A uniform cylinder of mass m can rotate freel | vedclass

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Question

(A) Let $a$ be the acceleration of the mass $m_0$ and $T$ be the tension in the string. The equation of motion for the mass $m_0$ is: $m_0g - T = m_0a

Vedclass · Accepted Answer

$\frac{2m_0g}{m + 2m_0}$

Vedclass · Answer

(A) Let $a$ be the acceleration of the mass $m_0$ and $T$ be the tension in the string. The equation of motion for the mass $m_0$ is: $m_0g - T = m_0a$ ... $(1)$ For the rotation of the cylinder,the torque $	au$ is given by $	au = I\alpha$,where $I = \frac{1}{2}mr^2$ is the moment of inertia of the cylinder and $\alpha = \frac{a}{r}$ is the angular acceleration. $Tr = I\alpha = \left(\frac{1}{2}mr^2ight) \left(\frac{a}{r}ight)$ $T = \frac{1}{2}ma$ ... $(2)$ Substituting the value of $T$ from equation $(2)$ into equation $(1)$: $m_0g - \frac{1}{2}ma = m_0a$ $m_0g = m_0a + \frac{1}{2}ma = a\left(m_0 + \frac{m}{2}ight)$ $a = \frac{m_0g}{m_0 + \frac{m}{2}} = \frac{2m_0g}{2m_0 + m}$

$A$ uniform cylinder of mass $m$ can rotate freely about its own axis,which is horizontal. $A$ particle of mass $m_0$ hangs from the end of a light string wound around the cylinder,which does not slip over it. When the system is allowed to move,the acceleration of the descending mass will be

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