An 8.0-cm-diameter, 400 g solid sphere is released from rest at the top of a 2.1-m-long, 25 incline. It rolls, without slipping, to the bottom(b). What fraction of its kinetic energy is rotational?

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Hey, everyone in this problem, a 500 g solid cylinder with a diameter of 12 centimeters is rolled down from rest on an incline plane which is 1.5 m long, makes an angle of 30 degrees with the horizontal. It reaches the bottom of the slope without slipping. And we're asked to determine the ratio of rotational kinetic energy to total kinetic energy. We're given four answer choices. Option A one quarter, option B half, option C A third and option D three quarters. So let's start by drawing out a little diagram of what we have going on here. We have our inclined that makes a 30 degree angle with the horizontal. OK? It is 1.5 m long, the length of the incline and we have our cylinder that's gonna start at the top and rolled up. Now we as the ratio of rotational kinetic energy to total kinetic energy. OK. So what does our total kinetic energy look like? I recall that our total kinetic energy which we're gonna call K, it's gonna be made up of two things. OK? It's gonna be made up of our translational or linear kinetic energy. K T our rotational kinetic energy K R right? Because this cylinder is gonna be rotating, but it's also gonna be moving down along that incline. OK. So we have these two pieces to consider. Now this translational or linear kinetic energy, this is that typical kinetic energy you're used to seeing and that's given by one half M V squared. OK. And also recall the K kinetic energy, the rotational kinetic energy is very similar and that's given by one half I omega squirt. Hm. So when we're thinking about the ratio, what we're gonna do is take the kinetic energy K divided, sorry take the rotational kinetic energy K R which is just this second piece divided by the total energy K. But we have a couple of pieces to simplify. So let's look at this ratio and see what we can simplify. OK. So we have this rotational kinetic energy K R divided by the total kinetic energy K. OK. Again, the rotational kinetic energy is one half I omega squared. We've written out the total kinetic energy one half M V squared plus one half I omega squared. OK. So this is the ratio we're looking for and we're just gonna start to simplify what we can. First thing off the bat these one halfs will all divide it. OK? All right. Now we need this moment of inertia eye. OK? So we have this moment of inertia I that we need to calculate. So let's move over to the side and look at what the moment of inertia eye is going to be OK. Now we have a solid cylinder. So the moment of inertia for a solid cylinder, you can look up in a table in your textbook or that your professor provided is going to be equal to one half M R squared. OK. So when we substitute this in, we now have our moment of inertia, I as a function of M and R squared. Now, the other thing we know is the relationship between the velocity V and the angular speed omega. OK. Or the speed and the angular speed recall that we can write and let me do this in green. OK? Because this is talking about V, we have that the speed of the center of mass is equal to R omega. OK. The radius multiplied by that angular speed. OK. And why am I doing this? Well, you can see that there are lots of omega terms already in this equation. And we also are writing I as a function of the radius. OK. So if we can write this velocity V in terms of R and omega, we're gonna be able to cancel some of these terms that so let's do. So, the moment of inertia I is gonna be one half M R squared, we're multiplying by omega squared in the numerator in the denominator, we have M multiplied by V which we're now writing as our omega all squared plus that moment of inertia. I again, one half M R squared and again, multiplied by omega squared. And now you'll start to see why we did what we did. OK. First things, first, the mass M is going to divide it. We have this mass M in all of our terms. You'll also notice we have R squared omega squared in the numerator as well as R squared omega squared in each term in the denominator. So again, we're gonna divide this entire equation by R squared omega squared. And all of those terms are going to divide out. Hm. So that's really incredible. And now we're left with just constant, our ratio has become one half divided by one plus one half. OK? And what that means all of these terms dividing out and canceling. It means that this result does not depend on the mass of the object. It doesn't depend on the radius of the object and it doesn't even depend on the speed that it's going. OK. This ratio of rotational kinetic energy to total kinetic energy does not depend on any of these factors. OK. It's going to be consistent. And if we simplify this, OK, let's keep working on this. We have one house divided by one plus one half, a one plus one half is gonna give us three halves. We have a fraction divided by a fraction. So we can multiply by the reciprocal of the function or the fraction in the denominator. Sorry. So we have one half multiplied by two thirds. And which means that this ratio is going to be equal to one third, these twos will divide it. And so what we found is that the ratio of rotational kinetic energy to total kinetic energy in general? OK. Without having to substitute any of the values we were given in the problem is going to be one third which corresponds with answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

An 8.0-cm-diameter, 400 g solid sphere is released from rest at the top of a 2.1-m-long, 25 incline. It rolls, without slipping, to the bottom(b). What fraction of its kinetic energy is rotational?

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Key Concepts

Moment of Inertia

Kinetic Energy

Rolling Motion