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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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Categories: 0877 Tags: , , , ,
1. 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. 10.06.2012 at 06:49

Thank you. It was helpful.

3. 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. 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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