Finding angular velocity for a rope to be cut

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The discussion focuses on calculating angular velocity for a rope under tension using Newton's equations. The user derived two equations for two masses, incorporating tension values and angular velocities, ultimately finding two different angular velocities: 20.56 rad/s for the second mass and 16.37 rad/s for the first. The conclusion drawn is that the first rope will be cut at a lower angular velocity. However, it is noted that the first equation indicates that the tension in the first rope will always exceed that of the second until one breaks. This analysis confirms the relationship between tension and angular velocity in the system.
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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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