Homework Help: 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.