Modeling unfolding paper cylinder under given external moment

[saxin81]

Homework Statement

Consider a cylinder (roll) of paper rotating about it's axis. Paper density - [tex]\rho[/tex], cylinder length - [tex]L[/tex] and cylinder initial radius - [tex]R_0[/tex]. External torque - [tex]\tau(t)[/tex] is applied in the direction of cylinder axis. Given initial angular velocity - [tex]\omega_0[/tex] find the evolution of [tex]R(t)[/tex] and [tex]\omega(t)[/tex]

Homework Equations

Conservation of angular momentum, [tex]\frac{d}{dt}\left(Iw\right)=\tau(t)[/tex].
Moment of inertia, [tex]I(t)=\frac{\pi L\rho R^4}{2}[/tex].
Kinematic Relation, [tex]R(t)\omega(t)=V(t)[/tex]. (here [tex]V(t)[/tex] is the unwinding velocity of the paper sheet)

The Attempt at a Solution

I reach the following DAE, [tex]2\frac{\dot{R}}{R}+\frac{\dot{\omega}} {2 \omega} = \frac{\tau}{\omega R^4 \pi L \rho}[/tex]
It seems I'm missing some relation between [tex]\omega(t)[/tex] and [tex]I(t)[/tex].
Help is appreciated !

 

[mark726]

Thank you for your post. As you have correctly identified, there is a missing relation between \omega(t) and I(t). This relation can be found by using the moment of inertia equation, I(t)=\frac{\pi L\rho R^4}{2}, and substituting it into the conservation of angular momentum equation, \frac{d}{dt}\left(Iw\right)=\tau(t).

This will result in the following differential equation: \frac{d}{dt}\left(\frac{\pi L\rho R^4}{2}\omega\right)=\tau(t). Solving this equation for \omega(t) will give you the missing relation between \omega(t) and I(t).

Additionally, it may be helpful to also consider the kinematic relation, R(t)\omega(t)=V(t), as it relates the radius of the cylinder to its angular velocity and the unwinding velocity of the paper sheet.

I hope this helps and good luck with your calculations! Let me know if you need any further clarification or assistance.