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I'm asked to find the linear acceleration of this object

enter image description here

for a given tension $T$ knowing that both discs have mass $M$ and we don't consider the mass of the bar.

The answer of the book is different than mine and I'm not sure where I have errors. What I did:

Considering $T= I\alpha$ being $\alpha$ the angular accelaration I have $a_{cm}=\alpha (R-r)$. I have two discs of mass $M$ which means $I=\displaystyle\frac{1}{2}(2M)(R^2) = MR^2$.

Then $T = (MR^2)\alpha = (MR^2)\left(\displaystyle\frac{a_{cm}}{R-r}\right) \implies a_{cm} = \displaystyle\frac{T(R-r)}{MR^2}$.

But the answer in the book is $a_{cm}=\displaystyle\frac{T(R-r)}{3MR}$. Maybe $I$ is not right?, but this is just a guess and I'm not sure why $3MR$ would be $I$ in this case.

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closed as off-topic by John Rennie, user10851, Brandon Enright, Kyle Kanos, tpg2114 Feb 18 '14 at 18:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

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There are a couple of problems with this. One is that $$\tau_{net}=I\alpha$$ is an equation that relates net torque to angular acceleration, but $T$ shown in the figure is a force (tension), not a torque. So use the definition of torque to convert $T$ into torque due to the tension ($\tau_{T}$). Also, the relationship between the linear acceleration $a_{cm}$ and the angular acceleration $\alpha$ is simply $a_{cm} = \alpha R$ (don't use the difference between $R$ and $r$). Also, does the problem state that $T$ is applied parallel to the ground? (It looks like it from the figure, so if it doesn't say so, I would assume it's parallel to the ground, but it does make a difference if it's not...)

This problem boils down to two parts. First, think about why the object might roll to the right, rather than sliding to the right. (I'm assuming also somewhere in the problem statement you must have been told that the object is rolling without sliding, right?) There must be some force preventing it from sliding (friction). So draw a free body diagram with all the relevant forces (two of them: $T$ to the right, and $F$ (friction) to the left). Then use Newton's 2nd Law to relate the net force to the linear acceleration. Then figure out what torques are present (two of them: one due to $T$ and one due to friction), add them up, and use $$\tau_{net}=I\alpha$$ to relate the net torque to the angular acceleration (and we already know the relationship between angular and linear accelerations). So then you'll have two equations (force and torque) and two unknowns (force of friction and linear acceleration). Be sure to define positive torque as the torque that will cause rotation in the same direction as the linear acceleration (i.e. to the right). That should solve the problem.

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