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I have this physics problem and I need to solve for $\alpha$, but I can't seem to figure out how. I have been trying for a while and I keep running into dead ends. This is how far I get before I start having problems.

$$R_2(m_2g-m_2R_2\alpha)-R_1(m_1R_1\alpha+m_1g)=\alpha\left(\frac{M_1R_1^2}{2}+\frac{M_2R_2^2}{2}\right)$$

After this point, I usually try to divide by what's being multiplied against $\alpha$ on the right side, but then I can't figure how to isolate $\alpha$ on the left side! Any help will be very appreciated!

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Following what was mentioned in the comments, let's first multiply everything out.

$$R_2m_2g-m_2R^2_2\alpha-m_1R^2_1\alpha+R_1m_1g= \frac{\alpha M_1R_1^2}{2}+\frac{\alpha M_2R_2^2}{2}.$$

Next, bring everything with an $\alpha$ in it to one side. Let's go with the right side.

$$R_2m_2g+R_1m_1g= \frac{\alpha M_1R_1^2}{2}+\frac{\alpha M_2R_2^2}{2} +m_2R^2_2\alpha+m_1R^2_1\alpha.$$

Now, factor out an $\alpha$ from the right side.

$$R_2m_2g+R_1m_1g= \alpha\left(\frac{ M_1R_1^2}{2}+\frac{ M_2R_2^2}{2} +m_2R^2_2+m_1R^2_1\right).$$

Finally, divide both sides by the parenthesis to find $\alpha$.

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Did you multiply out the left hand side of the equation, collect terms involving α and then move that term over to the right hand side of the equation and simplify?

I see some 3/2's in your future.

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