# Blocks on a Turntable

1. Feb 7, 2009

### Trentonx

1. The problem statement, all variables and given/known data
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 5 cm from the center. The coefficient of static friction between the turntable and the blocks is µs = 0.71, and the string is taut.
What is the maximum angular frequency such that neither block slides?

2. Relevant equations
a_c = (V^2)/R
w = 2pi/T
a_c = (w^2)R
F = ma

3. The attempt at a solution
Sum of force on inner block (known to be zero0
0 = µmg - T + m(w^2)R1
Sum of force on outer block (also zero)
0 = µmg + T + m(w^2)R2

I then took the two tensions to be equal, solved for one, then substituted in the other an solved for w, the angular frequency.
-T = µmg + m(w^2)R2
inserted
0 = µmg +(µmg + m(w^2)R2) + m(w^2)R1
masses cancel
0 = 2µg + (w^2)(R2+R1)
w = 1.24

Where did I go wrong?

2. Feb 8, 2009

### LowlyPion

Check your math again. Remember your units are cm = .01 m

Btw, you missed a sign along the way.

Another way to think about it is that the outward acceleration needs to overcome the frictional maximum resistance.

That implies directly that 2m*μ *g = ω²(r1 + r2)

3. Feb 8, 2009

### Trentonx

Yup, it was my units. Not sure where I dropped the sign, but I got it to work out. Thank you.

4. Feb 8, 2009

### Trentonx

So, continuing this problem, they give the mass of the blocks to be 32g (.032kg) and ask for the tension of the string connecting the blocks. I thought that I could take either of my equations, insert everything that I now have (w=12.43 rad/s) and just solve for tension. But that didn't work. It seems it should, but I must be missing something.

5. Feb 9, 2009

### LowlyPion

Look at the inner block in isolation.

You have the maximum frictional resistance μ*m*g and that is being offset by 2 forces
1) centrifugal acceleration m*ω²*r1 and
2) the Tension necessary to make up the difference.

6. Feb 9, 2009

### Trentonx

That did it. I had assumed that the wrong direction for the centripital force to act. That was also what confused my signs in the first part of the problem. Thanks again.