Kia ora koutou, co-Douglas Walker, toku ingoa. We're here tonight with Andrew Sargent for a bit of level three mechanics revision. I can see a number of you have already put some pretty good puns in the chat.
If you would like to ask any questions during the presentation, pop them into chat. Andrew's really good at monitoring those as well, but I'll try and deal with as many of them as I can. so that Andrew can focus on delivering the information that he's got prepared.
Just remains for me to say a massive thank you to Andrew. Andrew is head of physics. That's not quite the name of his role, but he's effectively that position for Takura.
And him and I worked for a number of years together at St. Patrick's College in Wellington. So without further ado, I do need to say thank you to Studyit, who have sponsored these webinars. Thank you, Andrew.
Over to you. Cool, kia ora tatou. Ko Andrew Sargent, Tākou Inga.
As Doug mentioned, I'm the kaiaka matua of physics at TKORA. I used to be, that's what I was, now I'm sort of kaiaka matua of science, but it's all the same thing. So today we're obviously going to go over the mechanics exam. I'm going to take a little bit of a different approach to what I've previously done. In the last year I just did over, like I just did a full exam review.
this year we'll probably do exam strategy for the first 15 minutes. Then I'm still going to go over, I've got a 2013 mechanics exam simply because all the exams up from there have already been done, no point redoing what's already been done. But I've written, I've pre-written some of my written answers and then I pre-wrote them.
I've just had COVID, so I had COVID on Wednesday so my brain's a little foggy. I wrote them when I was a bit elucidated, so then hopefully we can come back and I can sort of critique them. and look at them from a different perspective so you can sort of see from an as an out or yeah from an outsider in view Sort of what you should be looking for right better quickly share my screen Actually go full screen on the stone.
I share screen Screen one try that Cool. Yeah, and you can post in the chat if you've got questions I posted at the top to the link to the folder. Inside the folder is this slideshow that you should see in front of you now.
I'm actually just going to chuck that slideshow in the chat. I'll chuck that in there. Right, so I'll quickly make this full screen.
There we go. Right, so some great resources i thought i'd start off with that so you've got mr wibbly's youtube channel um he's done last year's physics exams i think the 2021s and maybe down to the 20 2018 his exam videos are awesome um i didn't bother making the last year's exams because his ones are awesome and there's no point to double up um he's also got notes for all the topics um they're pretty much the same as the folder notes you can buy um they're like Maybe 20 bucks or used to be a while ago and they're all color-coded. I highly rate them So if there's anything I sort of mention you can go to those notes and you've got to find like a summarized version Ish some of it's obviously quite summarized because they're just notes and I like the full explanation So you probably want to throw back to textbook.
I've got my channel which is like all nice and organized everything's like I'll put Mr. Will be videos on there as well. It's all I organize for level three you can have a look at that. There's a physics folder, that's what I put in the description of the video, so if you go into description video, you open that up, that'll take you to the year 13 physics folder, which you can like, that's where I've got all my resources, that's where this PowerPoint is found.
I've chucked in a simulator site, a circuit simulator, and then I'll put the link to physics realm as well. This realm's just, it's just all the like level two and three and one physics as well I think. Yeah, it's in the description.
It's literally someone led. This is linked to the slideshow. It's literally right above you in the chat Right, so it's not be walking about this search exam Strategy you should be sitting a minimum of two exams prior to Thursday. So this coming Thursday Is obviously your exam you should be sitting at minimum of two exams and you want to take what's called the KDR approach So if you know if you continue on in the aviation world um any of you want pilots there's what's called a knowledge deficiency report so if you search any aviation exam either in the ppl or the cpl or in the atpl i think as well so they're all like the different pilot exams um any question that you get wrong you need to write like a mini report and that ensures that when you've sort of finished an exam you've come away with a hundred percent of the knowledge because you know as a pilot you want to be up in the sky and think oh i only passed 18 of my tears i don't need you know i only know 80 of what i should doing we really need to know 100 so how that relates to physics like sit an exam any question that you get wrong find out what topic it relates to because all the questions i've segmented up into their individual topics um and then practice two or three uh different version or like different types of those same questions you'll get what i mean very soon um so that's that uh historically i went through all the last exams for the last i don't know you Seven or eight years to see whether there was any trends in terms of excellence Typically, it's usually C or D is the excellence question. A is typically always the achieved question.
The rest are either merit or excellence Very very occasionally question B has been an excellence question But it's only been maybe sort of one out of every five years and maybe question B in there YouTube guides for questions that you're stuck on if you are stuck just use the YouTube videos The assessment schedules that NZQA post aren't always the full answer. A lot of the time they're not the full answer. There's often more that you could sort of write or add that the schedule doesn't put in. Also, the schedule is, it doesn't, it's not the annotated schedule that the actual examiners use to mark. So when examiners are marking your work, before they actually go and mark, they'll have what's called a panel meeting.
So like any paper might have 10 markers. So. all 10 of them will go to a meeting, they'll review maybe 20 exams and they'll have a look at the pre-written exam schedule and they'll annotate any questions that might need clarification in the answers so maybe I can't think of any in particular, no I can actually, so for waves, level three waves obviously, there's like one of the questions in terms of if you have like a diffraction grating and if you change the size of the diffraction grating. Often students will use a level two explanation such as if the wavelength is related to the size of the sled you get greater diffraction.
Not on the schedule, but what the examiners will write down is like a clarification to say that's not acceptable and needs a different answer, etc. So yeah, a lot of the stuff on the schedule is not always there. Right, common excellence questions for level three mechanics.
Simple harmonic motion. is like the excellence question for simple harmonic motion they're always very similar they're always going to use either you have to use the shm formulas so either you'd be asked to find some position at some certain time you might be asked to find a velocity at a certain time or quite a good excellence question to ask is how far something has gone how far something has swung so i would definitely practice the SHM excellence questions over the other excellence questions as like a priority because they're usually more straightforward. The other like Kepa's law questions, sometimes they can be straightforward, sometimes not so much.
Right, so mechanics is split into your four sort of areas, not directly distinct. We've got translational motion, this is taken straight from the the achievement standard and notice i've highlighted uh center of mass in two dimensions it's been a long time um since there's been a center of mass question that's relied uh which is asked for two dimensions but it's in the standard so it could be asked and the only difference is you'll just get an x and a wide coordinate uh like you work out the center mass in the x position and then you just do it completely separate in the y position and that would give you the two center of masses so it's sort of like battle battleships you just have you two different points that would line you up for that one. You've got conservation momentum that can be in two dimensions as well where you have to have conserved momentum in the x direction and conserved momentum in the y direction or just horizontal and vertical. It's limited to two dimensions, you're not going to get any three-dimensional stuff.
Is it common for, I'm just reading the chat now, is it common for excellence questions to ask about conservation energy slash momentum? Yes. Sort of, but not really. Not to the degree that scholarship.
So if you know about the scholarship questions, scholarship questions heavily rely on the knowledge of conservation of energy slash momentum, and they can do some... Yeah, you can have some very tricky algebra related to that. Not so much in level three There's not really much further that you can go with it other than like a simple Like a simple collision that's either at right angles Where you've got an exchange of impulse but not the previous question Yeah, the previous exam questions in the last five years Most of them have only ever been merit Nothing really higher than merit. So it's not that common if that's what you're asking for. Right, so circular motion and gravity You've got Kepler's law.
If you want to know about that look up the 2015, in fact I'm going to go through the actual exam questions. So translation motion is fairly common. You can see here I've gone through and categorized every single time a translation motion question has been an exam.
You can see it was skipped out in 2015 and 2016. and it was been in there ever since. You can see it's either A, A or B, sometimes C or D, A to C. So I've been a bit lenient, sometimes there's been a little bit of translational motion mixed in with some other stuff, but it's not usually mixed in that much. It's usually just an achieved question.
I think the merit, yeah, from you've got C to D, that's definitely a momentum impulse question. So you've got a, so I think Question, this one here, 2018 is a good example of momentum in two directions where it's coming at right angles. And you have to use vectors to represent the momentum. You can't just use numbers because those vectors, then you add them up using Pythagoras. And then you can find out like the total change in momentum.
And then you can find out like, normally the two blocks will stick together. And then you have to use vectors and you add them together and find it. the final way to find the final velocity. You're not normally asked for the angle, but sometimes it can be.
Right, circular motion and gravity. Yeah, Kepler's law, this is usually always an excellence question. Excellence or like a moderately hard merit. 2015, question one. 1d is the actual derivation of Kepler's law, the full derivation.
We had to work out the mass of the moon, and I think you're only given the period of the moon. Yeah, I think you're also given it in days, which is also tricky. So you need to be able to convert different time scales as well. So you need to be able to convert days to seconds.
So there's obviously 24 hours in a day, 60 minutes an hour, and 60 seconds in a minute. You times those three together, you get how many seconds in a day. And that's how you get the period for a moon, if it's 30 days and around.
Right, so yeah, these are all examples. Rotating systems. So this is the...
angular kinetic equations with angular momentum as well. So you've got torque, rotational inertia, conservation of angular momentum, and conservation of energy. There are often conservation of energy questions relating to rotating systems where 90% of the time the rotational energy is not conserved, and the excellence question is realizing. that when you pull your arms in you're doing work so if you're spinning around in a circle and you pull your arms in you're doing work because you're like it's a force over a distance um so that that's usually the one factor that separates um students from being merit to excellence not realizing um where this extra energy comes from um right shm questions you can see question three question three question three question three question three question three yeah it's pretty repetitive um so that's yeah they're pretty straight not straightforward but they're pretty uh similar every single year um right so that is that for mechanics um i think that's it yep so what i'm going to do is we'll open up a mechanics the 2013 mechanics exam i'll go through and sort of sort of show my approach to it i just need to find where i put it where did i put it it's up down here Right, we'll open this up and I need to hopefully change some settings.
How I present a view, sweet. So yeah, so this is the 2013 exam. You can go into the folder to find it.
It's just in the exam, like I'll put it, it's in the folder, it's not hard to find. Right, so we have a hollow ball with mass of 0.310 kgs, the radius of 0.0340 meters. It's thrown upwards, it rises through a height of 1.4 meters then drops down again when it is released is moving upwards of 5.24 meters per second.
So that's its linear velocity and you can see here it's actually given like the linear velocity and that's also given you the angular velocity as well. So it's given it in a weird form. It's given it as 2.7, like zero, revolutions per second and if you remember back to, yeah if you remember, I'm in the chat, if you remember back to what is like what is the definition of frequency, the definition for frequency is how many cycles per second.
that's really it. So frequency is just how many cycles per second. Period is just how long it took something to rotate, a full rotation.
So that's rotations. And it's, yeah, rotation, without an s in the end, per time. So that's normally like periods for things can be anywhere between two seconds.
Well, your class period, most schools have a class period of an hour. That's an example of a period. All right, so this here is technically F.
and it's written in a confusing way so that's really the frequency. It could have been written as 2.7 hertz but I've written it as 2.7 revolutions per second because I'm trying to confuse you. It could also be written as revolutions per minute or revolutions per hour and then you just need to divide the whatever number.
So if it was 100 revolutions per minute you just need to divide that by 60 to get to revolutions per second. We'll go to the first question, show that the angular speed of the ball when it's released is 17 radians per second. So in your formula sheet, we'll have a formula that is the angular velocity is equal to 2 pi f. So we have the frequency even though it's sort of convolutedly written. That is going to be equal to 2 pi and then we're going to have our frequency is 2.7.
So multiply that by 2.7 and I'll see if my graphics calculator will overlay on top of this it might actually do menu one and then we go two shift pi multiplied by 2.7 2.7 equals and that's 16.9 i need to move my thing 966 so because this is a show question you must show the formula first so we have to have the formula oh no you have to have the formula then you have to have the working and then you have to have the it always pays to have the un is the calculator on radians i don't actually know it doesn't really matter at the moment because um it only matters if your calculator is in radians if you're using the trigonometric functions so if you're using sine or cos or tan then your calculator needs to either be in radians if you're doing the simple motion parts or like part because you obviously you'll be working in radians This is just some more arithmetic so I don't actually need to worry. So it should be 6, so it was 16.96. And then I'll say radians per second, negative 1. So I give the full answer, obviously a bit neater than mine. And then we obviously give the final answer and we'll write it down here. And then we have 3SF3SF3SF3, 3 significant figures, that's what I mean.
333, and there's 4 up here randomly, but whatever. So we round that down to 17.0. 17.0 radians per second, simply because that's what they've given it to us.
How do I find where the 2.7, how do I get the 2.7? It's because it's given to us. It's given to us in a convoluted way, and I think I hopefully already covered that. Right, B, show that the average, ooh, average acceleration, our angular acceleration of the ball before it is released.
is 67.9 radians per second per second. So, before it is released, so it's probably it's from rest. Yeah, very quickly from rest, so it started at rest. This is probably going to be quite straightforward.
The angular acceleration is equal to the change in, or just go, I'll put change in angular velocity over the change in time. I know I've already got the change in time. There it is there. I have the torque, though, but I don't have the rotational inertia.
It's a hollow ball. Okay, so I could work out the rotational inertia. They haven't given you the rotational inertia's formula. If you haven't been given the rotational inertia formula, you're not actually expected to memorize this.
So this is something in the formula sheet. And what I mean by that is the formula I is proportional to. I'll put K here, even though this is not in the formula sheet and you won't be familiar with this, but M squared. I'll put K because that K just refers to.
any fraction so for a for like Yeah, I is equal to M. Oh wait, you must be in V square. I'm just looking the chat.
No, no, it's in V squared This is the rotational inertia. So in in R squared Though it's meant to be an hour. I can see how to the confusing. That's just super messy. This is messy of me.
Yeah So yeah, this is a rotational inertia that k just represents a fraction depending on the shape of the object so in your textbook you'll have a like a list of all different shapes um if it's a hollow ball pretty actually i'm not sure i think it's a hollow ball it's two-fifths um if it's yeah it all depends on the shape you can look up there's like different forms all different shapes um yeah so you'll be given the fraction if you need it so for this one here because we're not giving that fraction we don't we could have used a rotational inertia formula torque is equal to the rotational added that angle of acceleration is equal to the torque multiplied by the rotational inertia i'm pretty sure that is the formula i better just check that but i can guarantee you sooner or later we're going to use that formula ah yes that is correct that's right it's this one here that's what i was looking at let's move this out of the way Right, so the change in angular velocity is just from 17 to 0, because it started at rest. So it's just going to be 17 divided by, and the time was 0.25, 0.25, and that is going to be equal to, I'll just go back to my calculator, I can guarantee it's going to be equal to 60, oh hold on, I'm going to divide this by 0.25. I do get 67.85 but this is probably enough.
I'm just looking at this formula here. There we go. That there, I could have blown up the change in angular velocity to equal to b.
I should have written that to omega final minus omega initial and then divide that by the change in time. That's probably what I should have written that as and then I should have gone 17 minus zero. divided by 0.25 and then you'd have, I'll write the answer that was there, where was the answer, 67.85, so 67.85, I'm gonna leave the units out to the very end and then underneath you'll have the angular acceleration is equal to 67.9 rads per second squared, per second squared. Right, takeaway is show questions. You must show the formula.
Then you need to put the numbers in. Then you need to calculate it out. And then, like, write it finally. Otherwise, if you don't, if you miss any one of those steps, it's not achieved.
Right, calculate the rotational inertia of the ball. I sort of have already alluded to having to do that. This is kind of like writing up this question. This A and B will probably only be achieved. But they've written these, like, they've written, if I was to write this question.
you'd have to write these two first parts so you can get to part c and have it be a merit question. So we've got the torque is equal to the angular acceleration multiplied by the rotational inertia. So the rotational inertia is equal to the torque. Hold on, I've gone backwards haven't I?
Quickly double check it. yeah that's right talk is equal to rotational initial exhalation um how do we know which inertia formula to use they will either give it to you or it won't matter and when i mean when i say it won't matter um if you have let's say an angular momentum question and say you've got you've got two shapes and they'll always be identical shapes so if you've got like a what was this i think the 2015 or 2016 exam had a cat and it had the front of the cap rotating one way the cat rotating the other way you didn't know the rotational inertia of the front of the cat or the back of the cat it just told you that they were the same so or that the front of the cat was modeled as a cylinder the back of the cat was modeled cylinders you just assumed that both cylinders started the same and then when you applied conservation of angular momentum to the front and the back um you had um you had uh i like the inertia of the front multiplied by the angular velocity of the front and that is equal to the inertia of the back. multiplied by the angular velocity of the back. Just remembering that the angular momentum is equal to the rotational velocity multiplied by the rotational inertia. Well the angular velocity or the angular inertia, rotational inertia, same thing.
So you can see here that if you had any fraction of the of the inertias, that fraction would just be cancelled out. So whatever that fraction is at the front, they're just divide both sides by that fraction and it would cancel out so you wouldn't actually need to know what that fraction is so this is an excellence concept um where it just eventually cancels out um right so yeah if you want to have a better example of that go to i'm pretty sure it's a either 2000 i'm sure it's a 2016 or maybe it's a 2000 i think it's 2015 question two because i think that whole question was based on movies or why not movies but like common like themes at the time i think one of the questions was on Total Recall which is a movie that probably none of you guys have any idea what it is. One of the questions was on Smarter Every Days video on cats, I think that's where it came from and there was another one on, I think it was on Gravity but I'm not sure where that was originally from.
Right, back to this question, it's the torque divided by the angular acceleration. We had the torque somewhere here, 0.48, 0.48 and we've got the angular acceleration which is 67 point nine and we'll just quickly work that out. I'll grab my calculator.
Here's the torque one where I would have used the fractional one. Okay yeah so because you don't know the shape of the object, well you do it's a ball, it's hollow, but they haven't given you the formula so don't use it because I mean you could you could if you've memorized them all. I think actually in the mark schedule it lets you away with it. Notice that like every exam will have mistakes from the exam, well from the writer.
And I'll go over the end when we have time, like writing approaches, how to approach writing an exam, what sort of mistakes you can sort of make if I have time. Right, I'll clear that, I'll quickly work this out. So 0.48 divided by 67.9, that should be delete, delete, 67. 0.9 and that equals 7.06. Yeah, I'll go 7.07. I'll just run that up.
7.07i is equal to times 10 to the negative 3 and it's rotational inertia so it's kg meters squared. Right, I better quickly check the chat. Nothing there at all. I'll be quick to check the sound. Capital rotation on the edge of the ball.
Talk over the angle of acceleration. Oh, I know what I went. Completely missed something. This is Newton's and needs to be done.
It needs to be a torque. So this should be 0.48 Multiplied by the radius. I didn't there was there was a force and not the torque.
Yeah, someone picked that up straight away I just realized I've done that so it should be multiplied by 0.034 Yeah, this seemed very very small. That's why I was a bit suspicious about that. Yeah, so yeah, I forgot to multiply by the force Right, let's just move on because we know I went wrong but everything else was fine and the rest of it is correct if I'd forgotten to, oh yeah, other than forgetting to multiply by the distance.
Right, well the following two situations explain whether the height to which the ball rises will be less than, greater than, or the same as 0.14 meters. You ignore the effects with air resistance. Okay, the first example is the ball was not rotating but given the same linear speed when it's released.
So I can think of quite a few different ways i could ask the same question so another way you could ask the same question is have a block uh like an ice block being getting pushed and given like a set amount of energy maybe a spring with yeah this would be a good question have a spring give an ice block some energy and it would shoot up a ring and you'd say it's no friction no yeah no tricky stuff no friction whatever um the block would just have purely linear kinetic energy and as it moves up the ramp that would just be directly translated to gravitational potential energy, forgetting out all the external forces etc because there's no friction. That's pretty much what this example is sort of asking so if there's no rotational energy, more energy can be put into linear translational energy. So it'll get higher.
The ball will end up higher up. And I'm pretty sure I've written something like that. So I see when the ball is done at the same speed, we'll have the same linear kinetic energy as previous example. That's all reached the same height as the other way.
I didn't read this yet. It's given the same linear speed. So this is me asking a completely different question, but if I was to write a question similar to this Um, that's like another version of the same question that you could write Yeah, so it's given the same linear speed not the same linear energy.
The next one's the same linear energy. I'm pretty sure Um, so it'll reach the same exact height because it's both shot at the same speed It doesn't matter that it's rotating rotation doesn't really affect anything because there's no air resistance. It stays up here Um ignore the effect of air resistance one thing I didn't so I just wrote this out without actually bothering to check the answer schedule and then i checked the answer schedule um and i've highlighted a red thing i missed so both balls experience a linear acceleration downwards of 9.8 meters per second per second and the angle velocity the ball won't change so i didn't put those things in there because i didn't really think it was important um but obviously it is so it seems to be just write out all the different things that affect the situation so obviously gravity is both going to affect them um and So you want to just say gravity accelerates from both.
And you want to probably note that the rotational velocity of the ball or the angle velocity of the ball doesn't change. Next question, the ball solid instead of hollow has the same mass and radius, same amount of total work is given to the ball. It's linear and rotational.
So solid instead of hollow, we instantly know that the rotational inertia is going to be greater because more mass is concentrated towards the centre. So... that means the rotational inertia is going to be greater which means it's probably going to be spun up with the same amount of energy or essentially continue reading the same amount of total work is done to give the ball its linear and rotational motion that is the same angular speed so if it's got the same angular speed it'll mean it's going to have more angular or rotational kinetic energy than the previous setup moving at that same speed so if it's got more rotational energy it's going to have less linear translational energy or more less kinetic energy going straight upwards and more kinetic energy in its spin um and i'm pretty sure yeah i'll put as a ball a solid or like greater rotational inertia as more mass is concentrated to the center so it's pretty much what like that was the first thought that you should be having when you read um so when you sort of read these questions answer the information given to you sequentially so they've said the ball is solid instead of hollow that should be the first point that you address, but have the same mass and radius, so that should be like the next point that you address.
Is it really necessary to write down the actual formulas in the discussion to get a higher grade? Not really. I didn't in this here because I was using PowerPoint and I really couldn't be bothered using that ad equation and I couldn't be bothered to write my pen as well because I'm kind of messy, but it will save you time probably. It'll probably make your explanation more succinct.
So I had to go about things in a bit of a roundabout way to explain it without having to bother to put formulas in. So I said, like, well, and I actually still put a formula in. I said work done is equal to the rotational energy plus the linear kinetic energy.
It's the same amount of work done to get the ball, the same amount of work is done to get the ball, the same amount of angular speed. This means I should have a comma there. This means the rotational energy will be greater than before, which is true, as the solid ball has less linear.
energy, its linear initial linear velocity will be less so it'll reach a height less than 1.4 meters, is linear kinetic energy is converted to gravitational potential energy. Things with these questions, if you get muddled between linear kinetic energy and rotational kinetic energy, you'll probably just get not achieved. You won't even lose marks.
If you muddle those two things together, which is a really common thing to do, Yeah, you'll just end up losing marks. Because that's what this whole question is trying to get you to do. It's trying to get you to distinguish the difference between things that spin and things that move, like, translationally or just, you know, in straight lines are completely separate. In order to get something to move in a straight line, all you need to do is provide a force.
In order to get something to spin, you must provide a torque. So a big thing that I get with students when they're explaining angular momentum questions. is they'll say blah blah blah angular measurements conserved if there's no external forces and they've instantly gone from excellence straight down to a merit or possibly a merit down to an achieved because angular momentum is only conserved if there's no external torques you can have things that spin and you can move them around without changing how fast they spin spinning tops is an example of that you can just move the board around and the top is translating through space but it's not slowing down or speeding up So that is a big takeaway for that one there. Right, next question.
Ball on the end of the chord, you've got 1.2 meters, mass ball is a square root of kg, the ball's position shown in the diagram, its speed is 4 meters per second, calculate the size of the centripetal force acting on the ball, the instant shown in the diagram. So that's just application of the centripetal force formula. So you're probably not going to be given one of these anymore. Um, don't solid balls have less rotational inertia? Yes, yes they do.
Hold up, solid balls have less rotational inertia. You're completely right. Did I completely muddle that up? I probably did. I did.
Yes, this is why I shouldn't be doing this when I had COVID. Um, so now I'm looking at it, the rotation, the... the solid ball actually have less rotational inertia because the mass is moved towards the center so this is actually backwards um there should be less it should be less which would hold on the ball is solid and standing hollow the same amount of total work does get warm so this whole question i'm pretty sure is backwards to what i had written i'm going to triple check that yes it is it's completely opposite to what i've written yeah this is what happens when i write when i have carbon um i write thinking the question is something different so this this question would work if i uh had paid attention to playing the ball solid so it's yeah it's the everything i've said here should be the opposite so it'll have more rotation the solid ball will have less rotational inertia because more mass is closer to the center that means less energy will go to having a rotational energy And then it'll end up reaching a higher height.
There we go So this is what happens if I write when I'm sick, and I think yeah Doug has put in the formula two fifths for a solid sphere Two thirds for a hollow. Yeah, so a hollow sphere actually has Two fifths for a solid sphere. Yeah, you don't even know that.
You're right. It's um because the hollow sphere has more rotation. I know I should have paid attention to that. I have no idea where I was going all about right.
Hopefully that has cleared up that question and you can see where people go wrong. Right next part. This we'll just quickly do this now.
fc is equal to mv squared over r. This is just level two. We're going to have what is the mass 0.25 0.25. It's not a show question so you actually need to show your work and you can just put the answer.
times four squared and it'll divide that by 1.2 and I'll quickly speed run this, can I quickly do this, clear 0.25 times four squared equals divided by 1.2, 1.2 equals 3.3 meters, 3.3 meters, 3.3. three newtons and you can see i'll just pop out another answer um explain why the ball moves fastest at the bottom of the circle um i don't like this question simply because in the real world you're like acting a force on it the whole time um so this is kind of a confusing it's a confusing thing to ask when you relate this to the real world um in terms of physics so it's going to be fast at the bottom because all the gravitational potential energy that it would have had is going to be converted to kinetic um that's as simple as that um We've got question C, diagram which shows the gravitational force acting on the top of the swing, assuming the tension force is non-zero at all points, you'll detect this to show the relative size of the tension forces top and bottom using the same scale to all the centripetal force on diagram two at these two positions. Right, so we need to see if it's going at a constant speed the whole way around. Ooh, relative size of the tension force top and bottom. Okay, so a really hard excellence question that can be asked with this, with regard to this, is if you swing something in a, ooh, if you swing something in a vertical circle, but you say it's moving at a constant speed.
So this is, it's mathematically easy to say. For Tubi, you can link that there is more tension FCs, so there's more speed. Yeah, you sort of can. um probably not the answer they're looking for um because yeah so what i'm trying to get to now is you Not, you could write up a question that's a little bit imaginary such that you just say it's somehow held at a constant speed all the way around.
Maybe a rigid rod. So a rigid rod, it's kept at a constant speed all the way around. And at the top, there's no tension in the rod, which means at the bottom, the tension ends up being double gravity.
Later on, when you get to some simple harmonic motion, I'll sort of explain that a little bit better. Right, so at the bottom of the swing the tension force obviously has to be greater than the gravitational force. So the net force when these two add together still points towards the center. Shows gravitational force in the top of the bottom.
Assuming the tension force is non-zero, all points to all the vectors show the relative size of tension forces at the top and the bottom. Okay, bottom of the swing using the, okay, so actually it's the bottom swing and then the top of the swing it's non-zero, so the tension force is non-zero, so maybe we'll make it, I don't know, make it maybe that big, doesn't have to say how big it needs to be, just it just means it needs to be non-zero. Right, using the same scale, drawing the centripetal force on diagram two at these two positions, hold on, centripetal force at these two positions, so the centripetal force is obviously going to be smaller, than the tension force?
Well this is a good question. Smaller than the center than the tension force because obviously you've got these two vectors like add together so one subtracts from the other. So you would just make that slightly smaller and you'd use a ruler as well and you'd label that Fc or you could label it Fnet either or is probably fine. I would call this Ft and you can call that Ft and then here they'll add together so maybe slightly bigger than gravity but not by much. and you try and get a ruler to try and measure that out and that is fc as well.
One thing to note, fc at the bottom should, well yeah it should be bigger than fc at the top because it's going to be moving faster at the bottom and that's just the way this question is set up. So this is actually quite a trickly like constructed question and the fact that the ball's changing speed throughout the loop in terms of the vectors. So this c could actually be Is FC the same size at the bottom, the same size at the bottom, the same as T at the top? No. So the centripetal force, oh, hold on.
No, it's not. It shouldn't be because the speed is changing. So the speed moves fastest at the bottom.
So that means the centripetal force. Yeah, it's a good question. So this is what makes this a very hard question because the speed is changing. So that means the net force or the centripetal force is going to be changing as well. Yeah, so this is like a diabolical question with a lot of pitfalls.
I think this has been asked quite a few times, but the speed's being kept constant. And what I just want to relate before is if the speed's kept constant and the force of gravity is just the force. at the top and there's no other forces acting and that's what's keeping it in like circular motion that means at the bottom of the loop it'll be exactly 2g because your the size of the centripetal force will be the same all the way around what's the difference between fr and ft i'm not sure what you mean by if i think they're meant to be both if t uh if t if tension force and that's meant to be tension force for it is me being super messy um they're both this this year's big big tension force and it is big big attention force just trying to write them up Yeah, it's just my writing. Right, show the minimum speed the ball must have during circular motion is 3.43 meters per second at the top.
So at the top the minimum speed is when gravity, when the force of gravity is acting as the centripetal force. If the centripetal force formula, or if that is less than gravity, then gravity will have a component which will actually accelerate the ball off the loop and then it'll detach. So that's one thing to sort of consider. So basically we just have it at the top.
We have Fg is equal to Fc and then we just got Mg mass times gravity is equal to the centripetal force formula, Mv squared over R, a little bit messy. We can cancel out the masses, just divide both sides by M. We can move the radius up and under.
So we have Gr and that would be equal to v squared, but we'll just square it by output, make that equal to v squared. it should be a square by the way, and then we have v is equal to square root gr and then from there we just go square root 9.81 move that square along multiplied by the radius which is 1.2 and that is going to be equal to, I'll quickly do a calculator. because we do have a little bit of time, we've got 15 minutes, we don't have that much time at all.
Shift, square root, shift, answer, bracket, no fleet, bracket, anywhere to go, 1.81 multiplied by 1.2, 1.2, bracket, 3.43, there we go, 3, oh no, let me move this up, 3.43. meters per second negative one. So explain your answer. So I'm not going to write it just because I want to save myself some time, but the reason for that is at the top the only force acting is the force of gravity, thus we can say the gravitational force is the centripetal force and then yeah we can just equate the two together like we've just done. Right last question we'll try and walk through this with a bit of speed, ball drops to its minimum speed 3.43 at the top.
Using conservation of energy, show the angle at which the tangential speed of the ball is four meters per second. So we're just trying to figure out, it's four meters per second here, from the top, what's the angle? So we're going to be using conservation of energy because that's really all we've got.
So it's lost gravitational potential and it's gained kinetic. So we're just going to write, it's like gravitational potential, mgh, is equal to the kinetic energy gain. So it's going to be half m.
And instead of saying you've got a final velocity and initial velocity, we're just trying to find that change in energy. So we can just put this in brackets, v squared minus initial v squared. So it's the final velocity squared minus initial velocity squared.
As you can see, we can sort of cancel out the masses. So just cancel out. We're trying to rearrange. How do you handle the vector arms? with an object on a slope.
I can't figure out which angle should be which in these diagrams. Typically, I know what you mean, typically the angle is always, you just look at the picture, like look at it, it'll always give a really small angle. That's how I'd describe this. Yeah, normally they always give the, yeah, Doug's right practice.
Normally like the diagram will give it away when you like move the triangle around. and the angle is usually always real small, it makes it obvious if you just look at it and go oh that's the smaller side of the triangle or that's a small side, that should be the small angle. Right, so quickly, h is going to be equal to bracket v f squared minus v i squared bracket and this will be divided by g and then we can just, I'm not going to put the numbers in, we'll just do it on my calculator, where is my calculator? right um final is four squared minus 3.43 squared bracket divided by 9.81 equals 0.43 so this is 0.43 meters yes must be hold on that doesn't seem right does it have i done something wrong ah but i need to No, that seems about right.
Surely, I'm going to quickly check myself. I feel like I've made a mistake somewhere. Have I made a mistake?
Where have I made a mistake? That doesn't seem right. Oh, the half.
I left the half out, that's why. I knew I made a mistake somewhere. That half should still be there. So it should be a half here.
So it should be 0.43 divided by 2, which gives me 0.215. There we go. 0.215.
I'll quickly do it on my calculator. So yeah, I could just divide that by 2. The reason why I was wondering, that's a bit suspicious, 43 centimeters, the whole radius of the of the circle. this 1.2. So when you're doing these questions, have a good look at like the picture and see if it physically makes sense because these questions are written to physically make sense. Going down 0.43 is going down quite a significant amount when you look at the picture.
So the picture kind of gives it away. So that should be 2.15. It should be 2.15. And then we have a, this is height here.
That's H, that's the adjacent. Do we have the hypotenuse? Yes, that's the hypotenuse because this is the right angle here. So we've got adjacent hypotenuse when you use cos.
So we say theta is equal to... cos inverse. And we do this in degrees. So it's the opposite, which is going to be always 0.215 divided by 1.2. And that is going to be equal to quickly grab my calculator.
We don't have much time to go clear that shift cos and I'll clear that shift cos go 0.215 divided by delete divided by 1.2 shift setup change the spec to degrees exit uh there you go 79 point um that was i think you've got you uh you've got your 0.215 it should be 0.7 something oh no wait 0.215 I have 1.2 minus 0.2155 on the numerator divided by 1.2 on the denominator. Cos 0.215 divided by 1.2. Opposite divided by, are they adjacent divided by the high point news? That seems right. Hold on, hold on, hold on.
That's the adjacent. That's the high point news. What is going on here?
That doesn't seem right at all. What's going on? Oh you're right, you're right.
It should be 1.2 minus. You're right, I completely missed that. Yeah it should just be 1. Yeah I know what you mean Doug.
I completely see. I still do have brain failure at the moment. That should be 9.9 I guess. 34 degrees, there we go. 30, 34. What was that?
point something um 34.4.4 degrees yeah so i had worked out the the height that it dropped um pays so yeah the reason why i made a mistake is because i didn't annotate my diagram so when you're doing these always take your diagram um so you can sort of follow what you're doing along um right so uh question a this is a show question so this is just like a show where you've got to put the numbers in um and this is a pendulum pendulum formula is always right at fault t is equal to 2 pi square root l over g, l over g you need to substitute in the values. So yeah just need to put it in so 2 pi square root 1.2 over 9.8. To show questions you've got to have that, you've got to have that and then you've got to have the answer. I'm not going to put it in right now time. Explain what must be done to ensure the motion of the ball approximates a simple amount of motion.
So this is not a very commonly asked question. Typically, it's always what is the definition of simple harmonic motion to some degree. And the definition is, in order to have something undergo simple harmonic motion, there needs to be a restoring force that's linearly proportional to displacement.
So what I mean by that is a spring, as you move the spring away, the restoring force is linearly proportional to displacement. So the formula is F is proportional to negative Kx. So springs will always, for the most part, undergo simple harmonic motion. if you have like a bungee jumper it jumps off when the bungee is under tension then you've got a spring force as soon as a bungee sort of like comes up and then is in free fall then it cannot be considered under simple harmonic motion because the force of gravity is the same it's the same like force wherever you are in space um so that is the best counter example i can think of with the best sort of analogy to explain what is not simple harmonic motion and what is so you need to like dedicate this to memory it's pretty much always a question a maybe question b uh yeah yeah maybe question b there's always usually just an achieved question um right so i just put in there you go oh so for for pendulums the ball needs to be swinging less than 10 degrees because if it swings more than that it starts getting close to free fall and when it's under free fall not a constant force so yeah it needs to be less than 10 degrees also when you derive the pendulum formula use a small angle approximation so that you're also that so small angle approximation uh what's also questions do we need to discuss energy i can think of rotational vertical circular transformations and oscillations how we discuss translational motion in terms of energy you don't really um no you wouldn't typically for translational motion questions i'd be looking at like velocities of center of mass um they wouldn't really um and if you had a whole bunch of things moving around that's more of a scholarship question that you'd ask. So if you have a whole bunch of particles, that's a CERAMAS question, that's more like a SOLAR question.
Right, access below, what happens to the ball's total energy? It just fades away, like so. So it doesn't, there's not a linear drop, you just need to remember the shape. So it's an exponential decay, a lot of things in physics are all exponential decays, the friction force is one of them.
Right, is it possible to get it swinging back and forth? Hold on, wait. swinging the bow on the top of the chord gently and shaking it backwards and forwards explaining how shaking the top end of the chord can make the ball at the bottom chord oscillate in a simple harmonic motion.
So this is a resonance question secretly sort of hidden in there. It's asking you how do things oscillate, essentially that. It's asking you how do things oscillate and the way to answer this question is you just need to say you need to drive it at its resonant frequency.
So I'm pretty sure in order to get a ball to oscillate in some kind of motion, you must swing the ball back and forth at its resonant frequency. You can maybe put brackets, the driving frequency must be the resonant frequency. This will cause the amplitude to build up, the amplitude to build up and make the oscillations visible. So if you don't drive something at its resonant frequency, it'll still wiggle but the amplitudes won't build up.
Yeah, and when you drive the chord of resonant frequency, the kinetic energy of your hand is providing energy to the system. And then the ball will cycle from gravitation potential to kinetic at the bottom. I don't know if I need to put that in there. I just chucked it in there because why not? Because it asks about energy transfer.
So it asks about energy transfer, so I just put in all the energy transfer that's in there. Last question. Simple harmonic motion requires a restoring force that changes in proportion to the size of the displacement.
Okay, so given the definition sort of, you just need to put in that it's linearly proportional to the size of the displacement. And it's negative. So if the displacement is one way, then the force needs to be the other way Yeah, resonant frequency is the same as the natural frequency.
They're just synonyms. They mean the same thing Right, so discuss what provides a restoring force when the ball is swinging in simple harmonic motion And your answer should describe all forces acting on the ball, explain how the forces change as the ball swings, tool vectors to show how the restoring force is produced. So the only forces acting on the ball are tension Well, start off with gravity Gravity is the thing that causes all the force to happen. Without gravity you wouldn't have tension because there'd be nothing pulling on the string. So the ball experiences the force of gravity, thus there is a tension force to obviously oppose that force of gravity because the string's not going to break.
When it's stationary by like just sitting there, they will be equal. When it's moving at the bottom, if it's moving it's changing direction because it was going from it was once at the top going down, now it's coming back up, so it's accelerating. Anything that's accelerating obviously has a net force greater than zeroes. The tension force at the bottom must be greater than the gravitational force. Has to be.
As it moves up to like this position here, the tension force has a vertical component which counteracts the force of gravity and it also has a component that points into the page. You can sort of break it down however you like. Not into the page. It points back the equilibrium.
like I've drawn in there. There's quite a few ways you can sort of break it up, but it doesn't really matter that much, as long as you sort of draw it like I've done there. And you can say that at the top here, you've got a force downwards of gravity, because gravity never disappears, it's always there.
So I've said the only forces acting on the ball are gravity and tension force of the swing. Force of gravity is constant by a swing, it is. However, the tension force depends on the ball's position.
bottom of the swing the tension force and the greatest is the ball considered undergoing circular motion i thought i'd chuck that in there just because it is at the bottom it's changing direction you could probably you could use the circular motion formula to work out the tension in the ball um the net force is non-zero so it's changing direction accelerating yeah at this point uh gravitational fg directly opposes the tension force so the tension force must be large the non-zero force so The only thing that I missed is you need to draw vectors of how this produces. You're going to have Fg down here and then you'd have obviously Ft here larger. At the top here you'd have, I can probably just get rid of that, that's just the component of tension that balances gravity. So you'd have that force up like so. That tension force doesn't actually need to be the same size vertically as a gravitational force because it's actually the ball is moving under its own inertia.
So it's decelerating downwards. You've got a bit sort of going on there. Do we need to draw a Vick diagram to support the answer? Yes, yes you do.
You need to have a component of the tension force. pointing back down towards equilibrium to show that there is a restoring force going on. Right, so that is the end of that.
Is tension force and gravity force smaller as the angle increases? So obviously the gravitational force never changes. The tension force should decrease.
Yes, it will decrease as the velocity decreases. So as it moves up the tension force should decrease. And then you've got components of gravity that are pulling down, which hold on, yeah it decreases to a point. It's actually kind of, I don't think they'd ask that in level three because that's actually kind of complicated.
Does a horizontal component of the tension provide the restoring force? Yes, yes it does. So in that case the tension force would remain the same.
Yeah, I'm actually sure. I don't think they could ask that in level three. I think that's a scholarship idea.
That's like above. I'd have to have a good think about that. Doug, you got anything to say on that? My brain is fried to try and think that one through. Sorry, I was just trying to type my answer.
At the top of my head, I would have said that tension would increase as the angle between vertical and the string increases. hmm it's going to have to provide a a vertical component but the ball is moving under its own inertia so it doesn't actually need to provide a vertical component because gravity is slowing it down so as it moves up gravity starts playing an effect um on the ball to decelerate it um moving upwards obviously and then there'll be a tension force there'll be a horizontal component well not as a horizontal component a component you uh perpendicular to the string which acts as a restoring force to bring the ball back in um yeah there's no way they'd ever ask you that um and uh because that's getting that's bordering into deriving the actual equation um and to derive the actual equation you've either got to do a little bit of algebra wizardry um or you've got to use calculus or you can use the brain gems um but that's like well above the scope of level three Right, so we've got maybe two minutes Doug, because we've got another stream running. Do you reckon we have time to quickly go over something? Yep, too late.
So this is an example of the actual exam I wrote a year ago, and why I wrote the questions I wrote. So when you're writing, like so I thought of just two simple ocean questions first, because the scene is the hardest. first question is always like it's sort of a show question so i want students to either show a spring formula um i'd give them one value and they have to like show another value um so you need to be able to rearrange the the pendulum formula or the spring formula because it's a really good question to ask it's like a starter and it's always usually achieved you can see here i put question b state the conditions were acquired for serious motion be considered simple motion of That's also a super common question.
It's a good one to ask because it's one of the takeaways you want your students to leave when they've done some vomit motion. They need to know the definition of it. That's probably, I think I've left that as an achieved. I think A was a merit.
C is an energy decay sort of question where they've got to draw the frequency decaying away. And what I'm looking for here is students knowing that the period of an object, or the frequency of an object, just depends, for a pendulum, it just depends on the length. For a spring, it just depends on the spring constant.
No other, like, friction won't change the period at all. And that's what I'm looking for, because it's somewhat counterintuitive. And then the last question, where you've got something that's released, where you have to use one of the trig functions to find out, like, where it's moved to.
Right, so, I think... Doug, can someone share? Yeah, right. So I think we're going to have to make that do. Thank you very much, Andrew.
I was just sketching a little diagram of the tension and weight question, and I think, yeah, I think you're right. Essentially, it's not going to matter for level three physics, but as a little exercise, I think the tension is greatest, like you said, in its central position of the pendulum. But that was a really cool question I had to think about for a little bit.
Massive thank you for your time and effort in putting this together and for the link to the resources you've shared in the description. And if you've got any other questions that you haven't had a chance to ask students, then please do head over to study it and pop the. questions in there and we'll get those answered before your exam.
All right, best wishes for your exam. Thanks again, Andrew. We'll catch you next time.
Ka kite.