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Franle Stud y (uid € Qr oblem 15 proc+-ic e0.6 KA40Physics QGQvestion Ccr 4 is hov led vp_6y alle along Curuko tvac k d ecriba x 2 / 40 C*y in e+evs ) Angy' (c V ve (octy 'Acreces 4+ Constant Ka + € when C0r Pa 5s es Poin + and 50 4h0 Va : 3.6 m ( $ OAd Ve 4.8 I ($ Sha w how O 50 (ve megnitvac Lo +0 | acceleration cov + WheM Ce " + 4 + which | be (ow fhe 4o p

franle Stud y (uid € Qr oblem 15 proc+-ic e 0.6 KA 40 Physics QGQvestion Ccr 4 is hov led vp_6y alle along Curuko tvac k d ecriba x 2 / 40 C*y in e+evs ) Angy' (c V ve (octy 'Acreces 4+ Constant Ka + € when C0r Pa 5s es Poin + and 50 4h0 Va : 3.6 m ( $ OAd Ve 4.8 I ($ Sha w how O 50 (ve megnitvac Lo +0 | acceleration cov + WheM Ce " + 4 + which | be (ow fhe 4o p



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(II) A particle is located at $\vec{\mathbf{r}}=(4.0 \mathbf{i}+3.5 \hat{\mathbf{j}}+6.0 \hat{\mathbf{k}}) \mathrm{m}$ .
A force $\vec{\mathbf{F}}=(9.0 \hat{\mathbf{j}}-4.0 \hat{\mathbf{k}}) \mathrm{N}$ acts on it. What is the torque, calculated about the origin?
components of the linear acceleration are:
$\mathbf{a}_{\mathrm{tan}}=\overline{\boldsymbol{\alpha}} \times \mathbf{r}$ and
$\mathbf{a}_{\mathrm{R}}=\vec{\epsilon} \times \mathbf{v}$

We know that Dark is given by cross product, off part and F on the used the dominant rule, which is given by a question 11.3 b to find this stock. So this will be goingto determining off the magics that parable. Right? So these are the directions. This will be the component off. The first factor it is, Sarge. And this will be the confidence off the second victim. It is force here. Make sure you include the signs on the directions. Now we find that the dominant off this metrics and this comes out to be 68 I gap this 16. Jake Gap, this 36 ski cap friend, the units don't hunger the units, so I forgot to mention them over here. So this will be last time. Snu done 30 meter times. New turn meter comes from the radius and Newton comes from the force. So this is the dark vectors, you know, I guess

So this is a little bit of an involved mathematical television. So strap in there are few patients here, acceleration equation, fluid equation and and uh general scale factor uh equation. Um And and so we start we start at equation 10. and the book which is Equation 10 which is one over R. Time's D. R. D. T. This uh oops This whole thing squared -8. Tiggy row. The three this whole quantity times R squared is equal to minus K. C squared. Alright. Row being the density D. R. D. T. Velocity of expansion of scale factor are of course scale factor and K. The curvature um term. And so this is um uh this this is D. R. DT um squared minus eight pi She row over three times R squared is equal to minus K. C squared. Now we're gonna multiply uh by our multiply by our um and then you get R. R. Cubed and our respectively for each of these terms. Okay then you take the time derivative, you take D. D. T. What you get is uh DDT of our times. D. R. Duty quantity squared minus eight Pi G row times three R cubed. You take the derivative of the left hand side walls to conservative on the right hand side. Dvt of minus K. C square times are. Um And so uh let's let's bring out these two terms on the left hand side, D. D. T. Of our time's D. R. D. T squared. Is it something minus D. D. T. Of a pi G row R cubed over three is its own thing. Note that got a lot of constants over here, write it down again. We've got a lot of constants that's eight pi G over three that are constant. That they go outside, they can factor outside this derivative and that's equal to minus K. C squared confidence time's D. R. D. T. Okay. Um and so uh let's write out these terms over here. Um First stop. You get so used to use the product rule obviously. So uh and the chain rule. And in this in the observations, the first stop. Um You have D R D. T squared. This thing comes out, time's D. R D. T. This will of course be the R. D. T cubed. Uh Plus R. Times D. D. T. Of D. R. DT squared. This is the more difficult term to work with but we know how to work with it. Um Plus A Pikey over three time's D DT of row are well are acute and that's equal to again minus K. C squared. Whoops. Casey squared the times times D. R D. T. On the right hand side. Ok, lots of terms here. Uh Let's uh that's simplify this further. Okay, Drd two cubed as a first term on the left hand side. Plus you have to which comes out because the chain rule two times are times uh D our duty two times our times of er duty that comes out of the chain. Roll the two times their duty stuff multiplied by DDT of D R D. T. Bear with me. Um Plus a pi G over three times DDT of ralph are cubed. Now, I'm going to replace this. All right, I'm going to replace this DDT of roe r cubed. Um Bye. Another term From the book. This is from equation 50. The fluid equation. This term is actually uh Oops, I um should have called us, I should have called us Well our are a cube inside and not outside DDT of real r cubed. And so this term from the fluid equation is just minus. Uh This time from fluid equations just minus P. C squared. Um And I realized that I have a side error. Told you this is a pretty involved problem. Gotta keep track of all these um All these signs minus pcs. This from the flood equation of minus P. C. Square times D. D. T. Of R cubed. All right. And that's equal to minus K. C squared times D. R. D. T. Okay. Um And then so finally we have the R. D. T cubed plus to our times. Uh D. R. D. T. Times D squared R. D. T. Square. That's this stuff here. Um Plus sure These these two these two minuses will multiply to give you a plus form plus eight pi G over three has p. Overseas squared times the DDT of R cubed and that is equal to minus cases square times D. R. T. T. Okay. All right. Now you do is you notice that there's a drd t roughly on every term except for this one. So you divide ah by D. R. D. T. By both sides by D. R. DT. What you end up with is uh d our duty squared um is equal uh the idea T squared plus to our T. Square are the T squared. This term drops out plus a pi G. P over three C square times DDT of R cubed divided by D. R. D. T. So that term uh doesn't cancel with anything. There is equal to minus K. C. Squared that the R. D. T. Drops up. All right. What can we do um From here? Further? So we first look at this term, We get eight pi gp over three C squared times You apply chain rule. Again, you have three hours squared times times D. R. If three are squares. uh Time's D. R. DT. And that's of course divided by D. R. DT. So what happens is that dif this term cancels out? Uh Yeah. And and then Uh we also know this term Casey from equation 10 in the book. This one K C squared minus K. C squared. Uh is equal two D. R. D. T. Quantity squared uh -8 Pi G over three. Time to roll our square. Okay, let's put it all together and change. Think now now you have um D. R. DT squared minus plus to our times. D squared R squared. Uh Yeah plus eight pi G. P over three C square times. We are squared Ups. two r squared. Not a whole squared just the R. Squared. And that is equal to uh D. R. D T squared -8 Pi G. Row R squared over three. As you can see this drd t square term cancels out. So we're getting somewhere obviously to our times D squared. R DT squared is now equal 2 -8 Pi G over three. This is a common term. Um Times row R squared. All right. Real R squared um minus the minuses equals this term can be pulled to the other side plus. So this becomes plus uh three R squared. She's over C squared. All right. So this is what we adopt with. And now you divide both sides by two. Are you get D squared? R DT squared is equal to -4 Pi G over three, divided by two times are mhm. By G divided by two times are ah times row plus three P over c squared. Um And that is the acceleration equation.

In this question, we have this raw assembly supported by buoyant soccer joins A and B. Um, the angular velocity omega, uh, we got is given to be five radiance. The second in a wind direction. The Alpha is given to be it. Um well, it really is this against square along the y axis. We once we need to determine the magnitude of the velocity and acceleration of point c at this instance. Yeah. So I'm not supposed to use some questions such as, uh, ee goes to we'll make a cross. Our and then we also have acceleration. Be equal to, uh, gotta across our plus. We'll make our cross, Omega Cross our He wants to use this to equation to determine the magnitudes off velocity and acceleration off point C. Yes. So four point c. We need the opposition factor for currency. So this is equal to tribe. My nest, your point fall, I plus your point tree. Okay, so you see is Omega Cross R c. But this is five J cross. Uh, 10. Yes. Well, I class tree K and something Softness, you get, uh, one quart size I class two okay. Meters per second so you can do with them by tone. For example, J Cross I j cross minus. I'd get K J Cross k. You get I so that many to off velocity. That's Quincy. You just be one point. Right square. That's two square square root. And this is equal to 2.5 m per second. Okay, so this is the many to form, you see. Okay, then my acceleration. So we'll be using a Christian, uh, across our C us. Oh, my God. Cross, We'll make a cross RC. So putting in the qualities. So this is it. Jay Cross one of the 10 minus. Well, I class tree k then, uh, that is that if this Ah, Okay, so, uh, put the Omega, which is by J Crosses, uh, 1.5 I plus two. Okay. Okay, so So this is easy to do it. Come back home, and then the finance. There will be a child point for by minus for country. Okay, you just once again square. And then the many to off the acceleration at point c will be chopped foursquare class 4.3 square. Swear it's and this is a 13.1 m per second squared. Okay, so this is the answer for no acceleration at point C. The magnitude

Hello students in this question we have to determine how to solve the study state function. Study state function statistic function of emotional forced oscillation. Okay, forced consolation. So we can take the system of governed by the equation acceptable or plus omega notice clear X. This is equal to if not sine omega t. We have not equals two amplitude of four. So this can be written as if not by them. Okay and substituting a solution X equals to a sign omega T plus Elka. Ok So we can write from here that -0mega Square same omega T plus alpha plus a omega square sine omega T plus alpha. This will be equals to the eggnog. Sine omega t. Okay so from this equation after solving it further so we will get that sign omega t course alpha and a omega not despair minus omega square minus f. More plus of course omega t. Sign alpha. And we're clear by a omega not squared minus omega is clear this is equal to zero. Okay, so the situation will be equals to zero when the coefficient of this sine omega T and cost omega t both will be zero. That is these values are equals to chill. Okay, these are the values which must be equals to zero. So hence we give we from here acting as and to a it is equals to have not developed by omega not square minus omega spare. An elf equals to zero or L five equals 25 So after substituting zero, we will get that equation ax equals to assign omega t. A sine omega T. And after substituting by here. So we will get that a sine omega t also, so we can write that the test solution both are uh for same and closing force both are same. Okay, so we can use either a sine omega t or we can also use a cost omega T. Okay, this f not cause omega T or f not sine omega t. Both can be used. Okay, thank you


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