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A tangential force of $20 \mathrm{~N}$ is applied on a cylinder of mass $4 \mathrm{~kg}$ and moment of inertia $\mid 0.02 \mathrm{~kg} \mathrm{~m}^2 \mid$ about its own axis. If the cylinder rolls without slipping, then linear acceleration of its centre of mass will be
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The correct answer is:
$6.7 \mathrm{~m} / \mathrm{s}^2$
Let friction force $=\mathrm{f} F+f=m a \mid$
(i) $(F-f) R=I \alpha \mid$
From $e q^n s .(i)$ and (ii),$
$\left|2 F=m a+\frac{I \alpha}{R}\right|$
Use $\left.\alpha=\frac{a}{R} \right\rvert\,$ (for pure rolling)
$2 F=m a+\frac{I a}{R^2}\left|40=4 a+\frac{0.02 a}{(0.1)^2} ; a=\frac{40}{6}=6.7 \mathrm{~m} / \mathrm{s}^2\right|$
(i) $(F-f) R=I \alpha \mid$
From $e q^n s .(i)$ and (ii),$
$\left|2 F=m a+\frac{I \alpha}{R}\right|$
Use $\left.\alpha=\frac{a}{R} \right\rvert\,$ (for pure rolling)
$2 F=m a+\frac{I a}{R^2}\left|40=4 a+\frac{0.02 a}{(0.1)^2} ; a=\frac{40}{6}=6.7 \mathrm{~m} / \mathrm{s}^2\right|$
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