Friction & Circular Motion Static and Moving Friction Centripetal and Centrifugal Forces. Friction was introduced in last lecture, expand Nature of Friction, its origins Static Friction Friction in Motion Circular motion using centripetal force ‘Equivalence’ with gravity
Friction & Circular Motion
Static and Moving Friction
Centripetal and Centrifugal Forces
This does NOT depend on the area in contact!
If the force BETWEEN the surfaces, R, is the same, then the area does not matter.
Total force needed = const × (F/A) × A
ms > mr
Easier to keep sliding than
to get moving
[ 1.0 is the biggest normal value. If >1.0, then ‘sticky’.
Often it is gravity that is creating the force
R = Force = mg
Brass-Steel ms = 0.51
W = 10kg x 9.81 ms-1 = 98N
So F = ms x R = 98N x 0.51 = 50N
This is the force it takes to get a
10kg steel block sliding over
a brass surface.
The reaction force between
the surfaces is given by
R = mg cos(q)
The force down the slope is
F = mg sin(q)
If F > msR block will slide
mgsin(q) > msmgcos(q)
ie if tan(q) > ms
kplane < kbus
(at same altitude offlight)
It takes a force to
change the direction
of a moving object
The force needed to make a circle is:
m is the mass of the circling body
v is its speed
r is the radius of the circle
the force is called the centripetal force
If the body is made to circle bya string, then the string will feela force pulling it into tension.
This is the centrifugal force.
Of course these two different namesare really describing the same force.
But from different perspectives
Riders feel thecentrifugal forceholding them inthe car.
The rail applies
a centripetal force
to make the carcircle.
but it is not enoughto make them fallout of the car
It is common to use w instead of v, where
w = v/2pr
This is the angular speed
- number of radians per second
Then the centripetal force is:
F = w2r
With F = mR (concrete-rubber m = 1.0) and F = mv2/r
can answer question:
How fast can you drive your Ferarri round a corner?
Answer: Depends on the speed limit
Answer: I cant afford a Ferarri
Answer: I cant afford a Ferarri, but in theory:
Consider a real car (Lotus Elise)
Condition is mv2/r = mmg => v = √(mgr)
Answer: v = √(mgr) (BUT remember m is much smaller in the wet!)