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) Angle subtended by the arc at center `2alpha=(L)/(R )`
At the highest point:
`mg-N=(mv^(2))/(R )`
`N=mg-(mv^(2))/(R )`
For no loss of contact at the highest point
`Nge0`
`mg-(mv^(2))/(R ) ge 0 implies v le sqrt(gR)`
`v_(max)=sqrt(gR)`
(b)
At P:
`mg cos theta-N=(mv^(2))/(R )=(m)/(R )(v_(max)^(2))/(2)=(m)/(R )xx(gR)/(2)=(mg)/(2)`
when the car loses contact,`N=0`.
`cos theta=(1)/(2)impliestheta=60^(@)=(pi)/(3)`
Distance traveled by the car `=AP=(piR)/(3)`
(c ) `N=mg cos theta-(mv^(2))/(R )`, as `theta` increase, `N` decreases
As the car moves downward, normal reaction decreases. `N` will be minimum at the lowest point,i.e., `theta=alpha`.
`N_(min=mg cos alpha-(mv^(2))/(R )`
For no loss of contact, `N_(minge0`.
`mg cos alpha-(mv^(2))/(R )ge0impliesvlesqrt(gRcosalpha)`
`v_(max)=sqrt(gRcos((L)/(2R)))`