A motorcyclist is describing a circle of radius $25 \mathrm{~m}$ at a speed of $5 \mathrm{~m} / \mathrm{s}$. Find his inclination with the vertical. What is the value of the coefficient of friction between the tyres and ground ?
Q 47.8
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Data : $v=5 \mathrm{~m} / \mathrm{s}, \mathrm{r}=25 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2$
(i)
$
\begin{aligned}
& \tan \theta=\frac{v^2}{r g}=\frac{(5)^2}{25 \times 10}=0.10 \\
& \therefore \theta=\tan ^{-1} 0.10=5^{\circ} \mathbf{4}^{\prime}
\end{aligned}
$
(inclination with the vertical)
(ii) $\frac{m v^2}{r}=\mu_s m g$
where $\mu_{\mathrm{s}}$ is the coefficient of friction.
$
\therefore \mu_{\mathrm{s}}=\frac{v^2}{r g}=0.10
$
(i)
$
\begin{aligned}
& \tan \theta=\frac{v^2}{r g}=\frac{(5)^2}{25 \times 10}=0.10 \\
& \therefore \theta=\tan ^{-1} 0.10=5^{\circ} \mathbf{4}^{\prime}
\end{aligned}
$
(inclination with the vertical)
(ii) $\frac{m v^2}{r}=\mu_s m g$
where $\mu_{\mathrm{s}}$ is the coefficient of friction.
$
\therefore \mu_{\mathrm{s}}=\frac{v^2}{r g}=0.10
$
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