In the given figure,a solid cylinder of mass | vedclass

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(A) For a solid cylinder of mass $m_2$ and radius $R$ undergoing downward motion,the equation of motion is $m_2g - T = m_2a_2$,where $T$ is the tensio

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$\frac{a_2}{R}$

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(A) For a solid cylinder of mass $m_2$ and radius $R$ undergoing downward motion,the equation of motion is $m_2g - T = m_2a_2$,where $T$ is the tension in the string.For the rotation of the cylinder about its center of mass,the torque equation is $	au = I\alpha_2$,where $I = \frac{1}{2}m_2R^2$ is the moment of inertia of the solid cylinder.The torque is provided by the tension $T$,so $TR = I\alpha_2 = (\frac{1}{2}m_2R^2)\alpha_2$.Thus,$T = \frac{1}{2}m_2R\alpha_2$.Since there is no slipping,the linear acceleration $a_2$ is related to the angular acceleration $\alpha_2$ by the condition $a_2 = R\alpha_2$.Therefore,$\alpha_2 = \frac{a_2}{R}$.

In the given figure,a solid cylinder of mass $m_2$ and radius $R$ is suspended by a string wrapped around it,which passes over a fixed pulley of mass $m_1$ and radius $R$. If the linear acceleration of the solid cylinder is $a_2$ and there is no slipping,what is its angular acceleration $\alpha_2$?

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