a car of mass M=1500kg travelling at 45.0 km/h enters a bankedturn coveres with

Question

a car of mass M=1500kg travelling at 45.0 km/h enters a bankedturn coveres with ice. The road is banked at an angle of=20o and there is no friction between theroad and the cars tyres. What is the radius of the turn (assumingthe car continues in uniform circular motion around the turn) Is this formula correct? R= v2/ gtan and should be in radians? a car of mass M=1500kg travelling at 45.0 km/h enters a bankedturn coveres with ice. The road is banked at an angle of=20o and there is no friction between theroad and the cars tyres. What is the radius of the turn (assumingthe car continues in uniform circular motion around the turn) Is this formula correct? R= v2/ gtan and should be in radians?

Explanation / Answer

The forces acting on the car are Fx = n * sin = m *arad and Fy = n * cos + (-mg) = 0 From the Fy equation,n =(mg/cos).Substituting this into the Fxequation gives an expression for the banking angle: tan = (arad/g) Finally,substituting the expression arad =(v2/R),we have tan = (v2/g * R) ----------(1) where is the banking angle,v is the velocity of thecar,g is the acceleration due to gravity and R is the radius of theturn. From equation (1) we get R = (v2/g * tan) -----------(2) where v = 45.0 km/h = 45.0 * (1000/3600) m/s = 45.0 *(5/18) m/s = 12.5 m/s g = 9.8 m/s2 = 20o Substituting the above values in equation (2) we get theradius of the turn. Substituting the above values in equation (2) we get theradius of the turn.

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