1) On a banked race track, the smallest circular path on which cars can move has
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Question
1) On a banked race track, the smallest circular path on which cars can move has a radius of 120 m, while the largest has a radius 155 m, as the drawing illustrates. The height of the outer wall is 18 m. (a) Find the smallest speed at which cars can move on this track without relying on friction. m/s (b) Find the largest speed at which cars can move on this track without relying on friction. m/s 2) A jet flying at 105 m/s banks to make a horizontal circular turn. The radius of the turn is 3810 m, and the mass of the jet is 2.04 105 kg. Calculate the magnitude of the necessary lifting force.Explanation / Answer
a.)The angle of incline of the track should be given by:
tan(theta) = (18m)/(155m-120m) = 0.514
=>theta=27.22 degrees
going around the bend the centrifugal force pushing the car horizontally to the outside of the track is given by:
Fc = mv^2/r
The component of this force pushing up the slope is:
Fc(up slope) = mv^2/r*cos(theta)
The component of gravity pulling the car down the slope is:
Fg(down slope) = mg*sin(theta)
You can equate the two at the maximum radius, and solve for v to get the maximum speed:
mv^2/r*cos(theta) = mg*sin(theta)
=> v^2 = r*g*tan(theta)
=> v= [r*g*tan(theta)]^0.5 = [(155m)*(9.8m/s^2)*(0.514)]^0.5
= 27.94 m/s maximum speed.
You can do the same at the minimum radius, and solve for the minimum speed:
v= [r*g*tan(theta)]^0.5 = [(120m)*(9.8m/s^2)*(0.514)]^0.5
= 24.58 m/s minimum speed.
b.)Thecentripetal force(lifting force) = m*v^2/r = 2.04x10^5kg*(105m/s)^2/3810m = 0.59x10^6N
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