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A car is moving along a circular path having coefficient of friction 0.5 and radius of curvature $16.2 \mathrm{~m}$. Then the maximum velocity of the car that can travel without falling outwards is
(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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The correct answer is:
$32.4 \mathrm{kmh}^{-1}$
Coefficient of friction, $\mu=0.5$
Radius of curvature, $r=16.2 \mathrm{~m}$
Along the radial direction, we have
friction provides the centripetal force, $\mathrm{f}=\mu \mathrm{N}=\mu \mathrm{mg}$
$\begin{aligned} & \frac{\mathrm{m}^2}{\mathrm{r}}=\mu \mathrm{mg} \\ & v=\sqrt{\mu \mathrm{rg}}=\sqrt{0.5 \times 16.2 \times 10} \\ & =9 \mathrm{~m} / \mathrm{s} \\ & =32.4 \mathrm{~km} / \mathrm{h}\end{aligned}$
Radius of curvature, $r=16.2 \mathrm{~m}$
Along the radial direction, we have
friction provides the centripetal force, $\mathrm{f}=\mu \mathrm{N}=\mu \mathrm{mg}$
$\begin{aligned} & \frac{\mathrm{m}^2}{\mathrm{r}}=\mu \mathrm{mg} \\ & v=\sqrt{\mu \mathrm{rg}}=\sqrt{0.5 \times 16.2 \times 10} \\ & =9 \mathrm{~m} / \mathrm{s} \\ & =32.4 \mathrm{~km} / \mathrm{h}\end{aligned}$
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