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Question: A circular road of radius 1000m has banking angle \(45^\circ \). The maximum safe speed of a car hav...

A circular road of radius 1000m has banking angle 4545^\circ . The maximum safe speed of a car having mass 2000kg2000kg will be, if coefficient of friction between tyre and road is 0.5:0.5:
A. 172m/s{\text{172m/s}}
B. 124m/s{\text{124m/s}}
C. 99m/s{\text{99m/s}}
D. 86m/s{\text{86m/s}}

Explanation

Solution

Find the correlation between the known and unknown terms. Here, use the formula vmax=grμtanθ(1μtanθ){v_{\max }} = \sqrt {\dfrac{{gr\mu \tan \theta }}{{(1 - \mu \tan \theta )}}} and place the given values and simplify using the mathematical operations.

Complete step by step solution:
Radius, r=1000mr = 1000m , Angle, θ=45\theta = 45^\circ , Mass of the car, mc=2000kg{m_c} = 2000kg.
The coefficient of friction, μ=0.5\mu = 0.5
Now, the maximum safe speed can be given by –
vmax=grμtanθ(1μtanθ){v_{\max }} = \sqrt {\dfrac{{gr\mu \tan \theta }}{{(1 - \mu \tan \theta )}}}
Place the given values in the above equation. Take g=9.8m/s2g = 9.8m/{s^2}
vmax=9.8×1000×0.5×tan451(0.5)tan45{v_{\max }} = \sqrt {\dfrac{{9.8 \times 1000 \times 0.5 \times \tan 45^\circ }}{{1 - (0.5)\tan 45^\circ }}}

Place the value of trigonometric function tangent and Simplify using the basic mathematical operations –

vmax=9.8×1000×0.5×11(0.5)(1) vmax=9.8×1000×0.51(0.5) vmax=98×10×50.5 {v_{\max }} = \sqrt {\dfrac{{9.8 \times 1000 \times 0.5 \times 1}}{{1 - (0.5)(1)}}} \\\ \Rightarrow{v_{\max }} = \sqrt {\dfrac{{9.8 \times 1000 \times 0.5}}{{1 - (0.5)}}} \\\ \Rightarrow{v_{\max }} = \sqrt {\dfrac{{98 \times 10 \times 5}}{{0.5}}} \\\

Now, 0.5=5100.5 = \dfrac{5}{{10}}
vmax=98×10×5×105{v_{\max }} = \sqrt {\dfrac{{98 \times 10 \times 5 \times 10}}{5}}
Same terms in the numerator and denominator cancel each other

vmax=98×10×10 vmax=9800 {v_{\max }} = \sqrt {98 \times 10 \times 10} \\\ \Rightarrow{v_{\max }} = \sqrt {9800} \\\

Taking square root
vmax=98.99m/s\Rightarrow {v_{\max }} = 98.99m/s
The above value can be equivalent and written as –
vmax=99m/s\therefore {v_{\max }} = 99m/s

Hence, the option C is the correct answer.

Note: Remember the trigonometric table for the reference values for different angles for sine, cosine and tangent functions for direct substitution.Since the friction and load are measured in units of force, they cancel each other as a result the unit of coefficient of friction (μ=FL\mu = \dfrac{F}{L} ) is dimensionless. Basically there are two types of coefficient of friction.
1. Static coefficient friction – It is applied to the objects which are motionless.
2. Kinetic coefficient friction – It is applied to the objects which are in motion.