Mass pulling another, rotating, mass

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SUMMARY

The discussion focuses on the dynamics of a rolling cylinder without friction, specifically analyzing the acceleration of mass 'm', the angular acceleration of mass 'M', and the tension in the system. Key equations utilized include Steiner's theorem for moment of inertia, torque and angular acceleration relationships, and the moment of inertia formula for a solid cylinder. The final expression derived for the acceleration of mass 'm' is 4mg/(3M + 8m), confirming the calculations presented by the user.

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Homework Statement


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The cylinder rolls without friction. what's m's acceleration, M's angular acceleration and the tension.

Homework Equations


Shteiner's theorem: ##I_c=I_{c.o.m.}+Mr^2##
Torque and angular acceleration: ##M=I\alpha##
Moment of inertia of a solid cylinder: ##I_{cen}=\frac{mr^2}{2}##

The Attempt at a Solution


I denote the center variables with c
Kinematics (or geometry? which discipline is suitable for this relation):
$$x_c=\omega r~~\rightarrow~~\dot x_c=\alpha r~~\rightarrow~~\alpha=\frac{\dot x_c}{r}$$
$$\left\{ \begin{array}{l} mg-T=m\dot x_A~~\rightarrow~~T=m(g-\dot x_A)=m(g-2\dot x_c) \\ 2r\cdot T=I_B\frac{\dot x_c}{r} \end{array} \right.$$
$$\rightarrow~\dot x_c=\frac{2r^2mg}{I_B+4mr^2}=\frac{2r^2mg}{I_c+Mr^2+4mr^2}=\frac{2r^2mg}{\frac{Mr^2}{2}+(M+4m)r^2}=\frac{4mg}{3M+8m}$$
 
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I assume you meant "rolls without slipping".
Your use of x for velocity instead of displacement is unusual, but quite OK.
Your working and answer look right to me.
 
Thank you very much Haruspex
 

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