Transcript Terminal Velocity - Northern Illinois University

Terminal Velocity
Drag

Kinetic friction is a constant force.
• If there is a net force an object would accelerate forever

Air resistance causes a friction called drag.
• At low velocity drag is proportional to velocity, Fd = bv
• At higher velocity drag goes as velocity squared, Fd = cv2

The direction of drag force is opposite to the velocity.
Terminal Velocity

An object may fall through
the air at constant velocity.

By the law of inertia the net
force is zero.

The force of drag must
balance the force of gravity.
Fd = cv2
Fg = -mg
Fd  Fg  0

This velocity is called the
terminal velocity.
cvt2 - m g  0
vt 
mg
c
Falling Leaves

The drag coefficient depends
on the surface area.
• Large surfaces – high drag
– Leaves
– Feathers
– Papers
• Small surfaces – low drag
– Stones
– Balls
– Bullets
Skydiving

Without a parachute:
With a parachute:
c = 0.25 kg / m
c = 28. kg / m
Terminal velocity for a 75-kg
skydiver without a parachute
is about 120 mph (53. m/s).
With a parachute the
terminal velocity is 5.1 m/s.
What are the drag
coefficients?
• Balance the weight and
drag
• mg = cv2
• c = mg / v2
Downhill Skiing


CBS Sports has invited you to be the special science
commentator for the Winter Olympics downhill ski
race.
You observe the following:
•
•
•
•

The downhill course is 2.5 km long with a drop of 800 m
The coefficient of kinetic friction is 0.052
The speed gun clocks the skier at a maximum of 130 km/h
An average skier is about 80 kg
What is the drag coefficient for a skier?
Force Diagram
FN = mg cosq

At constant velocity the
forces must all balance.

Friction doesn’t act in the
direction of the normal force.

The normal force cancels the
component of gravity.
Fd = -cv2
Ffr = -mkFN
Fg = -mg
q
Fgy = -mg cosq
FN  mg cosq
Downhill Run

FN = mg cosq
Fgx = mg sinq
• sinq = 800 m / 2500 m
 q = 19°
Fd = -cv2
Ffr = -mkFN
The slope of the downhill
course is

Drag force balances the
force of gravity and kinetic
friction.
Fgx  F fr  Fd  0
q  19°
m gsin q - m k m g cosq - cv2  0
m gsin q - m k m g cosq
c
v2
c  0.19kg/m