Discussion Problems
Thursday Feb 19
Circular motion
Circular motion relations
T=period (time to complete one revolution)
2
v
r
T
2


v
2
2
a     r 
v
r  T 
T
2

Amusement park fun
During the summer, you decide to visit
the an amusement park. You want
to ride the Ferris wheel, which has
seats on the rim of a circle with a
radius of 25 m. The Ferris wheel
rotates at a constant speed and
makes one complete revolution
every 20 seconds. While waiting,
you decide to calculate the
acceleration and forces (both
magnitude and direction) on a rider
at different positions on the ride.
What is the acceleration direction at
each position?
What must the forces (normal and
weight) look like at each position?
2
v
r
T
2


v
2
2
a     r 
v
r  T 
T
2
Ferris wheel physics
n
A
w
a
a
n
a
n
D
B
w
n
a
C
w
w
“Apparent Weight”
• A rider’s “apparent weight” is the magnitude
of the normal force on the rider. It would be
least
– A) at the top
– B) halfway down
– C) at the bottom
– D) halfway up
– E) back on the ground
Forces on a Ferris Wheel rider
• If the radius of the wheel is 25 meters
and the period is 20 seconds, what
would the normal force be on a rider
whose weight is 600 N at:
A) the highest point?
• B) the lowest point?
Forces on a Ferris Wheel rider
• If the radius of the wheel is 25 meters and the period
is 20 seconds, what would the normal force be on a
rider whose weight is 600 N at:
A) the highest point?
n
mac=w-n
n=w-mac=mg-mac
w
ac=(2/T)2r=2.5 m/s2
a
m=w/g=(600 N)/(9.8 m/s2)=61 kg
n=(61 kg)(9.8-2.5) m/s2=450 N
• B) the lowest point?
n
a
mac=n-w
n=w+mac=m(g+ac)=
n=(61 kg)(9.8+2.5) m/s2=750 N
w
Slippery road
• On a trip, you find yourself driving in your 3000-lb car along a
flat level road at 55 mph. The road makes a turn which you take
without changing your speed. The curve is approximately an arc
of a circle with a radius of 0.20 miles. You notice that the curve
is flat and level with no sign of banking. There are no warning
signs but you wonder if it would be safe to try to go 55 mph
around the curve in the rain when the wet surface has a lower
coefficient of friction. What is the minimum coefficient of static
friction between the road and your car's tires which will allow
your car to make the turn?
• Note: 1 mile=1.6x103 meter, 55 mph=25 meter/second
On a dry road µs is about 1.
Staying on the road
y
ac
n
fs
x
w
n-w=0: n=w=mg
fs=mac
fs,max=µsn=µsmg≥mac
µs≥ac/g
ac=v2/r
=(25 m/s)2/(320 m)
=1.95 m/s2
µs≥ac/g=1.95/9.8=0.20
Slippery road 2
• Curves are often “banked” to make
them easier to negotiate in poor driving
conditions. What banking angle would
be ideal (friction not needed!) for a
speed of 55 mph for the same curve
(0.2 mile radius)?
• Assume that the car rounds the curve
on a horizontal path.
Slippery road 2
• Without friction, the combination of
normal force and weight must produce
the necessary centripetal force:
y
n
n

ac
x
n cos( )  w  m g
w
w

n sin( )  m ac
ac
tan( )   0.2
g
  tan1 (0.2)  11.3