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