Suppose the moon were held in its orbit not by gravity but by a massless cable attached to the center of the earth. What would be the tension in the cable? Use the table of astronomical data inside the back cover of the book.

Video transcript

Welcome back, everybody. We are told that an object is attached to one end of a light rigid cord and the other end is attached to some sort of motor that essentially is going to make this thing rotate. We're told a couple different things about this system. We are told that the mass of the object Is kg and that the length of the cord is five m. We are told that it completes a rotation in 8. seconds and we are tasked with finding what the tension will be in this chord. Now, just as a reminder, the tension is going to be acting in the positive radial direction. So here's what I want to do. I want to start with Newton's Second law here. Let's take the sum of all forces in the radial direction and set it equal to the mass of the object times the radial acceleration. Now, what this gives us is that the tension is equal to mass times radial acceleration. And what I'm gonna do is I'm going to sub in for radial acceleration velocity squared over R we know what masses and we know what the radius is, but what is our velocity? Well, we know that velocity is equal to the radius times angular acceleration. And we know that the period is equal to two pi divided by our angular acceleration. Great. So what we can go ahead and do here is I'm going to look at this equation and I'm going to divide both sides by our radius and you'll see that this cancels out and we can plug in Omega to our equation for our period. So let's go ahead and do that. We have that. Our period is equal to two pi divided by our velocity divided by our radius, which really gives us two pi R over our velocity. But we need to solve for velocity. So I'm just gonna multiply both sides by V and divide both sides by P velocity cancels out. And we get that our velocity is equal to two pi R divided by the period. Now let me go sub this value back into our equation for tension. I'm gonna change the color here because all of these equations are getting a little cluttered. We have that attention is equal to the mass divided by the radius of two pi R divided by R period squared. If I take the R out, we then get that attention is equal to M R times two pi over our period squared. We finally have an equation for T. So let's plug in our values and that the mass is 25 kg times the radius of five m times two pi divided by a period of 8.5 seconds squared. Which when we plug all of this in our calculator gives us a final tension of 68. newtons in the cord corresponding to our answer. Choice of C Thank you all so much for watching. Hope this video helped. We will see you all in the next one.