While taking a turn, is it possible for a cyclist to lean at an angle of 45^(@) with the vertical?

While taking a turn, is it possible for a cyclist to lean at an angle

1.	While taking a turn, is it possible for a cyclist to lean at an angle of 45^(@) with the vertical?
Answer» Solution :We know that when a cyclist TAKES a turn along a circular path of radius r with velocity v , he leans at an angle of `theta` with the vertical. In this case, `tan theta = (v^(2))/(rg)""`…(1) Again, normal force on the cycle = mg, where m is the mass of the cycle along with the cyclist. So, if the coefficient of friction is `mu`, then limitingfrictional force = `mu` mg. this limiting FRICTIONAL force provides the necessary centripetal force. Hence, ` (mv^(2))/(r) = mu mg " " or, " " mu = (v^(2))/(rg)""`...(2) From the equations (1) and (2) we get, tan`theta = mu`. it is given that`theta = 45^(@) `, so , `mu = tan 45^(@) = ` 1. But the coefficient of friction between the WHEEL and the road can never be 1, actually it is less than 1. Hence, the cycle will skid when the cyclist leans at an angle of `45^(@)` with the vertical.