Achild is riding a merry-go-round and is s standing 4.65 m from its center: Atthis instant; his angular speed is 1.25 rad/s and his angular acceleration is 0.745 ra...

A child pushes a merry-go-round that has a diameter of $4.00 \mathrm{~m}$ and goes from rest to an angular speed of $18.0$ rpm in a time of $43.0 \mathrm{~s}$. (a) Calculate the average angular acceleration (in $\mathrm{rad} / \mathrm{s}^{2}$ ) of the merry-go-round. (b) Calculate the angular displacement (in rad) of the merry-go-round during this time interval. (c) What is the maximum tangential speed of the child if she rides on the edge of the platform? Example $8-11$

For part of the initial in world speed. Omega notice to buy upon the time to 80. And you can substitute the values to find the initial anguish Speed. That is to hire aliens Upon the time T. That is 12 seconds. So the nation and good spirits. 0.5-4 radiance for second. The four party the indexation Alfie's to finally understood. The minister initial in the state upon All right, that empty stopping time. Now we can secure the value then your acceleration and fights for stopping conditions. The fine industry will be zero minus the initial. That is 0.5 to four variants. Four seconds upon the stopping 10 that is 12 seconds. So the angular acceleration al files -010437 Radiance for 2nd Square. The negative sign means that the direction of the and the acceleration is opposed to the direction of the angular velocity part C retention acceleration. It is the radius are into the absolution And now we can substitute the value that unusual acceleration of the writer Mr Radius is nine m Into Alpha that is -0.04 37 radiance, four seconds. Sweat. All the 10 essential acceleration is Uh -0.393 m, or 2nd square. The negative center presents. The Kenyan acceleration is opposite to the direction of the, uh, politically means this radio scene government.

So for this problem will be looking at a wheel that starts at rest and increases its angular speed. Um, up 2.5 revolutions per second. So it has a constant angular acceleration during this period, and it rotates through an angle of two revolutions. So what we're gonna want to do is use one of our angular equations of motion for constant acceleration. Specifically, um, the one that will be convenient to use here is Omega Final square minus were made that initial squared equals soon Alfa Tilted Fator. So Omega Initial is going to be zero, so we can just knock that off. And then we console for Alpha, dividing both sides of the situation by two. Double figure to leave us with will make a final squared over to Delta Fada don't care. So at this point will probably want to convert um, Omega final and Delta theta into radiance since their revolutions right now, um, Delta theta equals two revolutions is also equal to one pie radiance since there are to Pirated Ian's periphery revolution. Similarly, with we'll make a spinal, we can apply the same conversion factor of two pi over reb here to get you on serial quaint file brides for seven? No, to get Hi radiance per second. Cool. So now we can play everything into our equation. So Alfa is equal to Omega. Final squared pi squared one over second squared. Fight it by two times for Hi. So it looks like our units or right? Just one of her second square. You can cancel pie here in here, and we just have all right over eight radiance per second squared. And there you have it. We determined our angular acceleration from knowing our initial and final angular velocity and the angle retread rotated.

In this problem, We're gonna talk about centripetal acceleration. Eso consider a particle. That's when they're going certain emotion as shown here, the bishop and the motion as the radius R Uh, and we know that the particle is being accelerated with a contributor acceleration, Alfa, we can decompose the total acceleration of this particle in two components. The first one is a radio components. So suppose that the party is here. The radio component called this interpretive exploration We'll be pointing towards the center off rotation, and it's equal to V squared over R where V is the linear speed off the particle or omega squared times are in this accounts for the change in direction off the particle and exists even when there is no change in speed. Okay, uh, the tangential exploration, on the other hand, accounts for only for the change in speed. So it's the derivative of the speed with respect to time and can be written as the angular acceleration times. The radios are okay. So in our problem, we have a car that is in a circular path of radius 23.5 m and at a certain point in the path the car is still accelerating and has a new angular speed of 0.5 71 radiance per second. And we have the information that the angle that the total speed does with the radius. So suppose that the car is right here on then the total Not I had big before, but it should be total exploration. This angle here is equal to 23.5 degrees. I'm sorry. It's actually 35 degrees. Sorry for the mistake on our goal is to find what is the value off the total exploration, the magnitude of the tour exploration. Well, notice that we have that the acceleration can be decomposed into two components. So we have one component That is the centripetal acceleration that points towards the center of rotation. And the other one that Is that the general exploration? Yes. So this is this attributable. This is the financial, and this is the total one. Uh, actually, I wrote it as a C. But it should be a t. Um, So notice that this angle here is theta. So a T is equal to a C divided by the cold sign of theta a c. The centripetal acceleration is omega squared times are and then we have to divide it by coastline of data. So 80 is equal to 0.5 71 radiance per second squared times are and that is 23.5 m and we have to divide it by the coal sine of the angle theta, which is 35 degrees mhm. Eso 80 is equal to 9.35 m per second square. So this is a total exploration and the answer to your question, Yeah, what?

There is a merry go round with a radius of two m and it has inertia of 275 kg meters squared. Um, a child stands one m from the axle, so mass of the child is 25 kilograms and its distance is one meter. Okay, and that child will in the merry go round rotates at 14 revolutions per minute. Okay, Child then gives the edge of the merry go round. What's the angular speed then? Okay, well I this is I of the merry go round is going to be I of the merry go round plus M D squared. But I final I'm going to call this initial I final is going to be I of the merry go round plus I am R squared. I initial omega initial is gonna be I final omega final. So omega final is going to be I initial over I final omega initial. So let's put this into the calculator. Um our is to I sub m is to 75 I am. It's 25. The is one omega initial is 14. Okay, I initial it's gonna be ice of em smd where'd I final is gonna equal I initial plus M R squared. So now I initial over I final times omega initial. That gives me 11.2 rpm And it is 60 so there's no answer in the back of the book. Thank you