Finding angular velocity for a rope to be cut

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SUMMARY

The discussion focuses on calculating angular velocity for a system involving two ropes under tension, using Newton's equations. The participant derived the equations for two masses, ##m_1## and ##m_2##, resulting in tensions of ##T_1## and ##T_2##, with ##T_2## given as 1100 N. The calculated angular velocities were ##\omega = 20.56 \frac{rad}{s}## for ##m_2## and ##\omega = 16.37 \frac{rad}{s}## for ##m_1##, indicating that the first rope will be cut at a lower angular velocity. The analysis confirms that ##T_1## will always exceed ##T_2## until one of the ropes fails.

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Homework Statement
The bodies shown in the picture spin on a horizontal plane, describing a circular path with constant acceleration. They're connected by ropes that resist ##1100 N## (each rope). Find the angular velocity when one of the ropes is cut.
Relevant Equations
Newton's equation
I wrote Newton's equations for each body (I took ##x## as the axis aligned with the tension)

##m_1##:
##x)f*_1 -T_1+T_2=0##
Where ##f*_1=\omega ^2 r_1##

##m_2##
##x)f*_2 -T_2=0##
##x)f*_2=T_2##
Where ##f*_2=\omega ^2 r_2##

I wrote that ##T_2=1100 N## and solved for ##\omega##, and I got ##\omega =20.56 \frac{rad}{s}##.

Then, I wrote ##T_2=f*_2## in the equation for ##m_1##, replace ##T_1=1100## and solved for ##\omega##. Doing so I found that ##\omega = 16.37 \frac{rad}{s}##.

So, the first rope will be cut with less angular velocity.

Is this right?
 

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This looks good to me. But note that your first equation (for ##m_1##) shows that ##T_1## is going to be greater than ##T_2## for any value of ##\omega## up until one of the ropes breaks.
 

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