A thin cylindrical shell of inner radius 1.5 m rotates horizontally, about a vertical axis, at an angular speed ω. A wooden block rests against the inner surface and rotates with it. If the coefficient of static friction between block and surface is 0.3, how fast must the shell be rotating if the block is not to slip and fall ?
Q 47.6
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Data : $\mathrm{r}=1.5 \mathrm{~m}, \mu_{\mathrm{s}}=0.3$
The normal force $\vec{N}$ of the shell on the block is the centripetal force which holds the block in place. $\vec{N}$ determines the friction on the block, which in turn keeps it from sliding downward.
If the block is not to slip, the friction force $\overrightarrow{f_{\mathrm{s}}}$ must balance the weight $m \vec{g}$ of the block.
$
\begin{aligned}
& \therefore N=m \omega^2 r \text { and } f_s=\mu_s N=m g \\
& \therefore \mu_s\left(\omega^2 r\right)=m g \\
& \therefore \omega=\sqrt{\frac{g}{\mu_s r}}=\sqrt{\frac{10}{(0.3)(1.5)}}=\sqrt{\frac{10}{0.45}} \\
& \quad=\sqrt{22.22}=4.714 \mathrm{rad} / \mathrm{s}=\frac{4.714}{2 \pi}=0.75 \mathrm{rev} / \mathrm{s}
\end{aligned}
$
This gives the required angular speed.
The normal force $\vec{N}$ of the shell on the block is the centripetal force which holds the block in place. $\vec{N}$ determines the friction on the block, which in turn keeps it from sliding downward.
If the block is not to slip, the friction force $\overrightarrow{f_{\mathrm{s}}}$ must balance the weight $m \vec{g}$ of the block.
$
\begin{aligned}
& \therefore N=m \omega^2 r \text { and } f_s=\mu_s N=m g \\
& \therefore \mu_s\left(\omega^2 r\right)=m g \\
& \therefore \omega=\sqrt{\frac{g}{\mu_s r}}=\sqrt{\frac{10}{(0.3)(1.5)}}=\sqrt{\frac{10}{0.45}} \\
& \quad=\sqrt{22.22}=4.714 \mathrm{rad} / \mathrm{s}=\frac{4.714}{2 \pi}=0.75 \mathrm{rev} / \mathrm{s}
\end{aligned}
$
This gives the required angular speed.
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