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Carnival merry-go-round rotates about vertica axis a constant rate_ man standing the edge has constant speed of 3.43 m/s and centripetal acceleration of gnitude 1.9...

Question

Carnival merry-go-round rotates about vertica axis a constant rate_ man standing the edge has constant speed of 3.43 m/s and centripetal acceleration of gnitude 1.96 m/s< , Position vector locates him relative to the rotation axis. (a) What is the magnitude of r What is the direction of r when is directed (b) due east and (c) due south?(a) NumberUnits(b)

carnival merry-go-round rotates about vertica axis a constant rate_ man standing the edge has constant speed of 3.43 m/s and centripetal acceleration of gnitude 1.96 m/s< , Position vector locates him relative to the rotation axis. (a) What is the magnitude of r What is the direction of r when is directed (b) due east and (c) due south? (a) Number Units (b)



Answers

A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of $3.66 \mathrm{~m} / \mathrm{s}$ and a centripetal acceleration $\vec{a}$ of magnitude $1.83 \mathrm{~m} / \mathrm{s}^{2}$ Position vector $\vec{r}$ locates him relative to the rotation axis. (a) What is the magnitude of $\vec{r}$ ? What is the direction of $\vec{r}$ when $\vec{a}$ is directed (b) due east and (c) due south?

So we know that for port A, the magnitude listen, triple acceleration would be pulling the linear velocity. Divided by are the radius and this would simply be the radius of whatever circular path one is following. And so this would be tor words the center of certain fueler. Uh, now we know that four part A, uh, the passenger is experiencing exit is experiencing an acceleration of one point 83 meters per second squared east, um, than the centre than the center means that this this means that the center is east of where of the location of the passenger. And so we can say that the distance are algebraic plea manipulating the magnitude of the centripetal acceleration. So we know that for port A, the magnitude listen, triple acceleration would be pulling the linear velocity. Divided by are the radius and this would simply be the radius of whatever circular path one is following. And so this would be forwards the center of certain fueler. Uh, now this would be equaling the velocity squared, divided by the acceleration. So this would be 3.66 meters per second quantity squared, divided by 1.83 meters per second squared and this would be equaling 7.32 meters. So we can say that the position is 7.32 meters east, so I'll be your answer for part. We know that four part A, uh, the passenger is experiencing it is experiencing an acceleration of one point 83 meters per second squared east than the centre than the center means that this this means that the center is east of where of the location of the passenger and so we can say that the distance are algebraic. Lee, manipulating the magnitude of the centripetal acceleration be so far apart for party. For part B. Relatives of the center, the passenger is located 7.32 meters tor words the West. This would be equaling the velocity squared, divided by the acceleration, so this would be 3.66 meters per second quantity squared, divided by 1.83 meters per second squared, and this would be equaling 7.32 meters. So we can say that the position is 7.32 meters east sob your answer for part B, and given this, it might be better to say that this would be 7.32 meters. That would be the magnitude of the distance between the center and the passenger. Um and we can say that uh center is 7.32 meters beast. So for part B, so far apart for party. For Part B relatives of the center the passenger is located 7.32 meters tor words the West relative to the center, the passengers located 37.32 meters towards the west. And for part C, we can say that if the direction of the acceleration experienced by the passengers now south, that's indicating that the center of the merry go round is south of him. So we can say that if the acceleration vector is directed, So uh, the and given this, it might be better to say that this would be 7.32 meters. That would be the magnitude of the distance between the center and the passenger. Um, and we can say that uh, center is 7.32 meters beast. So for part B, center of the beacon to say merry go round is south of the passenger and then of course, relative to the center. The passenger is 7.32 meters elective to the center, the passengers located 37.32 meters towards the west. And for part C, we can say that if the direction of the acceleration experienced by the passengers now south that's indicating that the center of the merry go round is south of him. So we can say that if the acceleration vector is directed So uh, the, uh to the north. So I'll be your answer for part C. That is the end of the solution. Thank you for watching.

To solve this question firstly to convert the final adversity into ingredients possible. The final speed is 0.6 revolutions or second. You can write this at 7.6 into two pi. The radiance for 2nd are the final and my experience. Uh 3.77 radiance or seven For 1st part A. The angular acceleration. Alfa is the final speed minus the initial speed upon the time taking tea and then groups regional violence 3.773 Aliens or 2nd -0 0 Upon packs in recent times. That is eight seconds. So we can write in the exhibition at 0.471 Radiance for 2nd Square for party. The centripetal acceleration A. C. Is omega squad F and do, but it is up the seat and you can substitute the value. Scientifically acceleration is 3.77 Radiance for 2nd Square into 2.75 m centripetal acceleration is 39.1 m for Second Square for part C. The tangential acceleration is the radius of the seat into the ongoing exploration Alpha. Now we can substitute the value that an initial acceleration is 2.75 m into the angular acceleration that is 0.471 ingredients Forsaken scrapped. Other dimensional acceleration is 1.3 m positive square. The total acceleration of the net acceleration squared off The square of the 10 year jail acceleration, that is 1.3, you just for a second square Plus the square of the centripetal acceleration, that is 39.1 m possible square. The net acceleration is almost equals true 39.1 m positive school the direction of the acceleration acceleration with the centripetal acceleration, tita is turning wards off That. An initial acceleration that is 1.3 we just go 2nd square Upon the centripetal acceleration, that is 39.1 m/s square. But the angle three times 1.9 periods

On a merry go round. The speed on the outside is 3.66 m per second. A well, first of all I'm going to figure out um Omega I think so V equals omega are but wait a minute, I don't even know the radius. Okay, so mhm. For each of these situations, how far they are from the center of the merry. Go round in what direction? So the acceleration which is V squared over R. Is 1.83 and it's to the east. So be this direction east. Mhm. Uh huh. Okay. Um so or would equal V squared over 1.83? Yeah. Where V. Is 3.66 that's going to be seven point 32 m. That's it's the carnival. Okay, passenger has an acceleration of that. Yeah, so the passenger would be accelerating outward and would have to exert a force uh to stay on the ride that would be exerted inward. So this would be outward. Which would be to the east. Second one B um same acceleration south. Well then itself because the acceleration would be outward it's still 7.32 Yeah. Mhm. Thank you for watching.

Hi driven here it is given off Ellenville with Really, it's 14 m is running about through the same good and now digital active. The union speed off a position on the rim is 7.0 m per second water, the magnitude and direction off passenger acceleration. You have to point magnitude under direction off passengers acceleration plus, but at the lowest point. Be at the highest point and see how much time it will take to complete. One revelation there could see it. It's a 10. Mutual acceleration is zero because it is moving with a constant velocity and having only a real component off acceleration on the real acceleration is the square by your no first kids at the lowest position we square by yacht velocity is given. Seven on a radius is 14 m, so it would be 3.5. Meet their particulate square the words a center, a pervert. We part at the highest point. Well, it'll be the same because disputed same. But watch the center don't work, see, but I'm petered off revolution. It's big, unusually beauties who by your appointee, So time period off revelation will be who by your upon his speed do in tow. 3.1 ft speed is 7 m per second. Sorry reduces 14 m per second on disputed civet, so it becomes 12.6 ticket. That's all for this problem. Thanks for watching it.


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