1. Sep 25, 2004

### justagirl

Two identical blocks are tied together with a string and placed along the same radius of a turntable that is spinning about its center. The inner block is 4 cm from the center and the outer block is 6 cm from the center. The coefficient of static friction between the turntable and the blocks is µs = 0.71, and the string is taut.

a) What is the maximum angular frequency such that neither block slides?

I've spent like 3 hours on this problem - drawing free body diagrams for both blocks, and trying to solve for angular frequency... no luck....please help!

2. Sep 25, 2004

### BLaH!

When you read the problem it sounds like there is some complicated physics going on. However, drawing a correct free body diagram for both blocks simplifies the problem. The key here is that the blocks are connected with a string that is aligned along the diameter (or radius in other words) of the turntable. Therefore, we know that the tension in the string is in the radial direction.

The next problem is to determine in which direction the frictional force acts on the blocks. To do this pretend for a second that the two blocks each of mass $$m$$ are really one big block of mass $$2m$$. We know that the big block is rotating along with the turntable at angular frequency $$\omega$$ (which is unkown obviously). Therefore, we know that there must be a radial inward force that is preventing the big block from slipping off the turntable. The only possible force at this point is the static frictional force. Thus, for the big block we have $$F_{radial} = -\mu_{s}2mg$$ where the negative sign indicates that this force is radially inward.

Let's go back to the picture of the two blocks connected with a string. Let's draw a free body diagram for one of them (say, the inner block). We know from the previous paragraph that the frictional force will be radially inward. Since the string is connected to the outer block, we also know that the tension in the string is acting radially outward. Thus, the sum of forces on the inner block reads $$F_{radial, inner} = -\mu_{s}mg + T = -mr\omega^2$$ where T is the tension and r = 4 cm. Similarly, for the second block $$F_{radial, outer} = -\mu_{s}mg - T = -mR\omega^2$$ where this time T has a minus sign because it is acting radially inward on the outer block and R = 6 cm. You now have two equations with two unkowns: $$T$$ and $$\omega$$. You can eliminate $$T$$ by adding the two equations together. Then simply solve for $$\omega$$. The answer I got was $$\omega = \sqrt{\frac{2\mu_{s}g}{r + R}}.$$

Note that throughout this problem there are NO tangential forces on the blocks because the blocks are moving at the constant angular frequency $$\omega$$.

Last edited: Sep 25, 2004