Turning on a straight, unbanked, frictionless road I was learning about circular motion when this question struck me:
In real life situations we are able to take a turn along a circular arc with our bike because friction provides us the necessary centripetal force for doing so. When we talk about cars, the road is banked which provides the centripetal force. But what if a biker wants to take a turn on an unbanked, frictionless road? Would he be able to turn? I think if he bends, then the normal reaction offered by the ground can possibly provide the centripetal force, but while bending, he will slip, and so this possibility might not be correct. I'm confused, and want a satisfactory answer to this question. 
 A: The only possibility I see is to use asymmetric air drag to turn. Or to create air flows - indeed, planes and helicopters do turn without any road friction.
A: Your biker cannot turn.   She needs a force parallel to the surface.  The normal force cannot provide that.  She needs friction, or something else to provide a horizontal force to turn.
A: In normal conditions, when a biker banks, the normal force can use friction from the road to balance the horizontal component of the biker's weight. This leads to a turn because to balance the horizontal and offset-vertical weight, friction from the road creates a force acting inward along a curve; a centripetal force. This counters a torque around the center of mass of the system and allows the biker to both not fall and turn.
On a frictionless road, there would be no horizontal component of force to balance the weight of the biker whenever they shift off of a perfectly balanced system. The center of mass of the bike-biker system would experience the full force of gravity minus the normal force. However, when offset (either in a banking turn or just a minor offset from perfect balance) the normal force becomes less than the force of gravity because of trigonometry. The ground contact point (the wheels) now exert a torque around the center of mass. The horizontal component of weight (again due to trigonometry) is no longer balanced at the contact point. This means a few things: First, the net force on the biker is now not zero; there is a net force down and so they will fall. Second, there is a net horizontal force on the wheels due to the weight of the bike and biker, which means the wheels will slip. And thrid, there is a net torque around the center of mass due to the normal force, which means the system will rotate such that the biker will move groundwards.
What this illustrates is not only could one not turn through banking, but biking in general on a frictionless road would be just as hard as sitting on a stationary bike without it falling over. One would be in an unstable state, any shift of weight away from a perfectly balanced state would cause the biker to fall.
A: of course it is possible.In my opinion the weight of cyclist is acting on his seat and as he bends the seat become an inclined surface.The body of thew cyclist is an object in the inclined plane and the friction and normal reaction offered by the seat can provide the necessary centripetal force.
Another method is to throw a stone in the opposite direction of the turn and in order to conserve the mommentum, we will be able to take the turn.
Again the method adopted by aeroplanes to turn can be used too.
