# Forces plus Rotational Motion

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1. Apr 11, 2015

### Cliff Bryant

Problem:
"Two blocks of equal masses m are attached by an ideal string. One mass lies at radial distance r from the center of a horizontal turntable rotating with constant angular speed of 6 rad/s, while the second hangs from the string inside the hollow spindle of the turntable.The coefficient of static friction between the surface of the turntable and the mass lying on it is 0.5. Find the maximum and minimum values of r such that the mass lying on the turntable does not slide."

My initial idea was
Tm - (Ffriction + Fcentrifugal) = 0
and then I would get the zeroes of the function, but it feels like I'm really wrong about this. Can anyone explain how to solve this?

2. Apr 11, 2015

### BvU

Hello Cliff, welcome to PF !

Please do use the template. It helps you to be complete in your post. Also, in PF its use is mandatory (guidelines) so the good spirits that watch over us will chastize you if you don't.

In this case I have no idea what you mean with $T_m$.

Write a force balance for the mass on the table that results in it following a circular trajectory. You only have to worry about the radial part of the forces. One of the forces comes from the wire. That one you can find from a force balance for the mass hanging from the thread.

3. Apr 11, 2015

### Cliff Bryant

Sorry, I was getting desperate.
Anyway, thank you for the reminder.

Sorry for the mistake, and I'll make it clear this time.

1. The problem statement, all variables and given/known data

Problem: (stated above)

Variables:
r = radial distance of the block above the turntable to the turntable's center
m = mass of the block

Tm = tension of the string pulling the block (on the turntable)

Given:
μ (static friction) = 0.5

2. Relevant equations

ac = v2 / r
Fc = mv2 / r
Ffriction = μmg

3. The attempt at a solution

My idea is that, to prevent slipping, the net forces acting on the block (or being the block on the origin of the free-body diagram) should be zero.
Tm - (Ffriction + Fc) = 0
mg - ( μmg + mv2 / r ) = 0
[ mg ( 1 - μ ) ] + mv2 / r = 0
r = v2 / [g ( 1 - μ )]

Then I would substitute the values.

4. Apr 11, 2015

### PeroK

This looks like a tricky problem. Your equation is going to give you a single value for r. Is that correct? Hint: perhaps think a little more about the frictional force.

Which value of r do you think you have calculated? The maximum or minimum?

Here's what I would do first:

a) Consider the problem where the table is not rotating. What happens?

b) Conside the problem where there is no hanging mass, but the table is rotating. What happens?

Last edited: Apr 11, 2015