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GR0877 Problem 99

PROBLEM STATEMENT:

A small particle of mass m is at rest on a horizontal circular platform that is free to rotate about a vertical axis through its center. The particle is located at a radius r from the axis, as shown in the figure above. The platform begins to rotate with constant angular acceleration \alpha. Because of friction between the particle and the platform, the particle remains at rest with respect to the platform. When the platform has reached angular speed \omega, the angle \theta between the static frictional force \mathbf{f}_s and the inward radial direction is given by which of the following?

(A) \displaystyle \theta = \frac{\omega^2 r}{g}

(B) \displaystyle \theta = \frac{\omega^2}{\alpha}

(C) \displaystyle \theta = \frac{\alpha}{\omega^2}

(D) \displaystyle \theta = \tan^{-1} \left(\frac{\omega^2}{\alpha}\right)

(E) \displaystyle \theta = \tan^{-1} \left(\frac{\alpha}{\omega^2}\right)

SOLUTION: (E) In inertial reference frame associated with the platform, Newton’s second law reads m \mathbf{a} = \mathbf{f}_s + m\mathbf{g} + \mathbf{N}, where \mathbf{N} is a normal force perpendicular to the surface of contact, m\mathbf{g} is the downward force of gravity. Acceleration \mathbf{a} has a tangential component \mathbf{a}_{\tau} and a radial component \mathbf{a}_n: \mathbf{a} = a_{\tau} \boldsymbol{\hat{\tau}} + a_n \mathbf{\hat{n}}, where \boldsymbol{\hat{\tau}} is a unit vector tangent to the path pointing in the direction of motion at the chosen moment in time and \mathbf{\hat{n}} is the unit (inward) normal vector to the particle’s trajectory, a_{\tau} = \frac{dv}{dt} = d(\omega r)/dt = r d \omega/dt = r \alpha; a_n = v^2/r = \omega^2 r^2/r = \omega^2 r. Projecting Newton’s second law of motion onto \boldsymbol{\hat{\tau}} and \mathbf{\hat{n}} gives m \alpha r= f_s \sin{\theta} and m \omega^2 r = f_s \cos{\theta}. From these two equations \alpha/\omega^2 = \tan {\theta} , so \theta = \tan^{-1} \left(\frac{\alpha}{\omega^2}\right).

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  1. PineApple
    15.10.2011 at 13:25

    Did you intentionally include mg here? it seems irrelevant to the motion. Or rather, you could include that and also N, both of which cancel each other out

    • 15.10.2011 at 14:09

      You’re right. I forgot to include \mathbf{N}. Thanks, fixed.

  2. Singh
    10.06.2012 at 06:49

    Thank you. It was helpful.

  3. Alex
    19.09.2015 at 06:55

    You can solve this by elimination. At very small w, the angle between the radial direction and the frictional force should be approximately 90 degrees (it is basically moving on a straight tangent line in this limit). So this leaves only E.

  4. Szwed
    05.04.2017 at 17:48

    Similar to Alex’s approach, if the angular acceleration is zero, then theta must be zero (just good old uniform circular motion). This leaves only C and E. Then looking at the limit as omega goes to zero, option C gives the ridiculous answer that theta is necessarily greater than 2*pi. This leaves only E.

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