a-mechanical-wheel-initially-at-rest-on-the-floor-begins-rolling-forward-with-an-angular-accelerat-3

Question

A mechanical wheel initially at rest on the floor begins rolling forward
with an angular acceleration of rad/s'. If the radius of the wheel is o.5
m, what is the linear velocity of the wheel after 5 ?
(A) 0.5 $\mathrm{m} / \mathrm{s}$
(B) 1 $\mathrm{m} / \mathrm{s}$
(C) 5 $\mathrm{m} / \mathrm{s}$
(D) 10 $\mathrm{m} / \mathrm{s}$

Salamat Ali · Accepted Answer

the linear velocity or for real, this region Daisy is a call to a radius Time made out. We&#x27;re where we need Ah, Mega R is given So the Omega is a call to initial maker. We&#x27;re gonna plus acceleration times t so many naughty zero within polar exploration we have ease into radiance. Toured in for a second square time is after five seconds. So this gives us on May go Which is it, then? 10 radiance for a second. So we&#x27;ll substrate this value where we have radius is little 0.5 meter offer well danced in radiance per second. This gives us the meaning of a lost year for five meters per second off the wheel. So option C is the right answer.

Kristela Garcia · Answer

So in this problem were given an angular acceleration of two pi radiance for second square, a time interval of three seconds and a radius of two meters and were asked to determine the distance traveled. So we&#x27;re going to start off by applying a rotational kin o Matic equation. And our initial angular velocity was zero. So Seda is the number of rotations that we made in degrees. I&#x27;m sorry in radiance, and we can go ahead, plug in our values. For this, you have the 1/2 in the to counsel to get a angular displacement of nine pie. And so, if we divide nine Pai bi to pie, it tells us how many rotations we made. And again, I&#x27;m gonna leave this something to get 4.5 rotations. So our circumference is two pi r. And so four times are circumference is going to tell us our total distance traveled around so we can say to pi times radius too. So this is 4.5. Yeah. Gives us a total distance. Traveled of 18

Averell Hause · Answer

So for part A, we can find the angular velocity in terms of radiance per second, this would be 2500 revolutions per minute. There are two pi radiance for everyone revolution. And then for every one minute there are 60 seconds and we&#x27;re getting that The angular velocity is gonna be equaling 260 to approximately radiance for a second. We can then say, for part B, the linear velocity would be equaling the angular velocity times the radius are and so this would be equal to 261.8 will round at the very end Radiance per second multiplied by 0.175 meters. This is equaling approximately 46 meters per second. This would be here Ah, one answer for part B and then to find the radial acceleration, this would simply be equaling Omega squared are and so this would be 261.8 radiance per second Quantity squared multiplied by 0.175 meters and we find that the radio acceleration is equally approximately 1.2 times 10 to the fourth meters per second squared. This would be your final answer for Part B and our final answer for party. That is the end of the solution. Thank you for watching.

Averell Hause · Answer

So here they want the angular velocity and radiance per second and then they want the linear velocity and acceleration. So we have a diameter equaling point three five meters and we can immediately say at the radius is going to be equal to point one seven five meters. We can say that for part A. The angular velocity is equal to twenty two hundred revolutions per minute rpm so we can convert and say to pi, radiance her everyone revolution. And then we can say for every minute there are sixty seconds. So this would be giving us two hundred and thirty point for radiance per second for Part B. If we wanted to find the linear velocity, we can multiply the angular velocity Omega by the radius are so this would simply be two hundred and thirty point four times point one seven five, and this is giving us forty point two five meters per second. To find the radio acceleration, we can simply take thie omega or the angular velocity squared times the radius. This will equal two hundred and thirty point for squared times again point one seven five, and this is giving us nine thousand two hundred and ninety meters per second squared. We have to round two Ah, force at least around three or four significant figures. We can&#x27;t really be any more accurate than that because again, we&#x27;re limited by our Constance. This is the end of the solution. Thank you for watching.

Katelyn Chen · Answer

with this question, we&#x27;re asked number things. So for part, they were asked to calculate angler sweet and units of radiance per second. So we know that there&#x27;s 1258 revolutions per minute which we&#x27;ll toss. That is 2500 radiance. So we know that 2500 Arabians for one moment is equal. Machar 2500 Pie radiance Is he cool too 125 pi over three Ingredients for second be were also fined earlier speed So you just multiply it by a radius of 125 pi over three house 10 centimeters which will get us 1250 pi over three centimetres for a second We&#x27;re next your ass to find if it was half our distance. So we have 1000 250 pi over three divided by to which will get us 625 pi or three centimeters per second. As an answer

Griffin Goodwin · Answer

So the question tells us that a wheel with a diameter of one meter accelerates at a rate of four radiance per second squared. And the first part of the problem wants us to figure out what the final angular velocity is of the wheel at a time T, which is 10 seconds later when were given that the initial angular velocity is to radiance uh, per second. And so to do this, we can use our Kinnah Matic formulas. The one that we&#x27;re gonna use for a part A is the one that states that the final angular velocity is equal to the initial angular velocity. Plus the acceleration are the angular acceleration times the time. So we know that the initial angular sorry here is the final angular velocity which we&#x27;re trying to find. We know that the initial angular velocity is to radiance per second plus the, uh, acceleration, which is four times the time, which is 10. So we find that the final, um, angular velocity is equal to 42 radiance per second. So this is the answer to part A. The second part of the problem wants us to figure out what the total angular displacement is equal to. And so to do this, we can just use another kid. A magic formula, which states that the total angular displacement is equal to the initial in your lost e times. The time plus 1/2 times the acceleration times time squares. So we&#x27;re trying to find this. Tilt the theater here. We know that the initial angular velocity is to times the time, which is 10 plus 1/2 times the acceleration, which is four times the time squared, which is tense, weird. And so if you plug this into a calculator or do you just you could probably just do it in your hand. You&#x27;ll see that Delta Theta is equal to 220 radiance. So that&#x27;s the answer to Part B. And lastly, Part C wants us to figure out with the after 10 seconds. What a point on the edge of the ah, the wheel. What it&#x27;s tangential speed is as well as what the acceleration is of the wheel. So to find a tangential speed, we can take our final, angular velocity, multiply it by the radius of the wheel, and this will give us a tangential velocity so the radius of the wheel is 0.5 meters, right, because the diameter is one meter, so it will just be 42 Radiance per second. Times 0.5 meters gives us 21 meters per second as the tangential velocity. Now, if we want to find the acceleration, there are two places where there&#x27;s acceleration. There&#x27;s a tangential acceleration as well as a centripetal acceleration. So to find the tangential acceleration, we can just do the same thing that we did appear for the velocity except will have the angular acceleration times. Uh, the radius is equal to the, um, tangential acceleration. So are angular. Acceleration is for radiance per second squared times the radius which is 0.5 meters and then I give us the conventional acceleration is two meters per second squared now for trying to find what the centripetal acceleration is remembered. This is defined as the, uh tangential velocity squared, divided by the radius squared. So this will be 21 meters per second squared, divided by point five, um, meters. And when you play this to a calculator will get 880 meters per second squared. So now that we have the 10 dental acceleration and the centripetal acceleration, which are perpendicular to each other. We can find the magnitude of the acceleration vector, which is equivalent to the square root of two squared plus 880 squared. And once again, playing this into a calculator gives us a. The magnitude of the acceleration vector is equal to 800 and 82.2 eaters per second squared, and these are your final solutions.

Problem

$f(x)=\frac{1}{x}, \quad \frac{f(x)-f(a)}{x-a}$

Problem 30 Medium Difficulty

Answer

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