STD 11 - 6. system of particles and rotational motion — NEET STD 11 Science — Question

$a=\frac{F+f}{m}$

for roational motional about $C . M$ of

$R(F-f)=I \alpha$

$\alpha=\frac{R}{I}(F-f)$

For pure rolling $a=R \alpha$

$\frac{F+f}{m}=\frac{R^{2}}{I}(F-f)$

$f\left(1+\frac{m R^{2}}{I}\right)=F\left(\frac{m R^{2}}{I}-1\right)$

$f=\frac{\left(m R^{2}-I\right)}{m R^{2}+I} F$

For ring//thin pipe $I=m R^{2},$ so $f=0$ therefore for rolling of ring//thin pipe does not need friction.