A racecar driver has to hold on tightly when going around a banked curve. How many "gs" does the driver experience in a 2200 kg car going around a circle with a radius of 95 meters at 25 m/s?
To determine the number of "gs" experienced by the racecar driver, we need to calculate the centripetal acceleration first.
Centripetal acceleration is given by the formula:
a = v^2 / r
Where:
a = centripetal acceleration
v = velocity of the car
r = radius of the curve
Given values:
v = 25 m/s
r = 95 m
Substituting the values into the formula:
a = (25 m/s)^2 / 95 m
Simplifying the equation further:
a = 625 m^2/s^2 / 95 m
a = 6.58 m/s^2
Now, to calculate the number of "gs" experienced by the driver, we can convert the centripetal acceleration to multiples of the force of gravity (g), where 1 g = 9.8 m/s^2.
Number of "gs" = a / g
Substituting the values into the formula:
Number of "gs" = 6.58 m/s^2 / 9.8 m/s^2
Number of "gs" = 0.67 g
Therefore, the racecar driver experiences approximately 0.67 "gs" while going around the banked curve.
v^2 / R is centripetal acceleration
divide that by g or about 9.81 m/s^2 to get the number of g s
Your question does not say if you are to include 1 g due to weight down.
If so it would be 1 g up and v^2 / gR inward
or sqrt (1 + v^2/gR)