So we have started chapter 22 of Hibler's book on vibrations. The first section was undamped free vibration and the resulting differential equations. So, I mean, if you want to give a complete definition of undamped free vibration, we can just say that the resulting differential equation can be written in this format.
One. Second order differential of the variable plus, like, multiplied by constant plus another constant multiplied by a third, a zero order differential equal to zero, or just a second order differential, second order linear homogeneous differential equation that does not include the first order term. So it does not include x dot. Homogeneous also means the right hand side is zero. So this is the resulting differential equation for this system was found and it was in this format.
So second order differential, second order linear differential equation which is also homogeneous. The right hand side is zero. This system, for this system we studied that the differential equation If the y is measured from equilibrium position, differential equation will just turn out to be exactly like this. But if we measure y. So this is the differential equation.
And as I mentioned in this book, it doesn't give you the method to find the solution. It just gives you the solution. It says that in the next book, in our main textbook, Ogata's book, we'll use Laplace method to solve all these types of differential equations.
So it says the solution is in this format. Constant a and b are determined by knowing the initial velocity and initial position of the mass. For example, you know what was the initial position of the mass, how much this, when before you release the, before your time zero. For example, you have pulled the mass to a specific position or like pulled it down and then released it or pulled it down and also gave it the initial velocity and released it. So by knowing the values of initial position x1, x0 x at time 0 position at time 0 of the mass x1 and knowing the value of initial velocity velocity at time 0 as v1 the solution would be in this format okay this is just this was just a review of last session any question okay so let us move on to the next slide Okay, so again, a review of last session.
This is our differential equation. Omega n can be any variable, any variable that is omega n, we can solve this differential equation. In our special case where we have a mass and a spring system, our omega n turned out to be square root of k over m.
We'll find out that this, why is it omega n squared? omega n will actually we'll find out later that omega n will be the natural frequency of the system We almost finally found out that when we look at this equation Because this is the solution of our system. We already can Guess that omega n is a natural frequency of the system harmonic motion But this is the summation of a sine and a cosine function Right now we are going to show that this solution can be converted into this format meaning equation 22.3 can always be converted by triagonometric relations into equation 22.9 and there we'll see that yes it is absolutely just a sine function just that a harmonic function sine function with um like a phase shift It could also be cosine function.
Sine and cosine function are the same. They are just these phi changes by 90 degrees to convert from sine to cosine. So it doesn't matter whether it's sine or cosine.
But if it was cosine, that phi would be a different phi, 90 degrees different from this phi. So here we are showing that this format of the solution, the solution that we have found can be converted to this format which is easier to analyze. So the general solution will be in the form of this.
Equation 22.3 may also be expressed in terms of simple sinusoidal motion. Just this simple sinusoidal motion. To show this, let us do this trigonometric calculations. Let us assume that A and we define two new variables C and phi. How do we define them?
We define C as this and phi as this, then A and B will be this. Or let's say that we define C and phi such that these two relations hold. We can always find phi, C and phi such that these two relations hold. And the way to find them is this.
You can consider C to be a square root of A square plus B square, phi being tangent inverse of B over A. a and b would be this. So we're introducing two new variables.
Instead of a and b, we use c and phi. And then replacing these, um, 22.7 and 22.8 into 22.3, um, it yields a would be c cosine phi, b c sine phi. So now we can factor out c. It would be cosine phi sine omega nt plus sine phi cosine omega nt. And this would be a summation of sine of the summation of two angles would be sine of omega nt plus phi because we have this relation in trigonometry sine theta plus phi sine theta cosine phi plus cosine theta sine phi therefore this would be equal to c sine of omega nt plus phi so now instead of that solution we are going to use this solution because it is simpler to talk about it just one sine function.
So any questions about this? Okay, now we are saying this is our differential equation. This is our solution.
Now let's analyze the solution and see how it behaves. We plot the equation. We plot the solution. If this equation is plotted, this equation, c sine omega n t plus this is a plot of c sine omega n t plus Okay, now we are saying this is our differential equation.
This is our solution. Now let's analyze the solution and see how it behaves. We plot the equation.
We plot the solution. If this equation is plotted, this equation, C sine omega nt plus phi, this is a plot of C sine omega nt plus phi. The graph shown is obtained. And we consider the horizontal axis here.
It is considered. to be omega n t normally we consider it to be t right and plot this but considering it omega n t would make uh some of these variables simpler for example if it was t then this angle would have been um for example phi over omega n here it makes it simpler this would be phi over omega n But the good thing would have been the cycle would have been just In this axis would have been just tau. I mean doesn't the book prefers to consider the horizontal axis being omega nt Okay, doesn't matter just a multiplication of but the constant multiplied by t 2t 3t anyway anything Okay, the maximum displacement of the block from its equilibrium position is defined as the amplitude of vibration So the you can easily see from this relation that the maximum value of X would be C because maximum value of sine is 1 minimum value of a sine function is minus 1 so this where X changes between plus C and minus C because maximum is positive 1 so maximum is plus C minimum is minus C so this is the amplitude so this is the range and X will not exceed positive C it will not drop below minus C from either the figure or equation 229 the amplitude is c the angle phi is called the phase angle since it represents the amount by which the curve is displaced from the origin when time equal to zero so when omega and t is equal to zero or let's say the other way when does this value in the parenthesis become zero like because this is the start of sine function sine of zero so when when does this thing become zero it happens when omega and t is equal to minus phi so this is the when omega and t is minus phi that is our zero this becomes zero or the way it says here when does what happens when t is equal to zero when t is equal to zero it will be sine of phi so the value here is c sine of phi yeah this is another way to say it so when t is zero becomes c sine of phi yeah the start of the curve. Note that the sine curve equation 22.9 completes the one cycle in time t equal to tau when omega n tau equal to 2 pi.
So when this value when the time increases from zero and goes to infinity how does what happens to the sine function? It's a cycle. It starts from Phi, sine of phi when the time is zero, c sine of phi.
If t changes by, if this value changes by 2 pi, it's same as being zero. If it changes to 4 pi, it's same as being zero because when you add 2 pi to this angle, it just plus 360 degrees. The sine of that angle will be the same.
So from here, or from let's say from here if you change omega and t by 2 pi you just repeat that same point you come to the same point or here if you move by 2 pi you just come to the same point so this is one cycle omega when omega t is 2 pi and we call that t tau so omega tau is 2 pi the t that makes omega n t equal to 2 pi is called tau. So tau is 2 pi over omega n. And I haven't talked about the frequency, the important one. The frequency is, so let's talk about it here.
The frequency f is defined as the number of cycles completed per unit of time, which is the reciprocal of the period, that is, So frequency is one over tau. So tau means how many seconds does it take to do one cycle. A frequency is opposite of that. It says how many cycles is completed in one second. So just reverse of tau.
So how many cycles per second. So how many cycles are completed for one second? Cycles per second.
Or for tau it would be second per cycles. Cycle has actually unitless. Cycle is 2 pi radian is unitless.
So tau is second or say second place per cycles. F is cycles per second. Unit is also opposite.
So this is frequency and you see that it is not omega n. F is 1 over tau is omega n over 2 pi. So why did we just say keep saying natural frequency natural frequency?
Why did we call omega n natural frequency? And we see that The frequency is actually different from that. The frequency, this one is also frequency, but it has a different unit.
They call it, sometimes they call it angular frequency. In some books here, in your book, and I think in the next book, it just uses the same name for F and omega n. It just calls both of them frequency. But normally they call omega n, I mean, it's going to be precise.
They say angular frequency, and the unit is radians per second, not cycles per second. So one cycle is 2 pi radian. One cycle is 2 pi radian. But in your book it gives them the same name although they have little different units. This one is frequency F.
The unit is cycles per second, means 2 pi radians per second, the unit for F. The unit for omega n is simply radians per second. So radians per second divided by 2 pi. Okay, so, um, omega n is the natural frequency of the system, and why do we call this frequency? Why do we call it natural frequency?
So, f was frequency, uh, is omega n over 2 pi. A good name for omega n would be maybe angular frequency, but we also call it the same frequency. Instead of saying how many cycles does it complete per second, we say how many radians does it move per second. Radiance in terms of the solution i mean the sine omega nt here we don't have i mean for the mass and a spring damper we don't have a rotation to call it radiance but we are talking about um the frequency of this sine function so omega n is also frequency of the system so it's the frequency of vibration okay let's say it's the frequency of vibration of the system um now why do they call it natural frequency? what is the term natural come from?
natural frequency means if you just have a free vibration the system for free vibration it vibrates at that frequency that means that is the meaning of natural frequency because for example so if i move this mass down and I release it, it will naturally vibrate at omega n, the frequency omega n. So it's the natural frequency of the system. But I can also apply a force to that system which has its own frequency.
For example, assume that, let's say, I somehow apply some force vibration a sine omega nt assume that i'm applying the force to that mass and making it vibrate at a specific frequency that is the frequency of the force that is different from the natural frequency of the system later we'll see that if the frequency of the force is equal to the natural frequency of the system we'll have something called resonance and which will call systems to break um as we will see later so this is natural frequency of the system is omega n here in this what i have written here omega would be the frequency of the force omega can be anything what omega n depends on m and k the square root of k over m if that omega later we'll see that if if i apply a force and that omega that i'm applying with the same frequency as the natural frequency of the system then the system will have resonance and It will break. Yeah, if there is no damping, theoretically it will definitely break because amplitude of vibration will go to infinity theoretically, but we always have some kind of damping which the system can survive the resonance and not break, but it will have dangerous vibrations at resonance normally. When the frequency of the force omega would be equal to the natural frequency of the system omega n, which would be a square root of k over m so we call it natural frequency because it is something that is inherent with the system and in free vibration the system will vibrate with that frequency any question okay Okay, we have another example for undamped free vibration. So we have already finished and gave you the solution for, gave you the complete solution for undamped free vibration, which is a very simple format, C sine omega NT plus phi.
Now you're studying just another system which also has the same governing differential equation. This system, which is a simple pendulum, if we write its governing equation, we'll see that the governing equation would be same as the governing equation for undamped free vibration. We had x double dot plus omega n square x equal to zero.
Here we'll see that. Yes, we also have x double dot plus. omega n square x equal to zero where omega n here is a square root of g over l so that's why we're studying this example so it doesn't have to be a spring and damper we have different types of system that have same governing or same differential equation let's derive the governing equation for the system so determine the period of oscillation for this Let's derive the governing equation for the system.
So determine the period of oscillation for the simple pendulum shown in this figure. The bob has a mass M and is attached to a chord of length L. This is M L. Neglect the size of the bob. Okay we'll talk about that a little bit later why it might be a little bit important. Not important but just a minor note that is interesting.
This book Hübler has a good emphasis on dynamic properties of systems and is very precise in regarding that. Okay so the book derives the equation of motion of the system by using tangential coordinates. I am using a different method to find that same governing equation. So let us first look at the way the book derives that. So it uses as a reminder we had, I'm just, this is, this slide is just a reminder of previous chapters in Hibbler's book, chapter 13. Let us first look at the way the book derives that.
So it uses as a reminder we had, I'm just, this slide is just a reminder of previous chapters in Hibbler's book, chapter 13, for tangential coordinates. I'm giving you another way to solve it because maybe you always don't remember that but this is the one I remember better normally. So first let us look at how the book is doing that. So I should look at what was in chapter 13.5, equation of motion for normal and tangential coordinates in uh Nibbler's book.
These are the um governing equations. Sum of forces in tangential direction is equal to mass multiplied by acceleration in tangential direction. Sum of forces in the normal direction. So you have a path. I mean for curvilinear motion you have a path of motion for your particle or your 3d object.
You have a path of or the center of gravity of your 3d object. You have a path for it. And on the path at any location you can define a tangent, you can define a coordinate system with one axis tangent to the path t and the unit vector in that direction ut and one direction normal to the path or normal to t and toward the direction of curvature of the path. direction n and the unit vector in that direction un and as you see this coordinate system is not fixed it moves based on the location of the particle wherever it goes the definition of t and n changes so but in the book it has shown that although the direction of t and n change all the time this would be the governing equation some forces in t direction would be mat some forces in n direction would be man sum of forces in a perpendicular to the plane would be zero when you have motion in the plane okay so we use this equation the book has used this equation here and write this governing equation.
Okay so we use this equation the book has used this equation here and write this governing equation. I will do it different in a different method the one that I remember I mean more easily. So sound forces in t-direction is m a t as the book says so what are sound forces in t-direction for this?
I have here drawn the free body diagram in t-direction If I consider this direction to be t direction in that direction, the same direction as theta, going that way, so the force would be minus mg sine theta equal to m a t. a t would be the second time differentiation of the, let's say, what do we call s, the position on the path, what is the name of s, what do you call it? like the position on the path like here also they have the same thing s they define some origin for it and it's just the position on that path it's just a scalar it's not a vector it's just when you say what is s you can know what is the position on that path and then it has been i mean a t is the second time derivative of s with respect to time so s double dot furthermore s can be replaced so what is s S in this picture is L theta. S is L theta, therefore, A T is L theta double dot.
Therefore, this equation reduces to ML theta double dot plus MG sine theta equal to 0. Dividing both of them by M and L, we have theta double dot. plus g over l sine theta equal to zero. So this is the governing equation for this system.
I'm deriving the same governing equation by using a different method. This is the one I remember more easily. So I'm using this Newton's second law for a three-dimensional object or a rigid body. So second law, this is Newton's second law, the acceleration of center of gravity of a rigid body.
Deriving the same governing equation by using a different method. This is the one I remember more easily. So I'm using this Newton's second law for a three-dimensional object or a rigid body. So second law, this is Newton's second law, the acceleration of center of gravity over rigid body is proportional to the vector sum of forces acting on it and it is in the direction of that vector sum. So sum of forces is MA but acceleration of center of gravity.
For a particle it's just MA. but when you have a mass that also rotates that a would be the acceleration of which point on the mass because the object is like an object is rotating and moving and so you apply a force so this force that you applied causes what is that a so f equal to ma but a is acceleration of which point because different points on the mass are moving differently that would be acceleration of center of gravity of the mass and we have a second relation for For particles, we don't have this relation. For particles, just sigma force equal ma.
For rigid bodies, it is F equal ma of center of gravity, and we also have additional equation about the rotation of the mass, which said that sum of moments about center of gravity would be the mass moment of inertia of the object about center of gravity multiplied by angular acceleration of the object, alpha. And I about CG alpha. So if you are applying some forces on this mass, you find the sum of moments of those forces above the center of gravity of the object, and you can find alpha from there, because you also need to compute I about the center of gravity of the object, and then you can find alpha.
Angular acceleration that is a result of applying all these forces. And there is also a note in your book, in every book, that this A moment equation can also be written about a fixed point. So either center of gravity or a fixed point.
If it is neither center of gravity or fixed point, there is a different relation that you had in your book. It had some additional terms here. But if it is center of gravity or is a fixed point, you should refer to those chapters. If a point O is a fixed point, not moving, it is exactly what I'm borrowing from Hibbler's book, previous chapter.
You can refer to those chapters. if point O is a fixed point and is not moving it can also be proved that moment about that fixed point if one of the point normally it's an object a point that is fixed on the object it can be any fixed point but normally it is one because you want to find the angular acceleration of that object so if O point O is a fixed point It can also be proved that sum of moments acting on the object about that fixed point is equal to mass moment of inertia of object about that fixed point O multiplied by angular acceleration of the object alpha. So I'm using this relation instead of using tangential coordinate that your book is using I'm using that relation sum of moments acting on this system about this fixed point O.
I could also write about center of gravity. So I'm using this relation. Instead of using tangential coordinate that your book is using, I'm using that relation.
Sum of moments acting on this system about this fixed point O. I could also write about center of gravity but that relation will also result in the same governing equation. But that sum of moments about the fixed point O is equal to I mass moment of inertia of this system or this mass and the string about point O the string has no mass so mass of mass moment of inertia of the Bob about point O mass moment of inertia of a pole not about the center of gravity but about point O multiplied by theta double dot its angular acceleration I'm using this equation I refer this one so what is the sum of moments of the forces acting on the bob about point O. about point O what are the moments of forces. so this force doesn't cause any moment about point O because it passes through point O.
this one causes a moment and is equal to minus mg sine theta multiplied by L. what is the I of this bob about point O? you can refer to the back of Hibbler's book you see that you are neglecting the size of the bob therefore you are considering one mass at this point m just one mass at this point it's ml squared it's ml squared i about o c g this should be actually let us check that m l square actually um i don't think it is in that table I don't think it is in this table and I should just use the um something called for for a particle that is like it so it says that neglect the size of the bob so you assume it's a particle it's just a one point so um for a point the i about its center is zero because it has no size and according to parallel axis theorem which i might have here i don't have parallel axis theorem here i have an included parallel axis theorem okay here it is a parallel axis theorem i have it here so you know parallel axis theorem from dynamics I is I about CG plus MD square.
So for any object, if you want to find the I of that object about any axis, you find the I of that object about its own center of gravity plus mass of the object multiplied by the distance of the two axes. So for a particle that has like a point mass, I is zero about its center of gravity, like a particle that has no size. It says neglect, that is what it said. It said, neglect the size of the bob in this problem. So I of this object about its center of gravity, it's assumed to be zero, assumed it has no size.
So I about CG is zero. So what is I about O? What is I of this object about O? That is what we need here.
It is I about CG plus M multiplied by parallax estuarii, M L squared. So I about O is equal to I about CG plus MD squared. Distance is here L.
So it's 0 plus ML squared. If it hadn't asked us to neglect the size of the Bob, you could also solve the problem. You could find the I about CG of this one and then add ML squared to it.
So it would be I of a sphere about its center of gravity, which we could have looked up from here. It would have been like a two fifth of m r square plus m l square. If we hadn't neglected the size of the bob it would be this value, I of this sphere about the center of gravity two-fifths of m r squared, and we should know also the radius of that bob, and then add m l square to it. Here it is m l square plus zero because I about Cg is assumed to be zero because we are neglecting the size of the bob.
Or You can, when it says neglect the size of the Bob, you can assume in other words. If it says neglect the size of the Bob, it means assume that the radius is zero. How about that? So I of this Bob about is CG, according to this formula, is zero because R is zero. If R is zero, I about its center of gravity is zero.
MR squared, M multiplied by zero, squared is zero. so it would be on zero plus ml square as shown here any question about so i'm just reviewing 0 theta double dot. So it should be about point O is M L square because I about any point is different. I about point O theta double dot M L square theta double dot and the resulting differential equation just canceling the L would be same as what we found here.
So this was found using the tangential coordinates. This was found using just simple Newton's law. four rigid bodies and using just the cartesian coordinates x y z not the tangential t and normal coordinates just cartesian x y z fixed coordinates without using tangential coordinates so anyway the resulting differential equation would be the same okay okay let's move on okay now we are here we are at this state So we have found this differential equation, but this differential equation is not in the format that we want for free undamped vibration. It should be theta double dot plus a constant multiplied by theta equal to zero.
This one is a sine theta. It's a more complicated differential equation. So, in fact, it is a more complicated differential equation.
However, for a small displacement, we can approximate this by this differential equation, because for a small displacement, theta... sine theta would be almost equal to theta in radians not in degrees so theta in radians is equal to sine of that theta for a small displacement if theta is in radians not in degrees of course how many radians are used for those small displacements how many normally i remember six i don't know there was a good i remember there was a good approximation about six degrees actually yeah but yeah theta is in radian yeah i remember it was about six degrees so very small displacement yeah it just starts to deviate so you cannot say which range of it that you have the more you have the more theta the more you are away from reality so uh for a small displacement however sine theta is almost equal to theta if theta is measured in radian Therefore, we replace sine theta by theta, and then we have the same differential equation as we had for our spring and mass system. This was our differential equation for mass and spring system. This is our differential equation for the pendulum, but for small angles of theta approximate, that one was precise.
So, Now we say that okay we have the same differential equation therefore the solution will be the same we just say if omega n is a square root of g over l this is our differential equation then the period then we just use the solution that we have found previously we say that tau is equal to 2 pi over omega n so omega n was the square root of g over l so tau would be 2 pi over a square root of g over l or 2 pi multiplied by square root of L over G. So this is our period and all the solutions also usable here. Of course for small values of theta and the theta that is found from this equation would of course be in radian whenever in degrees it's not even a standard unit for angles. Okay any question?
You asked for what angles theta? Is it a good approximation to consider it? to be equal to sine theta and I mentioned that it just starts to deviate from being precise as you move away from zero so I checked something I know if here at the google chrome okay google they have google chrome here there was there was an app on google chrome where it could actually plot let me see okay so here we can plot function in google chrome so I plotted sine of x so this is sine of x it plots it right away i can also oh i can do two plots tangent of x how about that what does it understand tgx or tan oh it doesn't understand tg so you probably tangent of x okay so these ones are tangent this is sine curve so if we zoom into it um you see that they are both So they both have a slope, they both pass through zero and they have a slope of 45 degrees. I mean, when you get close to zero, these two curves, sine and tangent, sine of x and tangent of function, they both approximate, they become close to a line which has a 45 degrees. So actually the horizontal axis X is equal to sine of X for small values of X or sine of theta and you can check how different they are for example if you zoom into it you see that for example when X is 0.01 sine of X is almost 0.01 as you move away the difference become greater so it just little by little it starts to get away from being precise okay we have a next example this is also rotation of a rigid body so it is anyway even if i wanted to you see even if i wanted to not in you may think oh i wasted a lot of time just giving you a new method but this method actually has to be used for the next this is normally more generally can be used in more problems so in the next example problem that we have it is using that method.
We have a vibration of a rigid body like this about like the horizontal rod so that horizontal rod is acting as a like a torsional spring you know that the 10 kilogram rectangular plate shown in this figure is suspended at its center from a rod having a torsional stiffness k equal to 1.5 newton meters per radian Determine the natural period of vibration of the plate when it is given a small angular displacement theta in the plane of the plate. So sum of moments about point this is for if O is a fixed point we have this relation right I just showed that if O is a fixed point sum of moments about point O is I O alpha so here point O is a fixed point so sum of moments about point O is IO alpha and we are our free-body diagram is just this mass so what are the forces acting on this mass this is our free-body diagram the only force that is acting on this mass of course there is a weight that is acting on it but some of you are talking about moments about point O and we are considering moment on this plane that W doesn't cause any moment about point O in this direction the only force causing or the only moment about point O is the moment due to the twisting of this rod. So this rod is twisting and that is actually that is acting as a torsional spring. So here I have talked about so what is linear spring. For a linear spring force is equal to kx for next session we'll talk more about it but just quickly go over it for a torsional spring Torque is just a matching equation.
In dynamics, we have for linear motion, we always have a matching equation for rotational motion. We have F equal to ma, matching that would be T equal to I alpha. Force, torque, mass, mass moment of inertia, acceleration, angular acceleration.
We also have relations for work and power that we'll talk about later, but for linear spring, force corresponding, I mean matching that is the torque. X, which is linear displacement, matching that would be theta, angular displacement. And matching K, which is the stiffness, the stiffness of a linear spring, this is the, let's say, let's call it capital K. It would be the stiffness of a torsional spring. So, a stiffness of that torsional spring was given to us in that problem here it says that it says this rod has a torsional stiffness of 1.5 newton meter per radian and the unit here is newton meter per radian or a torsional spring so if we sum up moments about 0.0 is IO alpha the only moment in this direction is the moment due to or the torque due to the twist of the torsional spring so if it is twisted by theta let's say it is twisted by theta then what torque or what moment does that rod exert on this mass it would be an opposing moment or torque in that direction right so m would be k theta and it is minus k theta because we are considering that direction to be positive so minus k theta is equal to i o theta double dot which is angular acceleration therefore the resulting governing equation would be this so completely without any approximation For a pendulum we have approximation but this one we have no approximation directly.
We come across a differential equation which is exactly in the format that we wanted for undamped free vibration. Omega n square, x double lot plus omega n square x equal to 0 and omega n being a square root of k over i about o. And all the solution apply but before that we are here computing i o. Io is the mass moment of inertia of this rectangular cube about its center of gravity that can be just picked up from the appendix on the back of the Hibbler's book which is when you find the mass moment of inertia when you look them up from the back of the book you should pay attention with respect to which point it is computed so I think because we have two examples here let me go and check it back cover of the hitler's book i think i should have it somewhere so um let's look up the value of i for that object this one is a thin plate is it perfect or not we don't have any better i don't think we have any do we have it for uh cube or rectangular cube it looks like this is the best that we have for a thin plate that is what we have here is it thin really i mean it's not giving us a thickness does it say it is thin we should ask whether it is thin or not thin this value would be the same because this one is about an axis so it doesn't for this value for ixx and iyy it might matter but for IZZ I'm pretty sure that whether it is teen or not teen that same IZZ is correct so I think it is precise whether it is teen or not that's my guess you can you can check it for on internet you can check IZZ for tick plate I'm pretty sure that it doesn't matter whether it for this one doesn't matter for IYY and IXX it might matter but for IZZ I'm pretty sure it shouldn't whether it is thin or thick and it says it is 1 12th of m multiplied by a square plus b square which was used in that example and you should always pay attention i is with respect to which point and which axis both the point is important and axis is important this one is with respect to center of gravity and this type of axis so you should always pay attention to the point and axis to pick up the correct one both point and both axis so both the point and the direction the other