A horizontal force F is applied at the centre of mass of a...

Let moment of inertia $= I$

For pure rolling, $a = \alphaR$

and $F - f = ma . . . \left(1\right)$

and $fR = I\alpha$

or $fR = \dfrac{Ia}{R}$

or $f = \dfrac{Ia}{{R}^{2}}$

Therefore, $F - \dfrac{Ia}{{R}^{2}} = ma$

$∴ F = \left(m + \dfrac{I}{{R}^{2}}\right)a$

$⇒$ For the same value of $F$, value of $a$ depends on $I$.

$⇒$ For a solid cylinder\left(I = \dfrac{m{R}^{2}}{2}\right)$(to calculate maximum$a$)</p><p>$f = \mumg$</p><p>$∴ \mumg = \left(\dfrac{m{R}^{2}}{2}\right) \times \dfrac{a}{{R}^{2}}$</p><p>Therefore,$a = 2\mug$</p><p>$⇒ f$is dependent on applied force.</p><p>$⇒$For thin-walled hollow cylinder\left(I = m{R}^{2}\right)$,

$F = \left(m + \dfrac{m{R}^{2}}{{R}^{2}}\right)a = 2ma$

$∴ a = \dfrac{F}{2m}$