A uniform solid cylindrical roller of mass m is being pulled on a horizontal surface with force F parallel…

1. $\frac{3}{2} ma$
2. $2 ma$
3. $\frac{5}{3} ma$
4. $ma$

Solution

From free body diagram of cylinder

$F-f_s=ma$ .....(1)

$\because \sum f_{ext}=m{a}_{cm}$

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

$⟹f_s R=I_{cm}\alpha$

$⟹f_s R=\frac{1}{2} m{R}^2\alpha$ ..... (2)

For rolling without slipping




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

$⟹ \alpha =\frac{a}{R}$

$\therefore f_s R=\frac{1}{2}m{R}^2\frac{a}{R}$

$⟹ f_s=\frac{1}{2}ma$

Put in (1)

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

$⟹F=\frac{3}{2}ma$

Asked in: JEE Main 2015 (10 Apr Online)

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