can someone help me with a centripetal force diagram?

Question

I have a tecaher who is supposed to be retired and has lost interest in his class. okay...A boy riding a bike leans into the corner, using a suitable diagram explain why he does this.
I know theres gravity down and the oposing force, then the horizontal component which supplies centripetal acceleration, and the aciuon and reaction of the road but im still stuck as to the purpose of leaning because it seems that all the forces are cancelled out. Please help

Answer 1

That's a nasty question to set you, because you can't actually draw a diagram showing all the force arrows cancelling each other out. This is because centripetal force, strictly speaking, acts TOWARDS THE CENTRE of rotation. Thus, if you draw all the force arrows, you'll have an imbalance, pushing the cyclist over, so if anything, it looks like he should lean the other way. BUT this is because a force diagram, inevitably represents a static system, but you are supposed to be representing a dynamic system, which in this case behaves very differently.

Okay, if you want to draw a diagram that has any meaning, you will have to invoke the imaginary concept of CENTRIFUGE. This you could represent as an arrow acting straight outwards from the effective axis of rotation of the arc the cyclist makes as he goes round the corner. You can use this force arrow to balance the other ones, BUT if you do that, you must emphasise that 'centrifugal force is a fiction, employed to represent the dynamic effect of centripetal acceleration, which cannot be meaningfully portrayed in a static diagram'. Use that exact phrase if you like, just as long as you understand its meaning (since even uninterested teachers seem to take a sick pleasure in catching out their pupils).
If you use CENTRIFUGE, it has an effective magnitude of force, F = ma, where m is the mass of the cyclist, and a is centripetal acceleration = (v^2/r), where r is the cyclist's velocity, and r is the radius of the circle he is making.

So, you now have a diagram, in which the cyclist has tilted, so that gravitational force on his centre of mass effectively has a horizontal component. This will be balanced by the horizontal 'centrifugal force' arrow.

Answer 2

I'll repeat Ian's answer in a quicker way:

You have two options: one is to do this in an inertial reference frame. In this case, you add up the true forces (gravity, normal force, friction) and they add up to the centripetal force, the force that keeps the bike going in a circle.

The other is to do this in the bike's reference frame. This is convenient, because now it is a static problem--the sum of the forces is zero. The downside is that fictitious forces can show up in non-inertial reference frames. In this case, it's the centrifugal force.

This centrifugal pseudoforce is equal and opposite to the centripetal force. You can see that either way you draw the diagram and write the equation, you get the same answer:

real forces = centripetal force
OR
real forces + centrifugal force = 0

Answer 3

Try these sites:

http://frank.mtsu.edu/~phys2010/Lectures/Part_2__L6_-_L11/Lecture_11/Centripetal_Force/centripetal_force.html

http://dev.physicslab.org/Document.aspx?doctype=3&filename=CircularMotion_CentripetalForceExamples.xml

http://www.duguid.org/physics/newtonslaws/centripetalforce.htm