### Video Transcript

A 40.0-kilogram solid cylinder is rolling across a horizontal surface at a speed of 6.00 meters per second. How much work is required to stop it?

In this problem, we’ll assume that there is no loss of energy due to friction as the cylinder rolls. Let’s start by highlighting some of the vital information we’re given in the statement. We’re told the cylinder is solid, and that its mass is 40.0 kilograms, and that its rolling on a horizontal surface at a speed of 6.00 meters per second.

We want to solve for the work, we’ll call it capital 𝑊, that’s needed to stop the cylinder from rolling. If we draw a sketch of the cylinder as it rolls, then we see that as this cylinder moves left to right across our surface, it is spinning with an angular rate 𝜔 and it’s also translating with a speed 𝑣 that we’re given as 6.00 meters per second. The critical fact to unlocking the solution to this problem is that our cylinder has two types of motion. It has rotational motion because it’s spinning and therefore has an angular speed 𝜔, and it also has translational motion as it moves left to right. Both of these types of motion contribute to the overall kinetic energy of the cylinder. We can write an equation for the total kinetic energy of the cylinder. KE is equal to KE sub 𝑟, the rotational kinetic energy plus KE sub 𝑡, the translational kinetic energy. Both these terms are needed so we can accurately represent the total kinetic energy of our system.

Now this will help us solve for kinetic energy but what we want to solve for is the work, 𝑊. These terms are connected through the work energy theorem. This theorem is simply stated as: The work done on an object is equal to the change in kinetic energy of that object. In this problem, we’re gonna change the kinetic energy of the cylinder from its initial kinetic energy to zero. That will be its change and that change will be equal to the work done on the cylinder.

So let’s solve for the rotational kinetic energy and translational kinetic energy of this cylinder. Recall that KE sub 𝑟 is equal to one-half 𝐼, the moment of inertia of the object, times 𝜔 squared, where 𝜔 is the angular speed of the object. And KE sub 𝑡, the translational kinetic energy, is equal to one-half the mass of the object times its speed squared. Now let’s substitute these terms in for KE sub 𝑟 and KE sub 𝑡 in our equation. So we see that the overall kinetic energy of the cylinder is one-half 𝐼𝜔 squared plus one-half 𝑚𝑣 squared. But we can go a step further in expanding this out. We can look up 𝐼, the moment of inertia of a cylinder rolling about an axis through its center. That value, which we’ll call 𝐼 sub cyl for the moment of inertia of a cylinder, is equal to half the mass of the cylinder times its radius squared.

So let’s insert that expression for 𝐼 into our equation. So we have that the total kinetic energy of our cylinder is one-half times one-half 𝑚𝑟 squared times 𝜔 squared plus one-half 𝑚𝑣 squared. We don’t know 𝑟, the radius of our cylinder, and we also don’t know 𝜔, it’s angular speed. But, recall the relationship between translational and rotational speed, 𝑣 equals 𝑟𝜔. This relationship means that we can take the 𝑟 squared 𝜔 squared in the first term on the right-hand side of our equation and turn it into 𝑣 squared because 𝑣 equals 𝑟 times 𝜔. When we do that, we now have an equation for total kinetic energy that’s given in terms of values the problem statement gives us, the mass of the cylinder and its linear or translational speed. We can simplify this equation a bit further before we plug in our values of 𝑚 and 𝑣. One-half times one-half is one quarter, and one quarter plus one-half is three quarters. So overall, the total kinetic energy of the cylinder is three-quarters 𝑚𝑣 squared.

Recall that 𝑚 is 40.0 kilograms and 𝑣 is 6.00 meters per second. When we enter these numbers in our calculator, we find that the initial kinetic energy of the cylinder is 1080 joules. As we look at this result and at the work-energy theorem, we realize that the change in the kinetic energy, ΔKE, must be the minus of this result we’ve just calculated. That’s because we’re taking an object with this initial kinetic energy and changing it so it has zero kinetic energy. Therefore, the change in kinetic energy that the cylinder experiences, ΔKE, is equal to negative 1080 joules. And by the work-energy theorem, that change in kinetic energy is equal to the work done on the cylinder.

So we find that the work done to stop this cylinder is equal to negative 1080 joules. The work is negative because we’re taking kinetic energy away from this object. We can also see that it’s negative by realizing that the motion of this cylinder as it comes to a rest is opposite the direction of the force we would exert on the cylinder to stop it. So the minus sign does make sense. And that is the total work needed to stop the cylinder from rolling.