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Question

A circular racing track of radius 300 m is banked at an angle of 17 with the horizontal. If the coefficient of static friction (μs) between the wheels of the racing car and the road is 0.2, what is the maximum permissible speed to avoid slipping? Take g=10 m/s2. Use the given data: tan17=0.30, sin17=0.29, cos17=0.96 and 0.500.940.5

A
1015 m/s
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B
335 m/s
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C
515 m/s
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D
50 m/s
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Solution

The correct option is A 1015 m/s
For the maximum permissible speed (v), static friction will act at it's maximum value fmax.

Horizontal component of friction and normal reaction will provide the necessary centripetal force for the car. fmax=μsN ,will act down the inclined plane to prevent slipping of the car.



Applying the equation of dynamics towards the centre of circular path:
Nsinθ+fmaxcosθ=mac=mv2r
N(sinθ+μscosθ)=mv2r ...(i)

For the vertical equilibrium:
Ncosθ=fmaxsinθ+mg
N(cosθμssinθ)=mg ...(ii)

Dividing Eq. (i) and (ii):
sinθ+μscosθcosθμssinθ=v2rg
tanθ+μs1μstanθ=v2rg
v=rg(tanθ+μs)1μstanθ ...(iii)

Putting the given values in (iii) r=300 m, tan17=0.30, μs=0.20
v=300×10(0.30+0.20)1(0.20×0.30)
v=3000×0.500.94=3000×0.5
v=1015 m/s
This is the maximum permissible speed on the track.

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