I guess we are doing feet and pounds and such.
force down on car = 3000 lbs = m g
car mass = 3000/g
normal force of weight on track = 3000cos 31
so max friction force down tangent to track = 3000 cos31*0.9
component of weight down tangent to track = 3000 sin 31
acceleration tangent to track = v^2/R * cos 31 down track
so
3000 * .9 * cos 31 + 3000 sin 31 = m v^2/R = 3000/g * v^2/R
.9 cos 31 + sin 31 = (v^2/R) / g = ratio of centripetal to gravitational acc
(of course in the end the weight does not matter :)
R = 1000
g = 32 ft/s^2
solve for v in feet/second
The Daytona International Speedway shown below has the following track statistics: Bank turn of 31.0o, a radius of curvature of 1000 ft, and a total length of 2.50 mi. Given that the typical weight of a NASCAR car is 3000 lbs, and that the friction coefficient between the tires and the track is around 0.9-1.3 (use 0.900), estimate the maximum speed of the race car in a banked turn.
2 answers
Forgot to continue component of centrripetal down slope
acceleration tangent to track = v^2/R * cos 31 down track
so
3000 * .9 * cos 31 + 3000 sin 31 = m v^2/R = 3000/g * v^2/R cos 31
.9 cos 31 + sin 31 = (v^2/R) cos 31 / g = ratio of centripetal to gravitational acc
(of course in the end the weight does not matter :)
R = 1000
g = 32 ft/s^2
solve for v in feet/second
acceleration tangent to track = v^2/R * cos 31 down track
so
3000 * .9 * cos 31 + 3000 sin 31 = m v^2/R = 3000/g * v^2/R cos 31
.9 cos 31 + sin 31 = (v^2/R) cos 31 / g = ratio of centripetal to gravitational acc
(of course in the end the weight does not matter :)
R = 1000
g = 32 ft/s^2
solve for v in feet/second
1 answer