention to the Miro board that we'll be using and can you just quickly scribble something for us just to confirm that you are able to connect and you are able to write because we will be using this board.
Okay, perfect, so from this point forward... I'm going to erase this, but from this point forward please do not erase anything on the Miro board. If you make a mistake or you want to, we will cross it out for you because we'll be using a different colour, but just keep going on ahead and please do not erase anything. And the other thing is, as one of us is going to be asking the questions, the other, my colleague, is going to be writing notes down, so don't pay any attention to that and don't worry about what they may or may not be writing, just- we'll just focus on the question. So, right, so I was looking through your UCAS form and I noticed that you did the UNIQ summer school programme and can you just talk more about that and maybe some of the things that you've learned? Yes, so I went on the UNIQ summer school programme in 2019. It was a hugely valuable experience and a big part of why I ultimately applied to Oxford for Engineering. We did a huge range of activities, we had some sample lectures and we did some practical work, for example, a robotics lab, but also some bridge testing - we did the calculations then we actually physically built a bridge, and also we had some experience working with the problem sheet style work, which I'm of the impression you get a lot more of when you're at Oxford. Excellent, and also I noticed that you did some work with or you enjoy modelling roller coasters? That is not my area of expertise, so could you maybe elaborate a little bit more on that and educate me on what the modelling of roller coasters would entail? Yes, so what I use is a piece of model software called No Limits, and it's a fantastic piece of software, it allows you to calculate the the g-force read-offs to render things in very good detail, so it really just allows me to create these very realistic and feasible designs, and I think it got me thinking a lot about real life engineering stuff as well, because as I got more and more experienced with this software I was able to cater to build things that were more accurate and based more closely off of real manufacturers' designs, in terms of things like how the elements are shaped and the progression and how I guess you'd orchestrate a ride layout. And, out of curiosity, I know sort of these loop-de-loops in roller coasters are very popular, but my understanding is that you can't really build a perfect circle with the loop-de-loop? Is that true, or is there something- Well, originally a perfectly circular loop was what they used on the first roller coasters and it was not a big success. One thing that modern roller coasters have that old roller coasters didn't is they have a thing called upstop wheels, so basically, on a modern roller coaster, if it experiences a negative g-force so the path would essentially be pushing riders upwards, then instead of just flying off the track there's the wheels underneath that can keep it conjoined and keep the ride smooth, so nowadays, ironically enough, after going to more like a teardrop shape, that keeps the forces a lot steadier throughout the course of a loop, they've now started to experiment with going back to more circular or even in some cases more stretched out loops. I think it's very interesting actually, quite interesting to see how the industry almost flips back and forth on itself. Well, they always say 'old is gold' right!? So with that we're gonna jump straight into the questions. For myself, I'll be asking you more of a math problem, so I want you to, at the Miro board I really would like you to sketch out a certain function for me. So I'm just going to write the function out here...
Okay and if you could for me, just somewhere nearby that, plot what you think this function would look like? But as you're doing so could you also just talk me through the process of what you are sketching out, and drawing our attention to any points of interest that you might think would be relevant for us to look at? Yeah, sure, so I'm just going to start off by drawing myself a set of axes... Now the first curve I'm going to draw is... so we have sine x over x that's... sine x is a sinusoid shape, so, on this side that is one full period, and on this side, again, one full period - albeit a little off when it comes to scale! So the next thing is the over x so that means that as x increases, because we're dividing by x, it's going to be decreasing in magnitude, so 1 over x as just a general kind of curve would look a little something like, well it would look like this, and there is an asymptote which I'm just going to try and draw on now at x is equal to zero, because at this point the value is undefined. Yeah, the one over x does not exist when x equals zero and as you tend towards infinity x tends towards zero from the positive and negative directions respectively. So when you combine these functions together the point where you have interesting behaviour is at the point x is equal to zero. Now something I learned in Further Maths is called L'Hôpital's rule, and that's basically that the limit of a function - in fact I'll just change that to an x - the limit of a function as it tends towards a certain value, let's say zero for sine x over x would be equal to the limit of the function
um
minus cos x over one as that tends towards zero. So that's, yeah, L'Hôpital's rule is something i would just use without proof. So because in the original one where we tried to put in zero, sine x of sine zero is just zero, but sine zero in the denominator, that's undefined. Whereas with this next one- Oh sorry, I've just realized I've made a sine error there... there we go plus cos x not minus... so, but with the second one, when we have this, it's going to cos of zero is just one, one over one, just equal to one so
now let's consider back to how this overall curve should look, combining those two things together. So okay
changing back to the pen
so
around zero obviously we have that rather unusual behaviour, so when x is greater than one it should be a smaller magnitude than we saw in that sine curve, but when it is less than one it should be a bigger magnitude, so what I would expect the curve to be looking like is
something to the effect of, it's a little bit more spread out, but since the effect of this where as we tend towards infinity the magnitude just gets smaller and smaller, and on the other side,
okay, so sine of x would be negative, but x itself is also negative, so negative divided by negative should be a positive, so I think that it would look more or less like this. Okay, I think this is a very good start, but I'd like to draw your attention really quickly to this section here...
Why would the sine function ever really go above one? So here you have one right, and if you look at your sine x over x, what happens to sine of x as x gets smaller and smaller and smaller? Not necessarily approaching... it's very small, let's just say, so, as opposed to practically zero, what would happen to sine of x- So when x is small, sine x over x is comparable to x over x so the absolute, okay, yes, yeah, I've done, I've drawn this side wrong, so basically the maximum value that the function can have is one, so I'll just move this down...
So it would come out looking more like, that's our zero, and then again that decaying oscillation. Okay I'm just gonna draw, I'm assuming, there's no x-axis on my screen, but this is where you're- Yeah, I was just having it, oh, perfect. Okay, excellent, that's absolutely right. It's also known as a sync function, just in case you've ever come across it, but that's all absolutely right. I guess I could quickly ask you where the zeros are, where are all the zero crossings, where would they be? So the zero crossings are, with the exception of x is equal to zero which would be on sine x but not sync x, they're going to be at 𝝅...
yeah, 𝝅, 2𝝅, 3𝝅, so n𝝅, where n is an integer but n is non-zero. Right, greater than zero, excellent. I think we have time, I'm going to ask another quick curve-sketching problem. So here before we were asking you to sketch out sine x over x but now I would like you to sketch out x factorial for me. Are you familiar with factorials? Yes, so factorials would be like one factorial's one, two factorial's two, three factorial's six, and so on. Okay,
so, first things first, I would think about the range over which a factorial is defined. Zero factorial does exist and it's just one, but minus one factorial is not a valid thing that works, so essentially the lowest... the minimum value on our graph is going to be
one in terms of the y-axis, and the minimum on the x is going to be zero.
So with the factorial function, so three factorial is one times two times three, four factorial is equal to one times two times three times four, and so on. Because the thing that you're multiplying by is always getting bigger I would expect it to always have an increasing gradient so
if I just draw a set of axis myself,
now on the left side and those two quadrants it's just going to be blank. At zero it's going to be one... oops, I did try to draw a straight line there... So zero, one it's still just just one, and then after that I would expect the curve to be going a little like this, where the gradient gets increasingly steeper and and steeper. Okay, excellent. Now I guess a quick question would be then, it's because you mentioned that factorials typically only work with integers, but really quickly could you estimate for me what 2.5 factorial would be?
Okay so the first thing I'd probably do would be what... jot down we've already got what three factorial is, that is six, and two factorial, just gonna add on one side there, is two. So the factorial function is not linear, so we couldn't just linearly interpolate between these two points as you might often do in a situation like this,
so
what... one possible idea I've had, I don't think I know how to make it work for this exact situation, would be just to sort of estimate a curve that would fit this factorial curve approximately, and then substitute 2.5 into that. But obviously I think that's requiring- We're only looking for estimates so I mean even in the, what you've drawn right here, do you think you could give a rough approximation? From what I've drawn here it looks similar- So if this is one,
and I'm gonna assume if this is two,
and then you would have three which you said would be six, scale-ish right-
In that case, just visually, I'd probably expect it to be about
about 3.25 or so? Sure, exactly, it's, as you mentioned earlier, it is non-linear, so it's not halfway between two and six, but it's actually slightly closer to two than it is to six. Great, well, thank you very much, and I'm gonna pass over the baton to my colleague Brian. Okay now we'll move on to something completely different, so you can forget everything you've just done with James and we'll start afresh. So now I'm just going to ask you a little question about rolling a marble. So to begin with I'm just going to give you a slope - what we have here is, that's supposed to be a straight line, and it's at an angle theta to the horizontal, and onto it I'm going to place a marble and that marble is round - even if I can't draw it terribly round! - and we'll say that it's a distance l measured along the slope down to where the slope becomes horizontal. I place this marble at rest, and let's say it's got a mass m, and then I let go. Could you work out for me how fast the marble will be going at the point it reaches the bottom of the slope please? So the first thing I just want to check before we get started - I'm assuming that this marble is quite smooth and the surface it's on is also quite smooth, so is it reasonable to assume that the effects of friction are negligible? Absolutely and I apologise, I was supposed to tell you that when giving you the question - we have no friction at all, no friction and no air resistance. Perfect, so that means, because in this situation the only force acting on the marble is the force of its weight, and that means that in this system, because there's no frictional air resistance, energy is conserved so energy total is constant. At the top, the marble is at rest, so its initial kinetic energy I'm gonna denote that's KEi is equal to zero, it's stationary. However, it has gravitational potential energy. Again we're gonna, there's a bit of an assumption but a reasonable one is that the distance is small enough that we can reasonably take gravity as constant, so GP is equal to mgh, so m g and the height h, is - it's not a very straight line but the effect is the same - is going to be l sin(theta), because l is the distance kind of on that hypotenuse and sin(theta) allows us to take the vertical distance, so mgl sin(theta). And at the bottom it's going to have no gravitational potential energy relative to where it started, so set that to zero, and its kinetic energy is going to be half m v squared. Now we can equate those two energies as energy's conserved, so mgl sin(theta) is equal to half mv squared. Just going to move this down a bit,
and subsequently we can cancel off some terms, so m, that's present in both, v squared is equal to 2 g l sin(theta), and v is equal to plus or minus square root 2 g l sin(theta). Although I think in this case we'd probably define, if we're just taking the speed, for example, you just take the positive of those answers. Good, I was just about to ask you what a negative speed would mean in the context. Okay, that's excellent, so we'll just move on to the next portion. So this ramp that the marble rolls down, I should probably just clarify because it's not clear on my sketch, at the bottom where it goes horizontal it's actually a nice smooth join so it's not an impact, it doesn't bounce, it'll just smoothly exit at that speed you've just calculated. So now I'm going to take a cone, and it's a hollow cone so picture like an ice cream cone or something, okay. And what's going to happen- sorry, I should say this cone has a half angle of alpha, and, at a certain height, let's say at a height from the base of H.
The slope that we rolled the ball down is going to feed the marble in, and imagine there's like a little trapdoor here, so the marble comes through the side of the wall and then it's going to roll around the inside of the cone, okay. So hopefully you can picture it - I've got just a hollow cone and a marble rolling around on the inside.
What I'd like for you to do eventually is to work out for me how high
I need to release the marble from initially, so that once the marble goes into the cone it traverses a perfect horizontal circle, okay?
Now that's a very big question, so perhaps I can give you a hint to get started, and there's our little marble on the inside of the cone, remembering that this angle was alpha, could you perhaps start by considering the forces acting on it?
Yes, so okay the first force is going to be the weight of the marble, so that's going to have a value of mg. The next force is going to be the normal reaction force which is perpendicular to the surface that the marble is sat on, and
now I know that this is going to be a question involving centripetal acceleration, so what we're going to want to find is the magnitude of force required that would allow the marble to move in this circular motion. So, the centripetal force required for that would be m v squared over r, where v is the speed it's travelling at, r is the radius of the circle. So
we can find what the value of r should be. Oh sorry, I'm inadvertently moving myself around. There we go, back to where we should be. So r is going to be equal to
H tan alpha, because again it's our height H up to this point and it's an angle alpha. v we can substitute in as our value from a little bit earlier which was... or rather v squared, that's going to be two...
sorry,
2 g l sin(theta). Was that v or v squared?
That is v squared, thank you very much, you're quite right.
So now what I can do is substitute those forces into that formula I've just said earlier, as kind of what our desired force should be to achieve that.
2mgl sin(theta) divided by H tan alpha, that would be our centripetal force. Now when we look at our marble actually free body force diagram
the direction we want to take the forces in is normal to the plane so in the direction that N is acting. So
if we can, we can almost like set up our marbles own little axes, x and y. Now the next thing you want to do, just so we can work with those axes, is just confirm that this angle here, this angle here is alpha, you can kind of see it because of how you've got that going through perpendicular through the same plane.
So the sum of the forces in the y direction is going to, taking again the axis as I've drawn them, would be- Sorry, Sam could I just interrupt you?
If we look quite very carefully at this, if I took the angle alpha that I've marked here,
that would also be alpha wouldn't it? Yes, yep. So, I think actually you've made a small slip and that- Ah, apologies, that would be 90 minus alpha.
Anywa,y that's fine, now that we've fixed that perhaps I could,
perhaps instead we can just draw in a right angled triangle, and that angle's alpha and that might be enough so please carry on. Okay so now that we have, yeah, reconfirmed where alpha should be, so that'll be N, and then because the gravity, the weight force acting downwards, mg
cos alpha
Okay so now we've got this far... So, sorry, Sam- I've gone the wrong way round. I think perhaps you're being misled by a drawing that I've made rather too crowded, so perhaps if I just draw above your work a clearer diagram and invite you to take another look
and just remembering that we are looking in the y direction, as you've labelled it.
Okay, apologies, in that case yeah it's definitely not cos, so
N minus mg, and then we can very clearly see that's the vertical component, yeah, sin(alpha).
Yep, that's okay. So now that we have that sum of forces
so we want to equate that with what we obtained earlier as our centripetal force. So that was 2 mgl sin(theta) over H tan alpha is equal to N minus m g sin(alpha).
Sam, sorry to interrupt you, we're just running out of time, so in the interests of time I'm not going to ask you to crank through all of the maths, but could you just explain to me, down here, so if I, let's see, if I draw a nice big arrow so you can see where I'm pointing, that was your expression for the, I believe the centripetal force you said it was. Could I ask you to go back up to the free body diagram and just indicate to me where that centripetal force is?
So the centripetal force...
I'm trying to recall whether it would act just perfectly horizontal...
yeah, I think it should act just perfectly horizontal to the plane in which it's kind of moving. Okay, and so can I just get you to draw very clearly one arrow onto that free body diagram to show me where that centripetal force is?
Apologies, can you just repeat that question? So I just want to be absolutely clear, so could you just draw one arrow onto that free body diagram showing the centripetal force itself?
Okay, so you're almost right there. What does the centripetal force do, how do you know there's a centripetal force? So, the centripetal force is what keeps an object in circular motion, what keeps it in circular motion, now I've said that out loud to myself I realise that, okay, it should be acting in the opposite direction to the directional- It's pushing it towards the centre of the circle. Good, so that's going in the wrong direction. You were heading in, broadly speaking, the right direction. Just down here, in the bottom right hand corner if you're, if you can see it? Yep. This line is- Yeah, no that doesn't work because it's, if I was equating these forces to be in equilibrium I should have taken them to be, instead of perpendicular to the plane the marble's moving on, I should have been taking them to be horizontal. Well, so what you could have done was, the line above, N minus mg sin(alpha) was the net force in the direction normal to the surface, but we know that the marble stays on the surface, so I think where you were going with it was you were trying to equate that to zero? Yeah, and then I could solve for N and take that further on. But I think you were heading, broadly speaking, in the right direction there. I'm afraid we've run out of time though, so this concludes the technical part of your interview. So if you have any questions that you'd like to ask us before we close now's your opportunity, but this does not form part of your assessment so if you don't have anything that you want to ask that's fine too. Yep, I have no particular questions I want to raise. In which case thank you for joining us. And best of luck! Thank you very much. Thanks.
Well we just concluded Sam's interview and we would like to just give some feedback on his performance, and, just to get started, we thought that he did an excellent job in sort of describing and discussing the things that he was interested in based on his personal statement. You know, it really did give a sense that he really was interested in engineering beyond just purely the academic sense. I don't know if you've got anything else from that Brian? I was particularly impressed that when we started talking about roller coasters he could actually demonstrate that understanding and the knowledge that backs it up, and that's what really makes me confident that it's genuine interest rather than something he thought might sound good on his personal statement. Yeah, and although we don't spend too much time clearly on the personal statements, it's those little touches that really, you know, make us realise how interested they are in the topic of engineering. So I guess on to the math questions, which were all about curve sketching, and again it's not always about, it's not- Every college is going to have different types of questions so it may not be a curve sketching question. One thing we did want to highlight as part of the feedback was that, you know, it was very evident that this particular, that Sam knew a lot more about certain mathematical principles that you may not be aware of, and that's perfectly fine. It's not that much of an issue that you know or don't know L'Hopital's rule, or have seen the sine x or over x function, or whatever, those are not really what we're looking for, if you could just jump straight in and solve it. But in fact it's really just his ability to talk about what he's thinking as he's writing it down, that was really what stood out to me as something that was interesting. I was not so, it was no extra benefit to me that he was able to pull out L'Hopital's rule, but the fact that he could talk about it as he's trying to draw what he's thinking about as he's doing it, that's quite critical. And you would have noticed that despite knowing L'Hopital's rule he was still unable to draft the very first sketch on the first go, so that's not a big deal, very few people are able to, and it's just once you point out 'well, what about this?' when you probe someone, we probe them 'have you thought about this part?', the fact that he was immediately able to recognise his error or think about it and come to the correct conclusion from when the mistake was pointed out, that's another good plus, basically being able to take direction or take criticism. And similarly with the factorial problem he happened to know what zero factorial was, I mean if you don't know that, that would not have made any real difference, as you can see in the problem, because ultimately the question was could you basically draw between the points and infer information, and that's what what we were able to see. You could have inferred what zero factorial would have been, and in theory you could have inferred what 3.5 or 2.5 factorial would have been, and again it wasn't about exact answers, it was about methods and how he approached the problem. But that was my takeaway from his, from my part, my question of his interview. On the physics side, on the marble question, I thought Sam actually did pretty well overall. I thought it was very clear, the manner in which he was approaching the problems, it seemed to me that the starting point of working out the exit speed looked like a problem that he'd seen before, and that's fine, and he clearly applied sensible reasoning and worked his way through it. What was interesting to me was actually as we went through the problem and it became more advanced and he started to make little slips. And one thing I would say is the really little slips like getting the angle in the wrong place are really not a big deal for us, it's very easy. What was beneficial was, just as with the maths, when we pointed it out to him, and we try to point mistakes out fairly gently to see if the candidates can work out why we're giving them that hint, he took the information on board and carried on, and that is one of the things that we're really looking for. The other thing was, I guess he got himself a bit confused towards the end, and you could see the nervousness building up, and of course that will come into play in an interview, but hopefully you could see that actually we're not here to make a candidate feel bad, and what we're trying to do is get the best out of a candidate. I think given a bit, a little bit longer time wise there, I've got no doubt that he would have made it through to the end of the question with a couple more nudges, but unfortunately we just don't have that freedom, we don't have that, the time available always. So, if you do find, just as we did here, that your interviewer has to end the question before you reach the answer, don't worry. A lot of the time we don't actually intend every candidate to make it to the end of the question and that doesn't necessarily reflect on your performance negatively. As I say I feel that Sam actually did pretty well overall on this question. So that's exactly right, so I think Brian and I would have discussed offline and probably would have given Sam pretty high marks and would have considered him moving forward. Now all that being said, keep in mind that the interview is part of a bigger package overall, and there are many factors that are considered for applicants of whether or not they make it in or not. So really there is context both in the interview itself, but even in the application overall. So don't worry, the interview isn't necessarily, you know, be-all end-all, but it is, you know, an opportunity for us to get a sense of who you are and and how you think and how you approach problems, and really that's, at the end of the day, what we do here at Oxford.