A circular turn on a racing track has a radius of 150 m what must be the banking angle if optimum speed for taking the turn is to be 108 kmph
Answers
Answer:
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Explanation:
Let the angle of banking be θ. The forces on the car are as shown in the figure.
(a) weight of the car Mg downward and
(b) normal force N.
For proper banking, static frictional force is not needed.
For vertical direction the acceleration is zero. So,
N cos θ = Mg .......... (i)
For horizontal direction, the acceleration is v
2
/r towards the centre, so that
N sin θ=Mv
2
/r .......... (ii)
From (i) and (ii), tanθ=v
2
/rg
Putting the values, tanθ=
(600m)(10m/s
2
)
180(km/h)
2
=0.4167⇒θ=22.6
∘
Answer:
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Hope it helps.