# Relating angular acceleration to translational acceleration when there is slipping [closed]

This is the statement of the problem:

Three identical cylinders with moments of inertia $I = βmR^2$ are situated in a triangle as shown in the figure. Find the initial downward acceleration of the top cylinder if:

There is no friction between the bottom two cylinders and the ground, but there is friction between the cylinders (so they don’t slip with respect to each other).

Here is the figure:

The following is my attempt at the solution:

Let $F =$ frictional force between the top cylinder and either of the two other cylinders, $N =$ normal force between the top cylinder and either of the two other cylinders, $a_y =$ initial vertical acceleration of the top cylinder, $a_x$ = initial horizontal acceleration of either of the two cylinders in contact with the ground and $\alpha =$ initial angular acceleration of either of the two cylinders in contact with the ground.

The centers of the cylinders form an equilateral triangle of side $2R$.

Applying Newton's second law to the top cylinder, we have:

$$mg - 2 F \cos 60 - 2 N \cos 30 = m a_y$$

And to either of the cylinders in contact with the ground:

$$N \cos 60 - F \cos 30 = m a_x$$

Using $\tau = I \alpha$ on either of the bottom cylinders, we get

$$R F = \beta m R^2 \alpha$$

And by the geometry of the problem,

$$a_x = \sqrt{3} a_y$$

We have five variables and four equations here. The problem is that I cannot relate $\alpha$ to $a_x$, because even though there is no slipping between the two cylinders, there is slipping between the bottom cylinder and the ground. How do I find this relationship?

## closed as off-topic by John Rennie, Kyle Kanos, Jon Custer, sammy gerbil, Rory AlsopJan 7 '18 at 21:02

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• You can relate $\alpha$ to $a_x, a_y$ by applying the no slip condition between the cylinders. This is again a geometrical constraint. As the upper cylinder falls the point of contact rotates around its circumference. To avoid slipping the lower cylinders must rotate through the same angle. – sammy gerbil Jan 6 '18 at 22:54