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A rope of negligible mass is wound around a hollow cylinder of mass $3 \mathrm{~kg}$ and radius $40 \mathrm{~cm}$. What is the angular acceleration of the cylinder, if the rope is pulled with a force of $30 \mathrm{~N}$ ? What is the linear acceleration of the rope? Assume that there is no slipping.
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Verified Answer
M.I of the hollow cylinder about its axis $=M R^2=3 \times$
$$
\begin{aligned}
&(0.4)^2=0.48 \mathrm{kgm}^2 \\
&{[\because M=3 \mathrm{~kg} ., R=40 \mathrm{~cm}=0.4 \mathrm{~m}]} \\
&F=30 \mathrm{~N} \therefore \tau=F \times R=30 \times 0.4=12 \text { N.m }
\end{aligned}
$$
But $\tau=I \alpha \Rightarrow \alpha=\frac{\tau}{I}=\frac{12}{0.48}=25 \mathrm{rad} / \mathrm{s}^2$
Linear acceleration $=a=R \alpha=0.4 \times 25=10 \mathrm{~m} / \mathrm{s}^2$
$$
\begin{aligned}
&(0.4)^2=0.48 \mathrm{kgm}^2 \\
&{[\because M=3 \mathrm{~kg} ., R=40 \mathrm{~cm}=0.4 \mathrm{~m}]} \\
&F=30 \mathrm{~N} \therefore \tau=F \times R=30 \times 0.4=12 \text { N.m }
\end{aligned}
$$
But $\tau=I \alpha \Rightarrow \alpha=\frac{\tau}{I}=\frac{12}{0.48}=25 \mathrm{rad} / \mathrm{s}^2$
Linear acceleration $=a=R \alpha=0.4 \times 25=10 \mathrm{~m} / \mathrm{s}^2$
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