What is the rotational kinetic energy of the earth? Assume the earth is a uniform sphere. Data for the earth can be found inside the back cover of the book.

Question

Accepted Answer

Hey, everyone in this problem, a solid cylinder of mass 2 kg and radius 0.5 m is rotating about its central axis at three revolutions per second. Assuming the cylinder to be a uniform object. We are asked to determine its rotational kinetic energy. We're given four answer choices. Option A 28.3 jewels, option B 88.8 jewels, option C 44.4 jewels and option D 82.3 jewels. So we're asked to calculate the rotational kinetic energy. And let's recall that the rotational kinetic energy Kr is given by one half I omega squared K where I is the moment of inertia and omega is at angular or rotational speed. OK. So this is very similar to the kinetic energy. In the linear case. In the linear case, we have one half MB squared. OK. So instead of m we have this moment of inertia, I instead of that linear speed, we have a rotational speed. OK. So let's just work out each part. We're gonna have to calculate I and we're gonna have to calculate omega. So we're starting with the moment of inertia. I we're told that this is a solid cylinder and that it has a, it is a uniform object. OK. So this is a uniform solid cylinder. So if you look up the moment of inertia for this object in a table in your textbook or that your professor provided, OK. It's rotating about its central axis. And so that moment of inertia is gonna be one half Mr squared. All right, substituting in our values, we have one half multiplied by the mass of this sphere which we're told is 2 kg multiplied by the radius of this sphere, which we're told is 0.5 m. And that radius is squared, we have one half multiplied by 2 kg multiplied by 0.5 m squared, which leaves us with a 0.25 kg meters squared. OK? And that is our moment of inertia. I we put a box around that. So we don't lose track of it. And the other thing we needed to calculate was omega and again, Omega is that rotational or angular speed? Now, we're told that this object is rotating at three revolutions per second and we have three revolutions per second. And we want this in radiance per second. OK. We know that in every revolution there are two pi radiant. So we're gonna multiply by two pi radiant divided by one revolution, the unit of revolution divides out, we're left with the unit of radiance per second, which is what we want and this is gonna be equal to six pi radiant per se. All right. So we have omega, we have, I now we can calculate a rotational kinetic energy. Kr Kr again is equal to one half I omega squared. This is gonna be equal to one half multiplied by 0.25 kg meter squared multiplied by six pi radiance per second. Oh, it's great. OK. If we deal with just the numbers first, the constants without the pi we have one half multiplied by 0.25 multiplied by six squared. This is gonna give us 4.5 and then we have our pi squared. So 4.5 pi squared, our units, we have kilogram meter squared multiplied by radiance per second squared. So we have kilogram meter squared per second squared, which is equivalent to a duel, which is what we want for kinetic energy. And if we use our calculator, we have 4.5 pi squared, which is equivalent to 44.4 Jews approximately. And so our rotational kinetic energy for this sphere is gonna be 44.4 jules. If we compare this to the answer choices we were given, we can see that this corresponds with answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

What is the rotational kinetic energy of the earth? Assume the earth is a uniform sphere. Data for the earth can be found inside the back cover of the book.

Verified Solution

Key Concepts

Rotational Kinetic Energy

Moment of Inertia

Angular Velocity