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Lecture 11
Ch6. Friction, Drag, and Centripetal Force
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
2012
Uniform Circular Motion
 Let us recall that when a body moves in a circle (or a circular
arc) with the radius R at constant speed v, it is said to be in
uniform circular motion.
 The body has a centripetal acceleration (directed toward the
center of the circle) of constant magnitude given by:
v2
a
R
(centripetal acceleration)
 As shown here, a hockey puck moves
around in a circle at constant speed v
while tied to a string looped around a
central peg.
 The centripetal
force is the radially
→
inward pull T on the puck from the string.
 Without that force, the puck would slide
off in a straight line instead of moving in
a circle.
Erwin Sitompul
University Physics: Mechanics
11/2
Uniform Circular Motion
 From Newton’s second law, we can write the magnitude F of
a centripetal force as:
v2
F m
R
(magnitude of centripetal force)
 Because the speed v is constant, the magnitudes of the
acceleration and the force are also constant.
 However, the directions of the
centripetal acceleration and force are
not constant; they vary continuously
and always point toward the center of
the circle (radially inward).
 The positive direction of the axis is
chosen to be radially outward.
Erwin Sitompul
University Physics: Mechanics
11/3
Checkpoint
When you ride in a Ferris wheel at constant
speed, what are
the directions of your
→
→
acceleration a and the normal force FN on you
(from the upright seat) as you pass through
(a) the highest point of the ride?
(b) the lowest point of the ride?
• The highest point
→
→
a downward, FN upward
FN  mg  ma
• The lowest point
→
→
a and FN upward
FN  mg  ma
Erwin Sitompul
University Physics: Mechanics
11/4
Example: “Allo Diavolo”
In a 1901 circus performance, “Allo
Diavolo“ introduced the stunt of
riding a bicycle in a loop-the-loop.
Assuming that the loop is a circle
with radius R = 2.7 m, what is the
least speed v Diavolo could have at
the top of the loop to remain in
contact with it there?
Fnet, y  ma y
FN  Fg  ma
 v2 
FN  mg  m  
R
Erwin Sitompul
• By the time Diavolo
remain in contact with the
loop, FN = 0.
 v 2  gR
v  gR
 (9.8)(2.7)
 5.144 m s
University Physics: Mechanics
11/5
Example: Riding the Rotor
A Rotor is a large, hollow cylinder that is
rotated rapidly around its central axis.
When the rider’s speed reaches some
predetermined value, the floor abruptly
falls away. The rider does not fall with it
but instead is pinned to the wall while the
cylinder rotates.
Suppose that the coefficient of static
friction μs between the rider’s clothing
and the wall is 0.40 and that the cylinder’s
radius R is 2.1 m.
Erwin Sitompul
FN
University Physics: Mechanics
11/6
Example: Riding the Rotor
(a) What minimum speed v must the
cylinder and rider have if the rider is not
to fall when the floor drops? (R = 2.1 m)
• Vertical calculations
FN
f s  Fg  ma y
s FN  mg  0
mg
FN 
s
• Radial calculations
v 
FN  m  
R
2
Erwin Sitompul
v
gR
s
(9.8)(2.1)

0.40
 7.173 m s
University Physics: Mechanics
11/7
Example: Riding the Rotor
(b) If the rider’s mass is 49 kg, what is the
magnitude of the centripetal force on her?
 v2 
FN  m  
R
(7.173) 2
 (49)
2.1
 1200 N
FN
If the Rotor initially moves at the minimum required speed for
the rider not to fall (7.173 m/s) and then its speed is
increased step
with (a) the
→
→ by step, what will happen
magnitude
of fs; (b) the magnitude of FN; (c) the value of
→
fs,max? Increase, decrease, or remain the same?
f s  mg
Erwin Sitompul
 v2 
FN  m  
R
f s ,max  s FN
University Physics: Mechanics
11/8
Example: Level Curves
A 1500 kg car moving on a flat, horizontal
road negotiates a curve as shown. If the
radius of the curve is 35.0 m and the
coefficient of static friction between the
tires and dry pavement is 0.523, find the
maximum speed the car can have and still
make the turn successfully.
f s  s FN
2
vmax
m
r
FN  Fg  0
FN  Fg  mg
Erwin Sitompul
vmax  s gr
 (0.523)(9.8)(35.0)
 13.4 m s
University Physics: Mechanics
11/9
Example: F1 Car
A modern F1 car is designed so that the passing air pushes
down on it, allowing the car to travel much faster through a flat
turn in a Grand Prix without friction failing. This downward push
is called negative lift or down force.
The following figure represents an F1 car of mass m = 600 kg
as it travels on a flat track in a circular arc of radius R = 100 m.
Because of the shape of→the car and the wings on it, the
passing air exerts a negative lift FL downward on the car. Take
μs = 0.75 and assume that the forces on the four tires are
identical.
Negative lift exerted
by passing air
Erwin Sitompul
University Physics: Mechanics 11/10
Example: F1 Car
(a) If the car is on the verge of sliding out of
the turn when its speed
is 28.6 m/s, what
→
is the magnitude of FL?
• Vertical calculations
FN  Fg  FL  0
FN  mg  FL
• Radial calculations
 v2 
fs  m  
R
• On the verge of sliding
f s  f s ,max   s FN
Erwin Sitompul
 v2 
s (mg  FL )  m  
R
 v2

FL  m 
g
 s R

Free body diagram
for the car
 (28.6)2

 (600) 
 9.8 
 (0.75)(100)

 663.68 N
University Physics: Mechanics 11/11
Example: F1 Car
(b) The magnitude FL of the negative lift on the car depends on
the square of the car’s speed v2. Thus, the negative lift on
the car here is greater when the car travels faster, as it does
on a straight section of track.
What is the magnitude of the negative lift for a speed of
90 m/s?
FL ,90
(90) 2
(90)2

FL,28.6
Fg  mg
2  FL ,90 
2
FL ,28.6 (28.6)
(28.6)
 (600)(9.8)
2
(90)
 5880 N

(663.68)
(28.6)2
 6572.21 N
• By the time the speed is 90 m/s, the
negative lift is already greater than the
gravitational force.
• Thus, the car “could have” run on a long
ceiling with that velocity (324 km/h).
Erwin Sitompul
University Physics: Mechanics 11/12
Example: Banked Curved Highways
Curved portions of highways
are always banked to prevent
cars from sliding off the
highway. When the highway
is wet, the frictional force
maybe significantly low, and
banking is then essential.
The figure below represents a normal car of mass m as it
moves at a constant speed v of 20 m/s around a banked
circular track of radius R = 190 m. If the frictional force from the
track is negligible, what bank angle θ prevents sliding?
Erwin Sitompul
University Physics: Mechanics 11/13
Example: Banked Curved Highways
• Vertical calculations
FNy  Fg  ma y  0
FN cos  mg
• Radial calculations
FNr  mar
 v2 
FN sin   m  
R
Erwin Sitompul
v2
tan  
gR
2


v
1
  tan  
 gR 
2


(20)
1
 tan 

(9.8)(190)


 12.12
University Physics: Mechanics 11/14
Example: Banked Curves
A car moving at the designated
speed can negotiate the curve. Such
a ramp is usually banked, which
means that the roadway is tilted
toward the inside of the curve.
Suppose the designated speed for
the ramp is to be 13.4 m/s and the
radius of the curve is 35.0 m. At what
angle should the curve be banked?
(13.4)2
v2

 0.5235
tan  
gR (9.8)(35)
  27.63
Erwin Sitompul
• Now, there is friction between
the tires and the pavement
with μs = 0.6?
• Using the calculated θ, what
will be the maximum speed of
the car by which the driver
does not have to concern
about sliding?
University Physics: Mechanics 11/15
A Puck on A Table
A puck of mass m = 1.5 kg slides in a circle of radius r = 20 cm
on a frictionless table while attached to a hanging cylinder of
mass M = 2.5 kg by a cord through a hole in the table.
What speed keeps the cylinder at rest?
• Answer: v = 1.81 m/s
Erwin Sitompul
University Physics: Mechanics 11/16
Homework 10A
Driving in a car with a constant speed of 12 m/s, a driver
encounters a bump in the road that has a circular cross-section,
as shown in the figure above.
a) If the radius of the curvature of the bum is 35 m, find the
apparent weight of the 70-kg driver as he passes over the
top of the bump.
b) At what speed must the driver go over the bump until he
feels “weightless”?
Erwin Sitompul
University Physics: Mechanics 11/17
Homework 10B
1. A car is driven at constant speed over a circular hill and then into a circular
valley with the same radius. At the top of the hill, the normal force on the
driver from the car seat is 25 N. The driver’s mass is 70.0 kg. What is the
magnitude of the normal force on the driver from the seat when the car
passes through the bottom of the valley?
2. The “Giant Swing” at a county fair consists of a
vertical central shaft with a number of horizontal
arms attached at its upper end.
Each arm supports a seat suspended from a
cable 5.00 m long, the upper end of the cable
being fastened to the arm at a point 3.00 m from
the central shaft.
(a) Find the time of one revolution of the swing if
the cable supporting a seat makes an angle of
30.0° with the vertical. (b) Does the angle depend
on the weight of the passenger for a given rate of
revolution?
Erwin Sitompul
University Physics: Mechanics 11/18