A 5.0-m-diameter merry-go-round is initially turning with a 4.0 s...

Question

A 5.0-m-diameter merry-go-round is initially turning with a 4.0 s period. It slows down and stops in 20 s. (b) How many revolutions does the merry-go-round make as it stops?

Accepted Answer

Hi, everyone in this particular practice problem, we're being asked to calculate the number of turns made by a disk. Before stopping, we will have a large disk with a diameter of three m spinning with a period of two seconds. The disc is slowed down to a complete stop in 15 seconds. And we're being asked to calculate the number of turns made by the disc before it finally comes to a full stop. The options given are a 23.6 revolutions, B 3.76 revolutions C 70.7 revolutions and D 11. revolutions. So we will assume that the disc is a particle that is in a non uniform circular motion. We're being asked to calculate the number of terms made by the disk before stopping or essentially what we're interested in finding is the angular distance where the disc is actually going through before it actually finally comes to a full stop. So the angle through which the disc turns is going to be given by this following uh formula where theta F equals to theta I plus omega I T plus A T divided by two R multiplied by T squared. So theta F equals to the initial angle plus omega I, which is the initial angular velocity multiplied by T, which is the time that it takes plus A P, which is a tangential acceleration divided by two R R is just the radius and the T is also just the time that it takes. So since the disk is starting from rest, then theta I will then equals to zero. And therefore, we can neglect this part. I'm gonna equal that to zero. And theta F will then equals to just omega I T plus A T divided by two R multiplied by T squared. Before we can substitute everything into this equation, we actually have some uh variables that we do not know. So first, I'm going to just list out all the variables given in our problem statement. So first is the diameter which is going to be three m, which is, which then leaves the radius to be just 1.5 m. The period is given to be two seconds and the time is given to be 15 seconds before or from the start up until the disk actually comes to a complete stop. So from the period, we can actually calculate the initial angular velocity omega I which is going to be used in this formula right here. So Omega I or if you recall angular velocity is equals to just two pi divided by the period or two pi divided by T which then in this case, Omega I will then equals to two pi divided by two seconds. That will then leave us with an initial angular velocity of pi radiance per seconds, which will equals to 3.14 radiant per second just like. So next, we have the initial angular velocity but we are still missing the tangential acceleration or A T. So that is what we have to find next. And we can get the tangential acceleration A T from the equation of Omega F equals omega I plus A T divided by R multiplied by T just like so, and we know that in this particular practice problem, the disc will comes to a complete stop, which means our omega F will then equals to zero radiance per seconds. We can then rearrange our equation in order for us to get an equation for A P or the tangential acceleration which will leave us with negative omega I multiplied by R divided by T. We know everything or every variable in this expression. So let's substitute all of them into this expression. Negative omega I is going to be negative 3. radiance per seconds multiply that by the radius which is going to be 1.5 m and divide everything with T which is going to be 15 seconds that will leave us with the tangential acceleration of negative 0.314 radiance per second squared just like. So, all right, So now that we have our tangential acceleration, we also have our initial angular velocity. We actually have every single information needed for us to calculate theta F. So then I'm gonna rewrite our equation for theta F one more time. So theta F is going to then equals to omega I multiplied by T plus A T divided by two R multiplied by T squared. So let's substitute everything into this following equation or expression right here, Omega I is 3.14 radiance per second. The T is going to be 15 seconds plus A T divided by two R A T is negative 0.314 radiance per second squared divide that by two R which is two multiplied by 1.5 m and multiplied that with our time, which is going to be 15 seconds squared just like so and it will give us a theta F value of 23. radius. So now that we have our uh theta F value which is 3rd 23. radiant, we want to convert that into revolution. So converting our radiance to revolution will then have to actually multiply this by one revolution for every two pi radiance. And that will give us a theta F value of 3.76 revolution, which will actually be the answer to this particular practice problem that will actually correspond with option B in our answer choices. So option B is going to be the answer to this particular practice problem with the total number of turns made by the disc before stopping being 3.76 revolutions. So that'll be it for this particular practice problem. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that will be all for this one. Thank you.

A 5.0-m-diameter merry-go-round is initially turning with a 4.0 s period. It slows down and stops in 20 s. (b) How many revolutions does the merry-go-round make as it stops?

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