Circular Motion: Friction, Centripetal Force & Polygons

[govinda]

hi guys .
i read that friction can be attributed to the centripetal force that causes circular motion . now suppose we consider our circle as a polygon , say an octagon and a bike or car was going along one of the sides of this octagon if i were to negotiate a turn of an angle of 45 degrees in order to go onto the next side of the octagon . so it urn the steering wheel and the tyres turn . now friction acts outwards opposing the change in directon and i think as the polygon approaches a circle this frictional force would be radial outward . something like a centrifugal force .. so where is the force directeed toward the centre of the circle . from my hand as i turn the steering wheel?
thanks
govinda

 

[FredGarvin]

You've got it backwards.

Friction is acting inwards. Friction is what pulls you through the turn. Newton says that your motion would continue in a straight line if left alone. The linear direction is the direction that friction is opposing, not your circular direction.

 

[quasar987]

Hi govinda, I too have been wondering about how exactly is friction the cause circular motion. Here's what I think. I hope someone could comment (i.e. validate or invalidate) on my drawing too.

In the drawing, the perpendicular component of friction is responsible for the circular motion. Because of inertia, the speed remains in the same direction, hence the direction of the force of friction.

( I drew bike tires as seen from above )

 

[govinda]

Hi govinda, I too have been wondering about how exactly is friction the cause circular motion. Here's what I think. I hope someone could comment (i.e. validate or invalidate) on my drawing too.

In the drawing, the perpendicular component of friction is responsible for the circular motion. Because of inertia, the speed remains in the same direction, hence the direction of the force of friction.

( I drew bike tires as seen from above )

nice explanation quasar .. so friction acts linearly .. friction while negotiating the turn can be ignored when the angle is small as the polygon approaches a circle ... also as the turn becomes less the friction opposing linear motion does infact become radially inward (look at the pic and reduce the angle).so in my opinion hypothesis validated.

 

[FredGarvin]

... so friction acts linearly ..

I'm not quite sure what you are implying with this statement, but I guess it is a moot point.

...friction while negotiating the turn can be ignored when the angle is small as the polygon approaches a circle

If you want to continue in a circular path of any kind, friction is not going to be ignored. Again, friction is what is responsible for the circular path.

... also as the turn becomes less the friction opposing linear motion does infact become radially inward (look at the pic and reduce the angle).so in my opinion hypothesis validated.

If you are doing a comparisson between differing circle radii, then you are somewhat correct, if not a bit misleading. You are referring to the resulatant frictional force, are you not? As the radius of the turn increases, the centripetal force (due to friction) decreases and the resultant frictional force approaches an equal value of the tangential friction (opposing the linear velocity). However, I still do not see how you can say that your initial notion of friction acting outward can be "validated"

now friction acts outwards opposing the change in directon and i think as the polygon approaches a circle this frictional force would be radial outward

 

[govinda]

fred,
by validated i meant that the theory of frictional force acting inward was validated which i had a problem with accepting initially ...

 

[quasi426]

When going about in circular motion, one will tend to want to leave the circular path on a tangent due to inertia. However in order to combat this, a radial frictional force is required. If the ground was slippery and there was little friction the bike would not be able to continue in the circular motion because it would skid out and leave tangent to the circular path since the frictional force directed towards the radius is not strong enough to compete with the inertia.