A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal…

1. $\frac{F}{3 mR}$
2. $\frac{3F}{2 m R}$
3. $\frac{2F}{3 m R}$
4. $\frac{F}{2 m R}$

Solution

$F-f_r=ma$ $\dots \left(i\right)$

$f_{r}R=I\alpha =\frac{m{R}^2}{2}\alpha$ $\dots \left(ii\right)$

For pure rolling

$a=\alpha R$ $\dots \left(iii\right)$

From $\left(i\right), \left(ii\right)$ and $\left(iii\right)$

$F-\frac{mR\alpha }{2}=m\alpha R$

$F=\frac{3}{2}mR\alpha$

$\alpha =\frac{2F}{3mR}$

Asked in: JEE Main 2019 (10 Jan Shift 1)

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