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3 - Newton'$ Second LawDifferent torques T are applied t0 solid object of moment of Inertia [. The following table contains the data about the torque applied ...

Question

3 - Newton'$ Second LawDifferent torques T are applied t0 solid object of moment of Inertia [. The following table contains the data about the torque applied and the relative angular acceleration of the solid object.FOrula (Nw 3.35 Nphaudsec)6.9693115.0320.5619.79224230.70Plot the torque- 20. Is the Braph linear? What is the numerical value of the slope? 22. What are the units of the slope? 23. What physical quantity does the slope e represent?

3 - Newton'$ Second Law Different torques T are applied t0 solid object of moment of Inertia [. The following table contains the data about the torque applied and the relative angular acceleration of the solid object. FOrula (Nw 3.35 Nphaudsec) 6.96 931 15.03 20.56 19.79 2242 30.70 Plot the torque- 20. Is the Braph linear? What is the numerical value of the slope? 22. What are the units of the slope? 23. What physical quantity does the slope e represent?



Answers

An object whose moment of inertia is $4.0 \mathrm{kg} \mathrm{m}^{2}$ experiences the torque shown in figure 12 26. What is the object's angular velocity at $t=3.0 \mathrm{s} ?$ Assume it starts from rest. (FIGURE CAN'T COPY)

In this question we have to find the talk. We are given the force here which is minus three X plus five plus five plus five which is a clear point A. have taken her to stay with someone Express three or crazy. Now we have to find in the first part we had to find it all which is acting above point 0 10 and zero. And I have considered your test point B. We have to find the distance from B to A. From B to a distance vehicle. Some kindness will be equal to A minus. We have taken it to the R. And the a minus three distance will be equal to simply 7 0. Someone next. Um I understand will be equal to -7 X -7. Y. And one man is zero will be able to one G. We have the force here no talk. It will be close to our process and we have E. G. Yeah you should know that our process condition if I had a determinant here 7- Salmon everyone and the thought is minus three one and fight yes the table predominantly and I let me confirm the value of the forces my issue line and if I will I endure I simply must take that days and minors this if you notice it will be equal to And 35 -36. And mine is J into jane too 25 years three years 28. Last day into seven minus 21 was 14 I'm just going to anyone else. We have the dark here at -36. I my interview gee minus 40. Yeah. And you were right there. You're not here. This is the answer for the first one in the second part. No. And the first part we have this point B as zero dan and zero. That this is done in the second part. They said that we are going to talk about origin. Origin is always 000. So if we take the art distance from origin 2.8 it will be simply I would write it. Yeah. I am traditional leader. You've actually because we have to subtract zero from it and it will recently and it is now I was asked the brain wants to I gave Yeah 71 71 minus To be one inside. I think I could hear because yes I would look it taking the determinant. I get sustained -1. It's 14 minus J Generally going to 35 plus three. Yeah. And Kay will be close to seven last night. Seven blocks. Nineties 16. We have the answer here. The dog is 49, 30 years 16 K. This is the answer and you will right the urine. Adults don't hit

Has been no turkeys to find that submission of I into adverse unit of torque. It's called the product of unit of Yeah, Moment of inertia and unit of angular acceleration. It is K G meter square into radial per second squared. So it is to me kg meters square for a second square so you can write K G meter for second square into meter. This is the neutral, so it will be new talent to meter, that's all. Thanks for watching it.

So we used the determinant rule equation 11.3 bait to evaluate the dog here. So Cisco and by the cross product off art and f on odd you called though I see your point to like I cap Let's zero point 335 meters. J Cap Andi resolved the components off fourth in these directions. So we're given that the magnitude of force is 250 Newton. So the confidence will be a magnitude times course off the anger plus the horse magnitude or force times sign off them. So this is X component and this is why company on We used these two equations to find the determinant. So this will be done equally too. I give you can get gap. I see your point. Wait, which is the component along I direction suit a 0.335 There is no confident in the K direction and this is to 15 goes 32 degrees. Do 15 scientist too terribly. Z component is zero on me that is the unit off and you don't miss a unical force on dhe solved. This dominate and we get the answer the dog, It will be kept This means that the dog has a magnet. Dude off 27.6. You don't dines me dead on Dhe the point in the negative Z deduction.

And this problem were given a position vector as a function of time. Um that describes the motion of a particle of mass three kg. So the position vector is a is a again a time dependent vector. And they asked us first for the um for the torque acting. Um Well they ask us for the torque acting on the particle. Alright, sub, there's a couple ways we could do that. Um We can either say that. Now. One thing you have to be a little careful about is that you can't just say the torque. Um Let's see here, I'll do this in red because this is not correct. The torque is not equal to our cross em eh That is not correct because what is correct is the torque equals DDT of our cross M. B. The angle madam. And then we have to use the product rule. So that is um um we can pull the em out if that's not a function of time. And so then we get V Cross with V. Plus M. R. Cross with a, well what we see here then is that we actually get V cross with V is zero. Right? So it turns out that in fact the torque is this, so this is actually correct. Um But you can't just you got to remember when you that you have to remember that you're taking this whole derivative here. So a couple of things we could do we could either calculate the acceleration across that with our or calculate the angular momentum first and then um and then take the time we were devoted and that's what I did. Again it's either way you should get the same answer. But again it's well this is correct. You've just got to remember that T. R. Is a function of time too. So it is correct just because it happens to be that you get re crossed with me here and that term drops out. But again, just writing this down naively, um might get you in trouble. Um So let's see here, we have um our angle momentum, our cross with the So we worked through the cross product and that turns out that the stuff cancels out here, which is nice. So it turns out that the angle momentum is simply in the K direction and it's 2040 squared in the Z. Direction and then the units are kilogram meters squared per second. So that's nice because the cubic term dropped out. Well it's not really nice just we could do the problem. That wasn't the case. But we just get the we just get the square term. So That's because we have 48 t. cubed here and then we have 48 here and then there's ones up canceling out. So that makes this fairly easy, so we just to take to find the time from time derivative of the angle momentum. Um that's just 4080 in the Z. direction. And that has a unit of Newton's nuclear tens, meters. Um Obviously we've converted some stuff here to Newton's, so let's see here. They wanted the magnitude, so the magnitude of the angular momentum Is 24 he squared. And so that's obviously increasing in time. So where whatever this whatever forces causing there's some kind of force or torque acting on the particle. Well, if it's a particle there's some kind of force acting on it that's generating a torque that's causing the angular momentum to increase with T. Squared.


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