Question 266 of 280

A uniform solid cylindrical roller of mass $m$ is being pulled on a horizontal surface with force $F$ parallel to the surface and applied at its centre. If the acceleration of the cylinder is $a$ and it is rolling without slipping then the value of $F$ is:

A$\dfrac{3}{2}ma$B$2ma$C$\dfrac{5}{3}ma$D$ma$

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Answer: (A) $\dfrac{3}{2}ma$

From free body diagram of cylinder

$F-{f}_{s}=ma$ .....(1)

$∵\sum {f}_{ext}=m{a}_{cm}$

$also\sum {\tau }_{ext}={I}_{cm}\alpha$

$⟹{f}_{s}R={I}_{cm}\alpha$

$⟹{f}_{s}R=\dfrac{1}{2}m{R}^{2}\alpha$ ..... (2)

For rolling without slipping



$a=R\alpha$ ...... (3)

$⟹\alpha =\dfrac{a}{R}$

$∴{f}_{s}R=\dfrac{1}{2}m{R}^{2}\dfrac{a}{R}$

$⟹{f}_{s}=\dfrac{1}{2}ma$

Put in (1)

$F-\dfrac{1}{2}ma=ma$

$⟹F=\dfrac{3}{2}ma$