Friction force in uniform circular motion

Question

A table spins around it's axis with the angular velocity of $\omega$, on the table there is an object with the mass of M connected to a weight with the mass of m through the center of the table (without friction between the string and the pulley)

It's given that the static friction coefficient is $\mu$.

The question is what is the maximum value of R (radius) for the object to remain still during the circular motion.

My attempt: I said that the centripetal force is equal to the tension minus the static friction . Because there is no movement in the y axis, the normal force equals Mg. From that we can say: $$R_{max} = \frac{m}{M-μ} \frac{g}{w²}$$

But the problem is that the book says the answer is: $$R_{max} = \frac{m}{M+μ} \frac{g}{w²}$$

I don't get it. It means that the direction of the friction force is in the center of the table. But how can that be? I thought that the acceleration is towards the center and because of that the direction of the friction force has to be in the opposite direction.

Answer 1

The necessary force needed by the mass to undergo circular motion is $MR\omega^2$.

As the radius $R$ increases so must that force.

Mass M wants to travel in a straight line.
For small values of $R$ it could well be that the tension in the string is too high to maintain that radius for a given angular speed and so the frictional force will be in the opposite direction to the tension in the string.

However as the radius increases the friction force and the tension in the string are trying to stop mass M travelling in a straight line ie they are helping each other to produce the necessary for for the mass M to undergo centripetal acceleration and so those two forces must be in the same direction.

Answer 2

The acceleration is indeed towards the centre of the rotating table.

But friction does not depend on acceleration direction; it depends on sliding direction.

You are asked what the maximum $R$ is for the block to stay stationary. Let's consider what would happen if $R$ was slightly large:

Then the block would fly outwards! Out of the circular motion, meaning further away from the centre of the rotating table.

Static friction prevents this sliding from happening by pulling inwards towards the centre. And if $R$ is larger than the maximum, static friction is not strong enough and let's go, and then the block starts sliding further away.

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Another view on this is to realize that the static friction isn't working against the centripetal acceleration; it is rather causing it! The constant string tension plus this static friction cause the centripetal acceleration. A larger $R$ in a circular motion means that a larger centripetal acceleration is required:

$$a=\omega^2 R$$

(because the angular speed $\omega$ is constant when you move further out on the rotating disc.) The string tension is constant and, thus, only the static friction can grow in order to cause this higher centripetal acceleration. When the static friction can grow no more (when it's limit $f_s\leq\mu_s n$ is reached), then it let's go.

This moment is when the block is at it's maximum $R$.