If k, m, and w are known, then r can be calculated.

[cameo_demon]

The following problem is take from Thorton and Marion's Classical Dynamics, 5th edition, p. 408, chapter 10, problem 3.

Given
A puck of mass m on a merry-go-round (a flat rotating disk) has constant angular velocity \omega and coefficient of static friction between the puck and the disk of {\mu}_{s}.Task
Determine how far away from the center of the merry-go-round the hockey puck can be placed without sliding.

Solve

here's my attempt at a solution:

we want the puck to have zero velocity, so that when we integrate velocity with respect to time we get out a constant k, which is our radius from the center.

for the general case:

F = ma_{f} = m\ddot{R}_{f} + ma_{r} + m\dot{\omega} \times r + 2m\omega \times v_{r}

the only force internal to the inertial reference frame is the friction force m\mu_{s}g

and are solving for r for the zero velocity case and constant angular velocity, we can throw out the first two terms as well as the last:

F = m\mu_{s}g = m\dot{\omega} \times r

which allows to say

m\mu_{s}g = mr\omega^{2} \hat{i}

solving for r:

\frac{\mu_{s}g}{\omega^{2}} = r \hat{i}

i feel really shaky about this result because i just don't feel confident about it because a) the thorton and marion book is too high-level and relies too heavily on mathematical formalism which b) i suck at.

any help or comments would be appreciated.

 

[Shooting Star]

You have made the solution too long.

The puck will not slide as long as the static frictional force is less than or equal to the centrifugal force in the rotating frame. The maximum frictional force is k*N. So,

mw^2r = kN = kmg =>
r = kg/w^2.