the acceleration of the plank and the center of mass of the cylinder

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Solution

We can choose any arbitrary directions of frictional forces at different contacts
In the final answer, the negative value will show the opposite directions.
Let $f_{1} =$ friction between plank and cylinder
$f_{2} =$ friction between cylinder and ground
$a_{1} =$ acceleration of plank
$a_{2} =$ acceleration of centre of mass of cylinder and $α$ angular acceleration of cylinder about is
$C O M .$ Directions of $f_{1}$ and $f_{2}$ are as shown here
Since, there is no slipping anywhere $∴ a_{1} = 2 a_{2}$ .......(i)
(Acceleration ofplank $=$ acceleration oflop point of cylinder)
$a_{1} = \frac{F - f_{1}}{m_{2}} . . . . . . . (i i)$
$a_{2} = \frac{f_{1} + f_{2}}{m_{1}} . . . . . . . (i i i)$
$α = \frac{(f_{1} - f_{2}) R}{I}$
$(I =$ moment of inertia of cylinder-about COM)
$∴ α = \frac{(f_{1} - f_{2}) R}{\frac{1}{2} m_{1} R_{2}}$
$α = \frac{2 (f_{1} - f_{2}) R}{\frac{1}{2} m_{1} R} . . . . . (i v)$
$α = \frac{2 (f_{1} - f_{2})}{m_{1}}$ .....(v)
(Acceleration of bottommost point of cylinder $= 0)$
a) Solving Eqs, (i), (ii), (iii) and (v), we get
$a_{1} = \frac{8 F}{3 m_{1} + 8 m_{2}}$
and $a_{2} = \frac{4 F}{3 m_{1} + 8 m_{2}}$

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