Maxwell disc linked to bar unwinds but stays at same height by raising bar

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The discussion revolves around a physics problem involving a Maxwell disc linked to a bar that unwinds while maintaining its height. The calculations begin with Newton's second law, leading to the conclusion that tension is half the weight of the mass. Torque analysis reveals that the angular acceleration is proportional to the tension and inversely related to the moment of inertia. A key point of confusion arises regarding the absence of time in the expression for initial angular velocity, which is attributed to constant torque resulting in constant angular acceleration. The participant acknowledges a mistake in equating angular velocity with linear acceleration, highlighting the complexity of the problem.
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Homework Statement
The ends of thin threads tightly wound on the axle of radius ##r## of the Maxwell disc are attached to a horizontal bar. When the disc unwinds, the bar is raised to keep the disc at the same height. The mass of the disc with the axle is ##m##, the moment of inertia of the arrangement relative to its axis is ##I##.
Relevant Equations
Find the tension of each thread and the acceleration of the bar.
Here is a pictorial depiction of the problem

1719178014410.png


From Newton's 2nd law we have

$$2T-mg=0\implies T=\frac{mg}{2}$$

Then, considering the torques created by the threads we have

$$\vec{\tau}=I\vec{\alpha}=(-r\hat{k}+\frac{l}{2}\hat{i})\times T\hat{j}+(-r\hat{k}-\frac{l}{2}\hat{i})\times T\hat{j}$$

$$=2rT\hat{i}$$

$$\implies \alpha_z=\frac{2rT}{I}=\frac{rmg}{I}$$

Then

$$\omega_z=\frac{rmgt}{I}$$

The velocity of the axle at the point of contact of the thread with the axle is

$$\vec{v}=\vec{\omega}\times\vec{r}=\frac{rmgt}{I}\hat{i}\times (-r\hat{k})=\frac{r^2mgt}{I}\hat{j}$$

This is a problem from the book "Problems in General Physics" by Irodov. The back of the book answer says that

$$\omega_0=\frac{gmr^2}{I}$$

Why is there no factor of ##t## in the expression above?

As far as I can tell, because there is a constant torque there is a constant angular acceleration. Thus, angular velocity and tangential velocity (which coincides with the velocity of the bar being raised) are increasing with time.

My question is if I have made a mistake in my reasoning.
 
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The question as posted asks for the linear acceleration of the bar. That matches the answer given. The only puzzle is that the LHS is ##\omega_0## instead of ##a##.
 
True what a silly oversight.
 
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