How to find the maximum angular speed from a bar which rotates about an axis?

Jun 2017
352
6
Lima, Peru
The problem is as follows:

A rigid bar of negligible mass has three particles whose masses are the same and are joined to the bar as indicated in the figure. The bar is free to rotate in a vertical plane about an axis with no friction perpendicular to the bar through point $P$ and is released from rest on the horizontal position at $t=0$. Find the maximum angular speed in radians per second attained by the bar. You may consider $g=9.8\frac{m}{s^2}$ and $d=4.2\pi$



The alternatives given are:

$\begin{array}{ll}
1.&1\,\frac{rad}{s}\\
2.&2\,\frac{rad}{s}\\
3.&3\,\frac{rad}{s}\\
4.&4\,\frac{rad}{s}\\
\end{array}$

What I've attempted to solve this problem was to equate the potential energy and the rotational energy of the masses assuming there's a superposition of the moment of inertia of the masses.

This is translated as:

$mgh=\frac{1}{2}I\omega^2$

$mgh=\left(\frac{1}{2}m\left(\frac{4d}{3}\right)^{2}+\frac{1}{2}m\left(\frac{d}{3}\right)^{2}+\frac{1}{2}m\left(\frac{2d}{3}\right)^{2}\right) \times \omega^2$

But after following the procedure I can't find a way to cancel the 4.2$\pi$. Supposedly, the answer is the first choice. How can I get to that result?
 

skeeter

Math Team
Jul 2011
3,293
1,778
Texas
$\omega_{max}$ will occur when the bar is at the vertical ...

$U_{g0}-U_{gf} = K_{Rf}-K_{R0}$

$mg \cdot \dfrac{4d}{3} + mg \cdot \dfrac{d}{3} - mg \cdot \dfrac{2d}{3} = mgd = \dfrac{1}{2} \cdot I \cdot \omega^2$

$I = m \cdot \dfrac{16d^2}{9} + m \cdot \dfrac{d^2}{9} + m \cdot \dfrac{4d^2}{9} = m \cdot \dfrac{7d^2}{3}$

$mgd = m \cdot \dfrac{7d^2}{6} \cdot \omega^2 \implies \omega = \sqrt{\dfrac{6g}{7d}}$

using the given value for $d$, which I suspect is mistaken, the result is close to 1 rad/sec
 
Jun 2017
352
6
Lima, Peru
$\omega_{max}$ will occur when the bar is at the vertical ...

$U_{g0}-U_{gf} = K_{Rf}-K_{R0}$

$mg \cdot \dfrac{4d}{3} + mg \cdot \dfrac{d}{3} - mg \cdot \dfrac{2d}{3} = mgd = \dfrac{1}{2} \cdot I \cdot \omega^2$

$I = m \cdot \dfrac{16d^2}{9} + m \cdot \dfrac{d^2}{9} + m \cdot \dfrac{4d^2}{9} = m \cdot \dfrac{7d^2}{3}$

$mgd = m \cdot \dfrac{7d^2}{6} \cdot \omega^2 \implies \omega = \sqrt{\dfrac{6g}{7d}}$

using the given value for $d$, which I suspect is mistaken, the result is close to 1 rad/sec
Can you include some drawing to accompany your solution? I'm having a problem at identifying where's the height to establish the potential energy. When you rotate the bar about the axis which is represented by the orange dot, the bar is vertical right? But from where to where you put the height?

Why Initially do we have

$mg \cdot \dfrac{4d}{3} + mg \cdot \dfrac{d}{3}$

and why finally do we have:

$mg \cdot \dfrac{2d}{3}$

@skeeter Can you please represent the justification of this in a drawing?

I got that part of adding the moment of inertia due the principle of superposition of them and the rest is just logical, but...

$ \omega = \sqrt{\dfrac{6g}{7d}}$

in this part is where I'm stuck. How on earth did you get an answer close to $1\frac{rad}{s}$?

If I do insert the given values, this would become into:

$\omega = \sqrt{\dfrac{6\times 9.8}{7\times 4.2 \times 3.14}} \approx 0.7980 \frac{rad}{s}$

Can you please explain this part? Or did I misunderstand something from your explanation? Any help please?
 

skeeter

Math Team
Jul 2011
3,293
1,778
Texas
0.798 is closest to 1rad/sec, given the available choices ... note that I stated disagreement with your given value for $d = 4.2\pi$

For $\omega = 1 \text{\rad/sec}$, $d$ would need to equal to 8.4

Full explanation of the method used to find the maximum angular velocity is given in the following diagram

Omega_final.jpg
 
  • Like
Reactions: topsquark