A plank of mass M is placed on a smooth surface over which a cylinder of mass m (= M) and radius R = 1 m is placed as shown in figure. Now the plank is pulled towards the right with an external force F (= 2Mg). If the cylinder does not slip over the surface of the plank, find the linear acceleration of the cylinder (in m/s²). (Take g = 10 m/s²).

Here we are given that cylinder does not slip over the plank surface, it is the case of pure rolling. The friction between the plank and cylinder should be static. We can assume friction on cylinder in any direction, let it is acting on cylinder towards the right. As ftiction is acting on the cylinder towards the right, it must be towards the left on the plank as shown in the force diagram of figure.

Let the plank move towards the right with an acceleration a₁. If we consider the motion of the cylinder w.r.t plank, the cylinder will experience a pseudo force ma₁ in the left direction, due to which it will roll towards the left with respect to the plank with an acceleration a₂. Let its angular acceleration during rolling be $\mathrm{\alpha }$, we have ${\mathrm{a}}_2=\mathrm{R\alpha }\mathrm{. }$

For translational motion of the plank, we have

$\mathrm{F}-\mathrm{f}={\mathrm{Ma}}_1$                 …(i)

For translational motion of the cylinder with respect to the plank, we have

${\mathrm{ma}}_1-\mathrm{f}={\mathrm{ma}}_2$             …(ii)

For rotational motion of the cylinder with respect to the plank, we have 

$\mathrm{fR}={\mathrm{I}}_{\mathrm{cm}}\mathrm{\alpha }\Rightarrow \mathrm{fR}=\left(\frac{1}{2}{\mathrm{mR}}^2\right)\left(\frac{{\mathrm{a}}_2}{\mathrm{R}}\right)$

Which gives $\mathrm{f}=\frac{1}{2}{\mathrm{ma}}_2$           …(iii)

From Equations (i) and (ii), we get

${\mathrm{ma}}_1-\frac{1}{2}{\mathrm{ma}}_2={\mathrm{ma}}_2$ or ${\mathrm{a}}_1=\frac{3}{2}{\mathrm{a}}_2$            …(iv)

Using equations (i), (iii) and (iv), we get $\mathrm{F}-\frac{1}{2}{\mathrm{ma}}_2=\frac{3}{2}{\mathrm{Ma}}_2$

$\Rightarrow {\mathrm{a}}_2=\frac{2\mathrm{F}}{3\mathrm{M}+\mathrm{m}}=10\mathrm{m}/{\mathrm{s}}^2$

From Eq. (iv), ${\mathrm{a}}_1=\frac{3\mathrm{F}}{3\mathrm{M}+\mathrm{m}}=15\mathrm{m}/{\mathrm{s}}^2$

As we have already discussed that the value of a₂ is relative to the plank, the net acceleration of the cylinder will be given as ${\mathrm{a}}_1-{\mathrm{a}}_2$.

Hence, the acceleration of the cylinder is

${\mathrm{a}}_{\text{cylinder }}={\mathrm{a}}_1-{\mathrm{a}}_2=15-10=5\mathrm{m}/{\mathrm{s}}^2$