# The Physics Philes, lesson 58: More on Yo-Yos

We’ve been talking for the past several weeks on the combination of translational and rotational motion. Now we’re going to check out the dynamics of what happens in this situation. Let’s go!

First, a little review. We’ve learned that for a body with a total mass M, the acceleration of the center of mass is the same as the acceleration of a point mass M acted on by all of the external forces on the actual body. The rotational motion about the center of mass is the moment of inertia with respect to an axis through the center of mass multiplied by the angular acceleration. The latter equation is true even when the axis of rotation moves, as long as the situation meets two requirements:

1. The axis through the center of mass must be an axis of symmetry.
2. The axis must not change direction.

We see this situation in the common, everyday bicycle. The axle of the wheel of a bike goes through the wheel’s center of mass and it is the axis of symmetry.

When you are trying to figure out a problem in which a body goes through rotational and translational motion at the same time, you need two separate equations for the same body; one equation would describe translational motion of the center of mass, the other would describe the rotational motion about the axis through the center of mass.

Let’s apply these concepts to last week’s yo-yo problem. Let’s find the downward acceleration of the yo-yo and the tension of the string. Here’s the diagram: And here is the free body diagram: Our target variables for this problem is T (the tension) and a_cm (the acceleration of the cylinder). We will assume that, in this case, the string does not slip.

First, let’s use the equation for translational motion of the center of mass: Only the tension force has any torque with respect to to the axis through the center of mass, we can use the equation for the moment of inertia to derive the value of that torque. Assuming the string unwinds without slipping, the velocity of the center of mass is represented by: The derivative of this equation is: We can use this equation to eliminate angular acceleration from the torque equation. Then we can solve for T and a_cm simultaneously. We get: and And there we have it! Two simple equations we can use to find the tension force and the acceleration of the center of mass. Easy peasy lemon squeezy. Sort of.

Anyhoo, let me know if I goofed like I did last week! It let’s me know that you’re paying attention.

Featured image credit: Yannis