# KINEMATYKA WZORY PDF

Wzory kinematyczne ruchu obrotowego. O nas. Transkrypcja. David explains the. mechanika dział fizyki zajmujący się ruchem, równowagą i oddziaływaniem ciał. mechanika klasyczna opiera się na trzech zasadach dynamiki newtona i bada. w opisie kinematyki oraz dynamiki ukła- dów korbowo-tłokowych . KINEMATYKA UKŁADU KORBOWO- jest od kąta α i dane jest wzorem (3), (5). The shift of. Author: Dourn Fehn Country: Denmark Language: English (Spanish) Genre: Politics Published (Last): 1 April 2005 Pages: 178 PDF File Size: 5.66 Mb ePub File Size: 11.71 Mb ISBN: 968-4-62358-208-8 Downloads: 33923 Price: Free* [*Free Regsitration Required] Uploader: Malahn And we know omega initial was zero. And if you solve this algebraically for alpha, you move the 40 over to the other side. So I’m gonna put a zero squared. We’re about to use these. We could say that omega final is gonna equal omega initial, that was just zero, plus the angular acceleration was 30, and now that we know the time we could say that this time was 1. So let’s say you had a velocity versus time graph and it looked like this.

So this whole term is zero. So that means that the area under the curve on a omega versus time graph, an angular velocity versus time graph is gonna represent the angular displacement.

We know the initial angular velocity was Let’s do a couple examples using these formulas, cause it takes a while before you get the swing of ’em. So what do we do?

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So that’s what we can say down here. So it said that it revolved five revolutions, that’s the amount of angle that it’s gone through, but it’s in weird units. Why is it negative? Because we solved for the time, we know every variable except for the final angular velocity. We’ve got a kinematuka that relates the speed to the angular speed.

Kinematykaa always need three in order to solve for a fourth. Similarly the area underneath the curve on a velocity versus time graph represented the displacement.

## Moment siły i moment pędu

So this time we know omega initial 40 radians per second. You too the square root. Instead of V initial, the initial velocity, I’d have the initial wzorj velocity. So that’s 40 pi radians. So what do we know? You ended up with seconds squared on the top.

Which variable isn’t involved? It tells us this 30 radians per second squared. In other words, if something’s speeding up, you have to make sure that your angular acceleration has the same sign as your angular velocity, and kinematykx angular velocity’ll have the same sign as your angular displacement. And it’s omega final. We wanna know the time. I don’t wanna use that one cause I wouldn’t know what to plug in here and I don’t wanna solve for it anyway.

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Let’s do another one. We could use those. And the next part asks, what was the angular acceleration of the bar? It was going meters per second. I’ve got my three knowns and my one unknown that I want to find. This is in units for revolutions. So let me get rid of all this and let’s tackle this problem.

That means that five revolutions would be five times two pi radians, which gives us 10 pi radians. I was neither give the time nor was I asked to find the time. So recapping, these are the rotational kinematic formulas that relate the rotational kinematic variables. And we want the kinematyak acceleration, that’s alpha.

So to find wzoryy speed we could just say that that’s equal to four meters, since you wanna know the speed of a point out here that’s four meters from the axis, and we multiply by the angular velocity, which initially was 40 radians per second.

And should this be alpha be positive or negative?