Question 264 of 280

a string is wound around a hollow cylinder of mass $5 \mathrm{~kg}$ and radius $0.5 \mathrm{~m}$. If the string is now pulled with a horizontal force of $40 \mathrm{~N}$, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)

A$20 \mathrm{rad} / \mathrm{s}^{2}$B$16 \mathrm{rad} / \mathrm{s}^{2}$C$12 \mathrm{rad} / \mathrm{s}^{2}$D$10 \mathrm{rad} / \mathrm{s}^{2}$

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Answer: (B) $16 \mathrm{rad} / \mathrm{s}^{2}$

From newton's second law

$40+\mathrm{f}=\mathrm{m}(\mathrm{R} \alpha)$ ...(i)

Taking torque about 0 we get

$40 \times \mathrm{R}-\mathrm{f} \times \mathrm{R}=\mathrm{I} \alpha$

$40 \times \mathrm{R}-\mathrm{f} \times \mathrm{R}=\mathrm{mR}^{2} \alpha$

$40-\mathrm{f}=\mathrm{mR} \alpha$ ...(ii)

Solving equation (i) and (ii) $\alpha=\frac{40}{\mathrm{mR}}=16 \mathrm{rad} / \mathrm{s}^{2}$