Friction and Rotation: Understanding the Ratio of Forces on a Rotating Disk

[lc99]

Homework Statement

A phonograph record is whiring around at 103 rpm. Two balls of mass 1 kg are sitting on the disk and are at rest with respect to disk. The first ball (1) sits at a radius 5 away from center. The second ball (2) sits at a radius of 10 away from center. What is the ratio of friction forces acting on them? Ratio = (Friction of 1/Friction of 2)

A)0.5
B) 1
C) 2
D) .25
E) not enough info

Homework Equations

T = Ialpha

The Attempt at a Solution

Okay, first,

I know that the disk is rotating at a constant velocity. This means that angular acceleration is 0, thus, the Torque is 0. If the torque is 0 for this disk that is rotating constantly, then that means that net torque is 0 because friction is acting on the balls. Therefore, some force is oppose this friction for net torque to be 0.

However, this doesn't really help me because Torque is just 0? I'm thinking the answer is either E) or could be B) or C) if I am thinking wrong

 

[berkeman]

The balls are stuck on the disk as it rotates, so the friction forces on each are enough to keep them from sliding. What is the equation for the centripetal acceleration on an object as a function of the radius?

 

[lc99]

The balls are stuck on the disk as it rotates, so the friction forces on each are enough to keep them from sliding. What is the equation for the centripetal acceleration on an object as a function of the radius?

well,

Ac = v^2/r --> w^2*r

Is Ac the torque?

If it is, then the friction force is twice the ball 2 for ball one because of Radius doubled for ball 1

 

[berkeman]

well,

Ac = v^2/r --> w^2*r

Yep. So what's the correct answer to the question then...

 

[lc99]

Yep. So what's the correct answer to the question then...

C) 2! Thanks :D

Unfortunately, i guessed ratio of 1 on my exam out of confusion and panic :(. Atleast I get it now!

 

[haruspex]

The balls are stuck on the disk as it rotates

That's unclear. If glued to the disk, why make them balls, and where does friction come into it?
A feasible interpretation is that the balls are only instantaneously at rest relative to the disc. But if so, we should have been told that there is sufficient friction that the balls do not, at first at least, slide.
Pretty sure the answer would still be 2, though.