1
$\begingroup$

I was given a question:

enter image description here

and this is how I would have normally solved it:

enter image description here

However, after learning about rotating frames of reference, I learned about the equation:

$$ \overrightarrow{a} = \left(\frac{\text{d}^2r}{\text{d}t^2} - \omega^2r\right)\widehat{e_{n}} + \left(2\omega\frac{\text{d}r}{\text{d}t} + \alpha r\right)\widehat{e_{t}} $$

So using this here,

In the $\widehat{e_{n}} $ direction,

$m(\frac{\text{d}^2r}{\text{d}t^2} - \omega^2L) = - \mu N$

and in the $\widehat{e_{t}}$ direction,

$m(2\omega\frac{\text{d}r}{\text{d}t} + \alpha L )= N$

I guess just when the bead starts slipping, $\frac{\text{d}r}{\text{d}t}$ would be zero, but we're not sure if $ \frac{\text{d}^2r}{\text{d}t^2} $ is zero, it could be finite (like how a body is momentary at rest at the highest point in a projectile, but still has acceleration)

..or is, $\frac{\text{d}r}{\text{d}t}$ actually zero?

Why does this not give the answer?

That equation should work for an any how rotating body, right?

$\endgroup$
1
$\begingroup$

Note that the value $\mu N$ is merely the maximum value that the frictional force can attain. The bead stays at rest before the friction reaches this maximum value. This becomes clearer if you rewrite the second equation as

$$m\frac{d^2r}{dt^2} = ω^2L − f \ge ω^2L − \mu N$$

Your second equation implies that $N$ is a constant before the bead starts moving. Also, the frictional force increases from zero initially (when the rod isn't rotating) to the maximum value $\mu N$. The increase occurs in such a way that the frictional force is equal to $min(\omega^2L,\mu N)$ (Think about a block at rest on a rough surface subjected to an increasing force, if this doesn't convince you). So, $\omega^2L-f$ is zero when $ω^2L − \mu N$ is negative, and equal to $ω^2L − \mu N$ otherwise. Using this, you can conclude that $\frac{d^2r}{dt^2}$ is zero before the bead starts moving, and then your set of equations reduces to essentially the same equations you used in your alternative method.

$\endgroup$
  • $\begingroup$ Thanks, by the way the normal force on the bead would be going inside the screen, right? and should have been represented by a cross I think. $\endgroup$ – Rick Nov 25 '17 at 14:11
  • $\begingroup$ Basically, friction is a self adjusting force, it won't cause the acceleration of the bead, just try to prevent it.. $\endgroup$ – Rick Nov 25 '17 at 14:30
  • 1
    $\begingroup$ @Rick Yes. The normal force on the bead in the lab frame would be in the direction opposite to that shown in your figure. In the rotating frame, it would be in the direction shown in your figure (you would need to consider the inertial force too). $\endgroup$ – Kishore Nov 25 '17 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.