Download Centripetal Force (Chapter Section 6.5)

PES 1110 Fall 2013, Spendier
Lecture 15/Page 1
Today:
- Centripetal Force (Chapter Section 6.5)
- Drag (Chapter Section 6.4)
- HW 4 given out due in a next Wednesday ( in one week)
Recap: friction
For a block on an incline plane we learned last lecture that the block will only start to
slide down the inline when the angle of the incline with the horizontal is large enough.
We decided that there must be some force preventing motion. We called this force
friction.
From another demo we saw that once the block is moving we need to apply less force
than right before it moved. Hence, the frictional force is different depending on whether
an object is stationary or moving.
When there is motion force of friction is in the opposite direction of motion. We called

this friction kinetic friction f k . When there is no motion frictional force varies with

applied force up to some maximum, which we called maximum static friction. f s ,max
We also learned that frictional force depends on how hard two objects are pushed
together; hence friction is depended on normal force:
f s ,max = μs Fn
(Coefficient of static friction μs multiplied by normal Force Fn)
fk
= μk Fn (Coefficient of kinetic friction μk multiplied by normal Force Fn)
Note: We will drop the subscript “max” on static friction from now on!
Example 1:
Find the shortest distance a car can stop in if μk = 0.6 and v0 = 89 km/hr (55 mi/hr). In
this problem we are using kinetic friction since the car is moving.
PES 1110 Fall 2013, Spendier
Lecture 15/Page 2
When you drive on snow, we have μk = 0.1. How much further would you go? You
would go 6 times further: x = 321 m.
This is how police can determine how fast a driver was going in an accident. They simply
work backward by measuring the length of the skid marks they can determine the speed
at which the car was going before the driver slammed the breaks.
Centripetal Force
We know that when a body moves in a circle (or a circular arc) at constant speed v, there
is a centripetal acceleration that is directed towards the center of the circle:
v2
a
R
We know from Newton’s second law that some force must be acting to provide

acceleration since  F  ma
net
Therefore:
v2
R
Forces which are directed inward when a body moves in a circle (or a circular arc) are
known as centripetal forces.
Fcentripetal  ma  m
Everyday Example:
When a car turns in a flat circle, what force gives rise to centripetal motion?
It is friction!
What kind of friction is the car holding in the turn? Where does it point to? In this case
we use static friction because the friction is acting perpendicular to the direction of
motion and it is this frictional force which prevents the car from slipping out of the turn.
What is the maximum friction?
f s = μs Fn
PES 1110 Fall 2013, Spendier
Lecture 15/Page 3
So if only friction is holding the car on the road in a turn, friction must point inward and
is therefore our centripetal force:
f s  ms Fn  ms mg  m
ms g 
v2
R
v2
R
From this expression we can now find the maximum velocity of a car before it slips out
of the turn:
vmax  ms gR
Let’s plug in some numbers: μs = 0.75 and R = 150m 
v max =33.2 m/s = 120 km/hr = 75 miles/hr
This estimate is large and typically turns are much sharper (R is smaller) and therefore
vmax is smaller.
What happens when there is any water or ice on the road? Then μs = 0.1 and vmax is
getting even smaller ( almost 3 times smaller) this is why we need to be careful driving
when the roads are wet or snow covered.
How can we improve the safety on the road, i.e. make less cars slip out of a turn? For
example can we build roads in such a way that the maximum speed can be increased?
(Nascars can go 200 mph (320 km/h) in a turn –From our previous calculation we
estimated vmax = 75 mph for a 150 m turn. So what is going on? How can a Nascar go
faster than this?)
Banked Turns:
So on flat roads we have seen that velocity is limited by friction since friction provides
the centripetal force to go around the corners.
By banking the track another force can also contribute to the net, unbalanced centripetal
force. Which force? The normal force!
PES 1110 Fall 2013, Spendier
Lecture 15/Page 4
Example 3: Banked turn ignoring friction between tires and road (in your HW you will
add this friction back)
This is the reason why turns on roads, specifically on highways, are banked so that cars
won’t slip as easily off the road when the road is wet or icy.
PES 1110 Fall 2013, Spendier
Lecture 15/Page 5
Drag:
Drag is simply kinetic friction caused by motion through a fluid (gas or liquid). It is a

force: drag force D
In order for an object to move through a fluid, it must push the fluid particles out of the
way.
Drag depends on:
- size of the object A: a larger cross-sectional area means more particles to push
away
- density of the fluid ρ: a denser fluid has either more particles of massive
particles, making it harder to push them away.
- Speed of the object v: the faster the object moves the more particles it must push
aside.
1
Considering each of these gives: D  C r Av 2
2
Where C is the drag coefficient
units [N]
Terminal speed

A drag force D opposes the relative motion and points in the direction in which the fluid
flows relative to the body.
Blunt object falling: When it falls from rest through air D is initially zero. As the object
keeps falling the drag force keeps increasing until it balances the gravitational force on
the object. The body now falls at its constant terminal speed.
To obtain this terminal speed we use Newton’s second law for the vertical y-axis:

F
 y  ma y
net
D W  ma y
When terminal speed is reached the body falls at constant velocity and therefore the
object will no longer accelerate:
D  mg  0
PES 1110 Fall 2013, Spendier
Lecture 15/Page 6
D  mg
1
C r Av 2  mg
2
Which gives a terminal velocity:
2mg
vter min al 
Cr A
Example 4:
A raindrop with radius R = 1.5 mm falls from a cloud that is 1200 m above the ground.
The drag coefficient C = 0.6. Assume that the drop is spherical throughout the fall and the
density of water is 1000 kg/m3 and the density of air is 1.2 kg/m3.
a) What is the drop’s terminal speed?
b) What is the drop’s speed just before impact if there were no drag force?