A light rope is wound around a hollow cylinder of mass $4 \mathrm{~kg}$ and radius $40 \mathrm{~cm}$. If the…

A light rope is wound around a hollow cylinder of mass $4 \mathrm{~kg}$ and radius $40 \mathrm{~cm}$. If the rope is pulled with a force of $40 \mathrm{~N}$, its angular acceleration is
  1. $0.40 \mathrm{rads}^{-2}$
  2. 0.25 rads $^{-2}$
  3. $25 \mathrm{rads}^{-2}$
  4. $40 \mathrm{rads}^{-2}$

Solution

Torque, $\tau=I \alpha$ $ \begin{aligned} \mathbf{F} \times \mathbf{r} & =M r^2 \alpha \\ 40 \times 0.4 & =4 \times(0.4)^2 \alpha \\ 16 & =0.64 \alpha \\ \alpha & =\frac{16}{0.64} \\ & =\frac{16}{64} \times 100 \\ & =25 \mathrm{rad} / \mathrm{s}^2 \end{aligned} $

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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