A car is moving at 20 m/s on a banked racing circular track of radius $\frac{160}{3} m$ , and there is no wear and tear of tyres. Height difference [in meter] between two rare or front wheels if width of car is 1m is $[g = 10 m / s^{2}]$ Write until two digits after the decimal point.

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Solution

$N s i n θ = \frac{m v^{2}}{R}$
$N c o s θ = m g$
$\Rightarrow tan θ = \frac{v^{2}}{R g}$
$tan θ = \frac{20 \times 20 \times 3}{160 \times 10} = \frac{12}{16} = \frac{3}{4}$
$\Rightarrow θ = 37^{\circ}$
$sin 37 = \frac{h}{1} \Rightarrow h = \frac{3}{5} = 0.60 m$

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A car is moving at 20 m/s on a banked racing circular track of radius 1603m, and there is no wear and tear of tyres. Height difference [in meter] between two rare or front wheels if width of car is 1m is[g=10 m/s2] Write until two digits after the decimal point.

N sinθ=mv2R N cosθ=mg ⇒tan θ=v2Rg tan θ=20×20×3160×10=1216=34 ⇒θ=37∘ sin 37=h1⇒h=35=0.60 m

A car is moving at 20 m/s on a banked racing circular track of radius $\frac{160}{3} m$ , and there is no wear and tear of tyres. Height difference [in meter] between two rare or front wheels if width of car is 1m is $[g = 10 m / s^{2}]$ Write until two digits after the decimal point.

$N s i n θ = \frac{m v^{2}}{R}$
$N c o s θ = m g$
$\Rightarrow tan θ = \frac{v^{2}}{R g}$
$tan θ = \frac{20 \times 20 \times 3}{160 \times 10} = \frac{12}{16} = \frac{3}{4}$
$\Rightarrow θ = 37^{\circ}$
$sin 37 = \frac{h}{1} \Rightarrow h = \frac{3}{5} = 0.60 m$