You decide to visit Santa Claus at the north pole to put in a ...

Question

You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While there, you notice that the elf Sneezy, when hanging from a rope, produces a tension of 395.0 N in the rope. If Sneezy hangs from a similar rope while delivering presents at the earth's equator, what will the tension in it be? (Recall that the earth is rotating about an axis through its north and south poles.)

Accepted Answer

Hey everyone in this problem. We're told that at the south pole a metallic sphere is suspended from a string. The measured tension in the string is 50 newtons. We want to know if we use the same device at the equator, what would be the tension in the string? And we're told to remember that the Earth rotates on an axis passing through the north and south poles and to take G as 9.81 m per second squared. Alright, so let's just draw what's going on here. Okay, so we have just to think about how the earth is rotating. So we have this axis, you can pretend that that's in the center. Okay, And this is the south pole here, A south pole and the Earth is rotating about this axis that goes through the south pole. Like this. More singing about the equator. The equator like this. So we're at a point over here and the earth is still rotating along this axis here. Okay, so we go from being on that axis of rotation to not Alright, so we want to find the tension in the string at the equator. Alright, so let's focus on the equator and this is the equator here. Okay, so let's focus on the equator. On the right hand side. So at the equator, the force of tension at the equator. Well this is going to be M times G plus M. A. Ok, recall that formula the tension mass times gravity plus mass times acceleration. And this is gonna be the acceleration at the equator. Alright, well we know G. We were given in the question, We don't know mass. We don't know the mass of the instrument or the metallic sphere. Mhm. Um And a the acceleration at the equator is something we also need to find. Okay, let's start with the mass. Okay, we have some information about what's happening at the south pole, so let's start there and see if we can find the mass. Okay, so at the south pole we have the same equation. Okay, the force of tension at the south pole is going to be MG plus M the acceleration at the south pole at the south pole because we are on that axis that the Earth is rotating about. The acceleration is actually going to be zero in this case. Okay, so this term is going to go to zero. We know that the force due to tension is 50 newtons. Okay, so we have 50 newtons Is equal to the mass times G which were given as 9. m/s squirt. This gives us a mess of 50, divided by 9.81 kg. All right, we'll just leave it like that until we use it in the equator equation so that we keep as many significant digits as we can. Alright, so, back to the equator equation. We know the mass. Now we don't know is that a E. Okay, the acceleration at the equator. Now let's recall we're talking about acceleration. We can write a so a. At the equator is going to be four pi square it R over T squirt. Where r. Is the radius and T. Is a period of rotation? Okay. So this gives us the acceleration when we're rotating and we can use these standard values. So we have four pi squared the radius. We're talking about Earth. So the radius of Earth is 6.37 times 10 to the six m In the period of Earth. Well we have 8. Times 10 to the four seconds. Okay? And that's gonna be all squared. And these are standard values that you look up either in your textbook or in some sort of formula sheet that your professor will provide. Okay. And this gives an acceleration at the equator of 0. meters per second squared. Okay. Alright. So we have this mass here that we found, we have this acceleration at the equator that we found. Now we can go ahead and find our tension. Okay so I'm gonna scroll up a bit so we have some more room to work And F. T. is going to be mass which we now know is 50. And again this force to detention at the equator. That's what we're looking for. That's what the question has asked us for. We have 50 divided by 9.81 kilograms Times G 9.81 m/s squared. Plus the mass 50 divided by 9.81 kg times the acceleration at the equator, 0.03369 m/s squared. Alright, so this 9.81. This will cancel. Okay. And that's one reason why we left it. How it was. It just makes it a little bit simpler. and working at the rest of the details we get 9.82 Newtons. So our attention force at the equator is 49. Newtons. And so that is going to correspond with answer B. That's it for this one. Thanks everyone for watching. See you in the next video.

Key Concepts

Tension in a Rope

Centripetal Force

Gravitational Force Variation

Watch next