At what angle of banking, speed at turn is = v √gγ?
At what angle of banking, speed at turn is = v √gγ?
Explanation
The speed at a banked turn is given by:
v = √(rg tan(θ))
Given v = √(rg), we can equate:
√(rg tan(θ)) = √(rg)
This simplifies to:
tan(θ) = 1
θ = arctan(1) = 45°.