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4. A speck of dust on spinning DVD has centripetal acceleration of ac 20.0 m/s2_(A) What is the acceleration of a different speck of dust that is twice as close to ...

Question

4. A speck of dust on spinning DVD has centripetal acceleration of ac 20.0 m/s2_(A) What is the acceleration of a different speck of dust that is twice as close to the center of the disk?(B) What would be the acceleration of the first speck of dust if the disk's angular velocity was halved?

4. A speck of dust on spinning DVD has centripetal acceleration of ac 20.0 m/s2_ (A) What is the acceleration of a different speck of dust that is twice as close to the center of the disk? (B) What would be the acceleration of the first speck of dust if the disk's angular velocity was halved?



Answers

4. A speck of dust on spinning DVD has centripetal acceleration of ac 20.0 m/s2_ (A) What is the acceleration of a different speck of dust that is twice as close to the center of the disk? (B) What would be the acceleration of the first speck of dust if the disk's angular velocity was halved?

This question covers the concept of the second of motion and to solve their problems. First, we need to find out the time taken by the object to the H. B. From it and the time to use the total distance that escaping upon the Stevie. The distance Abe's the angle that is fired by two into the radius from the speed week all the time tears by us upon to be this name. This as a question. Alberta average acceleration from it. To be, we need to uh we need to subtract the velocity of A from me so we can write the acceleration is but the city of B minus the velocity of A. Upon the little time taken to if we subtract ability A from B. Weird dream, develop velocity of Bs along the north direction and it is V. And velocity of a. Is along the each direction. Therefore the -1 is along the West Direction. And the result will be along the diagonal and its magnitude will be uh wrote two times a week, and angle tita will be 45 degrees. Best off not. So the acceleration is the change in velocity that is two times four m 4 seconds upon the time. And the time is given by uh bye Into our that is two m upon To enter the speed that is four m for a second. The average acceleration is 7.2 m four seconds And the direction two days 45 degrees west off. No, this is the solution of now. For part B, the centripetal acceleration is the square upon are uh, These four m for a second squared Upon the radius and there is two m or the centripetal acceleration is eight m four seconds square.

Solving party of this problem, so I will use the formula. Omega three square minus omega squared is equal to two Alpha Delta. So final value of angular acceleration alpha can be return edge Omega squared minus omega is square by two. Dairy free to so just putting the value I can like all physical to seven by 18 per second squared minus two pi radian per second squared by two multiplication 10 5. So on simplification I get angular Acceleration Al 52.25 2.25 pi radiant per second square. This is the answer for party now for part B, I can write. The value of guilty is equal to omega, minus omega by alpha. So just putting the value I can like dirty is equal to seven pi minus two pi by 2.255 So I'm solving I get dirty edge 2.22 2nd. This is the answer for part B. No tangential acceleration in parts. He can be calculated by the Formula 80 recalled to alfalfa. So just putting the value. I can write 2.25 by reading per second squared multiplication point 0.5 m. I just changed the unit from centimeters to meters. So finally I get 0.35 m per second. Squared edge, tendency, acceleration.

So over this problem, we're going to be looking at a spinning desk that is undergoing constant, angular acceleration. It begins at an angular velocity of two Iranians per second and speeds up to seven by radiance per second. And during this time, it rotates through 10 Pirie Deion's. So the first thing we want to determine is what is the angular acceleration? Then we'll want to determine how much time and took for it to do this to speed up and rotate through that angle. And then we're gonna look at what the tangential acceleration is of a point on the disk at a position, uh, five centimeters away from the center of the disc. Okay, so for the first part, we're gonna look at our equations of motion for constant, angular acceleration. I'm gonna try to find one where we have all the knowns we need. Besides what? So the equation for the job here is going to be Omega Final square minus omega Initial squared equals to Alfa Delta Beta. So we know about the final big initial and Delta data. So all we have to do is solve for health. So if we divide by two Delta data on both sides. We'll have will make a final squared minus Omega initial square. Oh, my to don't data. And then we're gonna plug all these values in. So if we do that, we'll make a final. Is 75 So last 49 I squared begins per second squared minus for pi squared radiance per second squared allure two times, 10 times 20. So the end result will be it looks like 45 pi over two. Since we could cancel a pie from everything for 45 pi over. Sorry. 20. If we want to simplify that eating Moeller, um, it looks like we can five is a common factor in both of those numbers on the numerator and denominator there. So get out for in the bottom in 19 talk nine pi over four. So now the way of Elsa finding the amount of time it took to accelerate from two Pomeranians per 2nd 75 radiance per second checks appreciate, for we have another equation of motion for this, for constant, angular acceleration, you remember, is just gonna be a mega final equals omega initial waas Alfa Times don't t so we have everything here we need besides delta t which we consult for. So this will end up being seven pi radiance per second. You could subtract or make a initial from both sides Schools every year when she is not for my two pi radiance per second equals Al Fidelity in l fa we determined waas nine. I was four when was going flat in there. Nine by those four radiance per second squared. So if we multiply both sides by 4/9 power, we will have Delta t Delta T sequel. Teoh, we're over nine pie times seven minus two will end up being just 59 And in the end, this will all just have units of seconds. So we get a counselor, Pozo, and we should just have four times five is 20 20/9 seconds, which coming out to be two points to two seconds. All right, so there is the amount of time that elapses. Okay, so now what we want to do is take a point on our disk here. Say right here. Maybe that is five centimeters out. So that distance there is five centimeters or if we convert two meters, it's 0.5 meters. We don't want to determine what the translational acceleration is. Not the angular one I'm going to. Your tangential also works. So this is actually related to the angular acceleration just by a tangential equals R I saw. So we've solved for Elsa and we have our So all we have to do this plug these values there so are is your 0.5 meters and also, uh is I already forgot. What? Iwas something with the nine 90 i over four radiance per second squared. So the end, this ends up coming out to 3.53 meters per second squared. Awesome. So we were able to determine the constant, angular acceleration of a spinning disc in this problem. Also, the amount of time it took to speed up from to buy radiance per 2nd 7 pi ratings for second and also to figure out the tangential acceleration of a point on that spinning things

In this problem. We want to find the angular acceleration at a point on a DVD, given the night's angular acceleration at a different point with a given radius and given angular acceleration to solve. This thing to do is rewrite RV in terms of our radius, and we could do this using the equation for period. Now this equation is t equals two pi over Omega and his omega is equal to V over r. This can be written as two pi r already. Then that could just be rearranged. We have to play our over tea, giving us an equation forever lost in any given point in terms of the radius. So l for one can be rewritten as four pi squared r squared bober t squared times are canceled that r squared and that are down here to get four pi are one over t squared. Now we can do the same thing for our acceleration or angular acceleration at our second radius, which is just gonna be given by the same equation except with our two. Now it's important to note that tea is gonna be a constant everyone. The disc is just the amount of time that it takes for the disc to complete one full rotation. So we can say that all for two over. Alpha one is four pi r to over t squared. It looks like tea too. T squared all over four pi r one over t squared and that just simplifies down. Two are two over our one. From there we can rewrite Alfa Two to be equal to our two over our one time's awful one. And these are all qualities that we know so we can plug in our values which are I'm not mistaken. 0.3 and 0.5 times how for one which is equal to ah 120 meters per second squared. Did you realize that I actually wrote these in the order? Alfa two is Are you Put your five on that zero points. You were three And now Oh, if you do assault and weak find that at 00.2 we have an angular acceleration of 200 meters per second squared, which is our final answer


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