A curve of radius 27 m is banked so that a 1000 kg car traveling at 48.4 km/h ca
ID: 1437791 • Letter: A
Question
A curve of radius 27 m is banked so that a 1000 kg car traveling at 48.4 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. You are commissioned to tell the local police the range of speeds at which a car can travel around this curve without skidding. Neglect the effects of air drag and rolling friction. If the coefficient of static friction between the road and the tires is 0.300, what is the range of speeds you tell them? v_min = 32 km/h v_max = 48.4 km/h A car could either skid up the bank, toward the outside of the curve, or down the bank, toward the inside of the curve. Draw a free-body diagram showing the three forces acting on the car skidding to the outside, and the car skidding toward the inside of the curve. (A head-on view of the car may be the most useful perspective.) The static friction force acts in opposite directions for these two cases. Apply Newton's second law to both cases to determine the range of speeds for which no skidding will occur.Explanation / Answer
here,
radius of curve, r = 27 m
mass of car, mc = 1000 kg
velocity, v = 48.4 km/hr = 48.4*0.105 = 5.082 m/s
Coefficient of static friction, us = 0.300
For banked curve angle is given as:
A = arcTan(V^2/rg)
A = arcTan(5.082^2 / (27*9.81))
A = 5.569 degrees
The maximum which car can drive, Vmax = sqrt(r*g*us)
Vmax = sqrt(27*9.81*0.3)
Vmax = 8.914 m/s = 8.914/0.105 = 84.895 km/hr or 85 km/hr (Rounded off)
Minimum Velocity is when there is no friction on track,
Vmin = sqrt(rgtanA)
Vmin = sqrt(27*9.81*0.098 )
Vmin = 5.095 m/s = 5.095/0.105 = 48.524 km/hr or 49 km/hr(Rounded off)