A motorcycle has to move with a constant speed on an overbridge which is in the form of a circular arc of radius R and has a total length L. Suppose the motorcycle starts from the highest point.
a. what can its maximum velocity be for which the contact with the road is not broken at the highest point?
b. If the motorcycle goes at speed `1/sqrt2` times the maximum found in part a. where will it lose the contact with the road?
c. What maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge?

Explanation

a. `theta = ("Arc length")/("Radius") = (L)/(R )`

From free body diagram at top position equation of motion is `mg - N = (mv^(2))/(R )`
` rArr N = mg - (mv^(2))/(r )`
If the constant at highest point does not loose .
` N gt 0, mg gt (mv^(2))/(R )`
`v^(2) lt g R rArr v lt sqrt(g R)`
b. If the velocity of motorcyclist `v'^(1) = (1)/(sqrt2) v sqrt(gR)/(sqrt(2))`

At position `A`
`mg cos alpha- N' = (mv'^(2))/(R )`
` rArr = mg cos alpha - (mg)/(2) (sqrt((gR)/(2)))^(2)`
`N' = mg cos alpha - (mgR)/(2R) = mg cos alpha - (mg)/(2)`
If he does not loose contact with road `N'gt 0`
`mg cos alpha - (mg)/(2) gt 0 rArr cos alpha gt (1)/(2) rArr alpha lt (pi)/(3)`
Hence are length `l' = (R pi)/(3)` from top.
c. The equation of motion of motorcyclist in the function of `alpha` measured from verticle,
`mg cos alpha - N^(")= (mv^(''2))/(R )`
The normal reaction will be minimum at the end of are Hence, critical position is at the end are , where
`alpha = (theta)/(2) = (L)/(2R)`
`mg cos ((L)/(2R)) - N^('') = (mv^('' 2))/(R )`
`rArr N^('') = mg cos ((OL)/(2R)) -(mv''^(2))/(R)`
If the motorcycle does not loose contact even at the end of are `N^('') gt 0`
`mg cos ((L)/(2R)) - (mv^(''2))/(R ) gt 0 `
`rArr g cos ((L)/(2R)) gt (v^(''2))/(R )`
`rArr v^('') lt sqrt( gR cos((L)/(2R)))`