# The Physics Philes, lesson 57: Let’s Roll

Howdy! I’m back and ready to do some physics! Are you? Good. Let’s get it.

I know I’ve been on a bit of a vacation the past couple of weeks, but think back. What were we doing? Oh right. We had just learned a little about what happens when a rigid body rotates around a moving axis. Now it’s time to do a simple problem, and it involves a yo-yo.

The yo-yo we have is very simple; it’s just a string wrapped around a solid cylinder. The cylinder has a mass M and a radius R. Now, you’re holding the end of the string stationary while releasing the cylinder. There is no initial motion. As the cylinder falls and rotates, the string unwinds, but does not slip or stretch. Let’s find the velocity of the center of mass of the cylinder after it drops a distance h.

OK, let’s make a diagram. Since we’re not pulling up on the string at all, the hand holding the string doesn’t do any work. Let’s identify the energies we’re dealing with. The potential energy at point one is equal to the product of the mass, the force of gravity, and height h from the ground. The potential energy at point 2 is zero. We flip this script when it comes to kinetic energy. The kinetic energy at point 1 is zero, while the kinetic energy at point 2 is the product of the mass and velocity of the center of mass squared divided by 2, added to the moment of inertia at the center of mass times the square of the angular speed. Actually, that’s confusing. The kinetic energy at point 2 is expressed in this equation: The moment of inertia is half of the mass times the square of the radius, and the angular speed is the velocity of the center of mass divided by the radius. (I neglected to mention this last time. Suffice it to say that this is so because the cylinder doesn’t slip on the string.)

OK, we need to find the speed of the center of mass, we need to figure out the kinetic energy at point 2. We can replace ω in the above equation with v/R:  The kinetic energy is one and a half times as treat as it would have been if the yo-yo were falling without rotating.

Conservation of energy says that the sum of the kinetic and potential energies at point 1 is equal to the kinetic and potential energies at point 2.  The equation for the speed of the center of mass after the cylinder has dropped distance h is: This is less than the speed than a dropped object would normally have because one-third of the potential energy released is manifested as rotational kinetic energy.

Alright, people. That’s it for me this week. see you next week with some other physics thing. See you then!

Featured image credit: Steve Koukoulas

1. I think that you may have forgotten to include the outside 1/2 in the equation that contains the substituted moment of inertia and angular velocity. But, the next equation seems like it is correct.

What I think is kind of interesting is that as the angular momentum of the yo-yo decreases, without a change in total mass (say, we have a yo-yo with a metal core and a light plastic casing), both the angular velocity and the translational velocity increase, while kinetic energy will actually go down. Rotational dynamics are quite not what you’d expect if only considering translational dynamics (for how indeed can you get a lower kinetic energy while one of the two terms in K = 1/2Mv^2 increases while the other remains unchanged?); it’s cheating, in a way.

• Oh you’re right. I goofed. Can I blame extreme procrastination combined with being a n00b at Latex for that?

• No it’s quite all right, I made a mistake myself in saying “as the angular momentum” decreases – the moment of inertia, I mean. And since procrastination and being a n00b at LaTeX certainly aren’t amongst any possible excuses I can use, I think you’re on solid ground.